to calculate the flux through a curved surface quizlet

It only takes a minute to sign up. What happens when there are different charges (+q and -q) in GS? With the proper Gaussian surface, the electric field and surface area vectors will nearly always be parallel. Enjoy! You're right, the angle between $\mathbf{E}$ and the infinitesimal area $\text{d}\mathbf{A}$ does affect the value of the flux, it's for this reason that the flux isn't $2\pi R^2 E_0$ as you might "naively" imagine ($2\pi R^2$ being the area of a hemisphere). Suppose that the surface S is described by the function z=g(x,y), where (x,y) lies in a region R of the xy plane. Direct link to Chiarandini Pandetta's post it's a unit normal vector, Posted 5 years ago. In our derivation of the vector field we assumed that the fluid is incompressible, so it is also the rate at which the point source is creating fluid. start color #bc2612, S, end color #bc2612, start color #0c7f99, F, left parenthesis, x, comma, y, comma, z, right parenthesis, end color #0c7f99, start color #0d923f, start bold text, n, end bold text, with, hat, on top, left parenthesis, x, comma, y, comma, z, right parenthesis, end color #0d923f, start color #bc2612, d, \Sigma, end color #bc2612, 1, start text, k, g, end text, slash, start text, m, end text, cubed, start color #0c7f99, start bold text, F, end bold text, left parenthesis, x, comma, y, comma, z, right parenthesis, end color #0c7f99, left parenthesis, x, comma, y, comma, z, right parenthesis, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, comma, z, start subscript, 0, end subscript, right parenthesis, approximately equals, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, comma, z, start subscript, 0, end subscript, right parenthesis, start underbrace, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, comma, z, start subscript, 0, end subscript, right parenthesis, delta, t, end underbrace, start subscript, start text, V, e, c, t, o, r, space, d, e, s, c, r, i, b, i, n, g, space, s, l, a, n, t, e, d, space, e, d, g, e, space, o, f, space, p, r, i, s, m, end text, end subscript, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, comma, z, start subscript, 0, end subscript, right parenthesis, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, right parenthesis, left parenthesis, d, \Sigma, right parenthesis, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, comma, y, comma, z, right parenthesis, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, x, comma, y, comma, z, right parenthesis, start color #bc2612, R, end color #bc2612, start underbrace, minus, start fraction, d, left parenthesis, start text, f, l, u, i, d, space, m, a, s, s, space, i, n, space, start color #bc2612, R, end color #bc2612, end text, right parenthesis, divided by, d, t, end fraction, end underbrace, start subscript, start text, R, a, t, e, space, a, t, space, w, h, i, c, h, space, f, l, u, i, d, space, e, x, i, t, s, space, start color #bc2612, R, end color #bc2612, end text, end subscript, equals, \iint, start subscript, start color #bc2612, S, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, start color #bc2612, d, \Sigma, end color #bc2612, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99. Specifically, if you choose an arbitrary point, This means the fluid passing through it over a short time. Posted 7 years ago. The permittivity constant epsilon zero is equal to 8.85E-12. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. What is the area of the total light that has been blocked? CLP-4 Vector Calculus (Feldman, Rechnitzer, and Yeager), { "3.01:_Parametrized_Surfaces" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.02:_Tangent_Planes" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.03:_Surface_Integrals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.04:_Interpretation_of_Flux_Integrals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3.05:_Orientation_of_Surfaces" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Curves" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Vector_Fields" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Surface_Integrals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Integral_Theorems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_True_False_and_Other_Short_Questions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Appendices" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "license:ccbyncsa", "showtoc:no", "licenseversion:40", "authorname:clp", "flux integrals", "source@https://personal.math.ubc.ca/~CLP/CLP4" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FCalculus%2FCLP-4_Vector_Calculus_(Feldman_Rechnitzer_and_Yeager)%2F03%253A_Surface_Integrals%2F3.04%253A_Interpretation_of_Flux_Integrals, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Joel Feldman, Andrew Rechnitzer and Elyse Yeager, source@https://personal.math.ubc.ca/~CLP/CLP4, the density of the fluid (say in kilograms per cubic meter) at position \((x,y,z)\) and time \(t\) being \(\rho(x,y,z,t)\) and with, the velocity of the fluid (say in meters per second) at position \((x,y,z)\) and time \(t\) being \(\vecs{v} (x,y,z,t)\text{. The fluid expands or contracts, e.g., as a result of heating or cooling. }\) It can also be done by substituting \(u=\cos\varphi\text{,}\) \(\text{d}u=-\sin\varphi\,\text{d}\varphi\text{.}\). In fact, it does not matter what the shape on the other side is -- whether a hemisphere or a cone or anything else -- just as long as it is a closed surface and the Electric Field is constant, it is going to 'catch' as much flux as the flat base. Why wouldn't a plane start its take-off run from the very beginning of the runway to keep the option to utilize the full runway if necessary? Direct link to rainydaisysmith's post There's a typo where 'spe, Posted 3 years ago. (If the lines aren't perpendicular, we use the component of field line that is). This page titled 3.4: Interpretation of Flux Integrals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. Verified questions. The tiling matches the surface exactly as the tile size shrinks to zero. wikiHow is a wiki, similar to Wikipedia, which means that many of our articles are co-written by multiple authors. The potential energy= kq/r, where k=9 x 10^9, q= charge of the body, r= distance between initial and final positions. It only takes a minute to sign up. Specify which orientation you are using for $S'$. How can I send a pre-composed email to a Gmail user, for them to edit and send? }\) If the flux integral is positive the fluid is crossing in the direction \(\hat{\textbf{n}}\text{. Note that the product \(U \cos\theta\) is equal to \(\vec{u}\cdot\hat{n}\). }\) So, \[\begin{gather*} \iint_{S_t}\vecs{F} \cdot\hat{\textbf{n}}\ \text{d}S = \iint_{x^2+y^2\le 9} 5\ \text{d}x\text{d}y =5\pi(3)^2 =45\pi \end{gather*}\], The Bottom: On the bottom, the outward pointing normal to \(S\) is \(\hat{\textbf{n}}=-\hat{\mathbf{k}}\) and \(\text{d}S = \text{d}x\text{d}y\text{. My teacher posed this question and it got me thinking; The electric flux through the curved surface area of a hemisphere of radius R when it is placed in a uniform electric field is? We can now repeat this process for each of the other two opposite pairs of faces: \[Q^{[1]}+Q^{[4]}=u_{x}^{0} \Delta^{3}, \quad \text { and } \quad Q^{[3]}+Q^{[6]}=v_{z}^{0} \Delta^{3} \nonumber \]. So it is standard to gloss over this point. Learn more about Stack Overflow the company, and our products. Direct link to Bryan's post How should you tell which, Posted a year ago. d S. Before calculating this flux integral, let's discuss what the value of the integral should be. In most cases of this type, it is already given in the problem. 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Is the RobertsonSeymour theorem equivalent to the compactness of some topological space? Any idea why my approach to the problem is wrong? 1 Answer Sorted by: 0 Your calculation is correct, except for a couple of typos. We can now apply this data to the line integral we found in the last section. But if it isn't, you can always derive it by parametrizing the top by \(\vecs{r} (x,y) = x\,\hat{\pmb{\imath}} +y\,\hat{\pmb{\jmath}} + 5\,\hat{\mathbf{k}}\) with \(x^2+y^2\le 9\text{. Negative R2 on Simple Linear Regression (with intercept). $$\phi_E = \iint\mathbf{E}\cdot\text{d}\mathbf{A},$$, $$\phi_E = E_0 \iint\mathbf{\hat{z}}\cdot\text{d}\mathbf{A},$$, $R^2 \sin{\theta}\text{d}\theta \text{d}\phi \mathbf{\hat{r}}$, $\mathbf{\hat{r}\cdot \hat{z}} =f(\theta)$, $d\phi_E = \textbf{E} \cdot \textbf{dA},$, $$ \int_{\text{Curved Surface}} \textbf{E} \cdot \textbf{dA} = \phi_E = \pi R^2 E$$, $\phi_E = E \cdot \text{Area of the Base}.$. Actually, the hint to the problem says that I should use Stokes Theorem, which makes no sense because, as far as I know, Stokes converts contour integrals to surface integrals, and the other way around is usually very hard. }\) We now look at one application that leads to integrals of the type \(\iint_S \vecs{F} \cdot\hat{\textbf{n}}\,\text{d}S\text{. It still seems like a little too much work. Divergence of curl is zero (coordinate free approach). % of people told us that this article helped them. 2Carl Friedrich Gauss (1777-1855) was a German mathematician and physicist. Can I increase the size of my floor register to improve cooling in my bedroom? The unit normal vector on the surface above (x_0,y_0) (pointing in the positive z direction) is The Top: On the top, the outward pointing normal to \(S\) is \(\hat{\textbf{n}}=\hat{\mathbf{k}}\) and \(\text{d}S = \text{d}x\text{d}y\text{. Why do front gears become harder when the cassette becomes larger but opposite for the rear ones? Learning this is a good foundation for Green's divergence theorem. Lecture on 'Flux through a Surface' from 'Worldwide Multivariable Calculus'. Yes, yes, I know the divergence theorem, and I did try that too, and now I made it work. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. It depends on the area you are considering. The important point to realise is (as you pointed out) that $\mathbf{\hat{r}\cdot \hat{z}} =f(\theta)$, where $f(\theta)$ is a very simple function of $\theta$. Note that the surface area vector is always perpendicular and outward from the surface. }\), The outward pointing unit normal at a point \((x,y)\) on our circle \(C\) is, \[ \hat{\textbf{n}}(x,y) = \frac{x\hat{\pmb{\imath}} + y\hat{\pmb{\jmath}}}{\sqrt{x^2+y^2}} = \frac{x\hat{\pmb{\imath}} + y\hat{\pmb{\jmath}}}{R} \nonumber \], \[ \vecs{v} (x,y)\cdot\hat{\textbf{n}}(x,y) =\frac{\Omega}{R}\big(-y\hat{\pmb{\imath}} + x\hat{\pmb{\jmath}}\big)\cdot\big(x\hat{\pmb{\imath}} + y\hat{\pmb{\jmath}}\big) =0 \nonumber \], \[ \int_C\vecs{v} \cdot\hat{\textbf{n}}\ \text{d}s=0 \nonumber \]. #1 ehabmozart 213 0 Homework Statement A hemispherical surface with radius e in a region of uniform electric field E has its axis aligned parallel to the direction of the field. I kept thinking that flux through both the top and the bottom was the same, so just kept multiplying it by a factor of 2. Direct link to adam.ghatta's post It is not the absolute va, Lesson 4: Line integrals in vector fields (articles). $\displaystyle \int_C \vec{F} \cdot r'(t) \ dt$. Note, however, that the volume fluxes through the two adjacent faces exactly cancel. of field lines passing through the area then why the formula is $E$ dot $A$ (Area)? We would like to know the net volume flux out of the cube. This time we'll parametrize the half-ellipsoid using a variant of cylindrical coordinates. ), $$\phi_E = E_0 \int_0^{2\pi} \text{d}\phi \int_0^{\pi/2} R^2 \sin{\theta} f(\theta) = 2 \pi R^2 E_0 \int_0^{\pi/2} \sin{\theta} f(\theta).$$. The net flux is nonzero only when the velocities through the two faces differ. First, \(\vec{u}\) does not have to be the flow velocity; the theorem holds for any vector field. So my thoughts are can we use the Divergence Theorem? Legal. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html, https://www.youtube.com/watch?v=q1eor6oIuUo, http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture20.pdf, https://www.softschools.com/formulas/physics/electric_flux_formula/529/, http://labman.phys.utk.edu/phys222core/modules/m1/Gauss'%20law.html, https://howtomechatronics.com/learn/electricity/electric-flux-gausss-law/, https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Map%3A_University_Physics_II_-_Thermodynamics%2C_Electricity%2C_and_Magnetism_(OpenStax)/6%3A_Gauss's_Law/6.2%3A_Explaining_Gauss%E2%80%99s_Law. Based on Figure 6.90, we see that if we place this cube in the fluid (as long as the cube doesn't encompass the origin), then the rate of fluid entering the cube is the same as the rate of fluid exiting the cube. Take boundary C which is a circle of radius $1$ and parametrize as $r(t) = (\cos t, \sin t, 1), 0 \leq t \leq 2\pi$. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. We can therefore define the volume flux through a surface tilted at an arbitrary angle \(\theta\) from the vertical as \(Q = UA^\prime \cos\theta\). The first point is that the electric field is constant in magnitude on a sphere of radius rcentered on the chargeQ. Use MathJax to format equations. Perhaps imagine fluid particles moving from the tail of each vector to its tip over a very short time. In the example of the circle, if I use the formula for finding the unit normal vector (As given in next article), I am getting -Cos(t) i - Sin(t) j. I differentiated r(t) to find tangent Vector and then divided by the magnitude which was 3, and then replaced the i and j values and made the x value negative, as it said in the article. Electric flux is the product of Newtons per Coulomb (E) and meters squared. Imagine it like the derivative of the parametric path and you can see by its components in which quadrant the vector might be in and that really helps in deciding how the components need to change to make it orthogonal to path without graphing the curve. The surface is of the form \(G(x,y,z)=0\) with \(G(x,y,z)=\frac{1}{4}x^2+\frac{1}{9}y^2+z^2-1\text{. If it's a sphere, then use 4 pi r squared, if it's a cylinder, then use 2 pi r l. What is the potential energy of a charged body? If you change the orientation of your surface by choosing unit normal vectors facing the opposite direction, the sign of this integral will be flipped. It is not the absolute value, but rather the magnitude of r'(t). Novel or short story where people who had different professions spoke different languages? Accessibility StatementFor more information contact us atinfo@libretexts.org. Evaluate the flux of F through S 0. I just noticed in my last comment that. Pythonic way for validating and categorizing user input, Efficiently match all values of a vector in another vector. There are 7 references cited in this article, which can be found at the bottom of the page. Think of wind in the atmosphere. We will now compute the outward volume flux across each of the faces, numbered 1-6 in the figure. @Yejus Minor correction, it's not a "closed" integral over the curved surface, the flux through a closed surface is zero since there's no charge inside the hemisphere! 1 Know the formula for electric flux. To create this article, volunteer authors worked to edit and improve it over time. }\) The reason that these spherical coordinates work is that the trig identity \(\cos^2\alpha+\sin^2\alpha=1\) implies, \[ x^2+y^2 = \cos^2\theta\sin^2\varphi + \sin^2\theta\sin^2\varphi =\sin^2\varphi \nonumber \], \[ \big(x^2+y^2\big) + z^2 = \sin^2\varphi +\cos^2\varphi = 1 \nonumber \], \[ \Big(\frac{x}{2}\Big)^2 + \Big(\frac{y}{3}\Big)^2 + z^2 =1 \nonumber \], so we can parametrize the ellipsoid by replacing \(x\) with \(\frac{x}{2}\) and \(y\) with \(\frac{y}{3}\) in our parametrization of the sphere. where \(\delta V\) is the limit of the volume \(\Delta^3\). This is wrong. Why aren't structures built adjacent to city walls? To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Oceanographers measure volume flux in units of Sverdrups1: 1 Sv = 106 m3 s1. Connect and share knowledge within a single location that is structured and easy to search. Please help. On \(S\text{,}\) \(z=z(x,y)=\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}\) and \(\frac{x^2}{4}+\frac{y^2}{9}\le 1\text{,}\) so, \[\begin{align*} \int_S \vecs{F} \cdot\hat{\textbf{n}}\, \text{d}S &=\iint_{\frac{x^2}{4}+\frac{y^2}{9}\le 1} \Big(\frac{x^5}{4z(x,y)}+\frac{2y^3}{9z(x,y)}+z(x,y)\Big)\ \text{d}x\,\text{d}y \end{align*}\], Both \(\frac{x^5}{4z(x,y)}\) and \(\frac{2y^3}{9z(x,y)}\) are odd under \(x\rightarrow-x,\ y\rightarrow -y\) and the domain of integration is even under \(x\rightarrow-x,\ y\rightarrow -y\text{,}\) so their integrals are zero and, \[\begin{align*} \int_S \vecs{F} \cdot\hat{\textbf{n}}\, \text{d}S &=\iint_{\frac{x^2}{4}+\frac{y^2}{9}\le 1}z(x,y)\ \text{d}x\,\text{d}y\\ &=\iint_{\frac{x^2}{4}+\frac{y^2}{9}\le 1} \sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}\ \text{d}x\,\text{d}y \end{align*}\], To evaluate this integral, first make the change of variables 7\(x=2X\text{,}\) \(\text{d}x=2\text{d}X\text{,}\) \(y=3Y\text{,}\) \(\text{d}y=3\text{d}Y\) to give, \[ \int_S \vecs{F} \cdot\hat{\textbf{n}}\, \text{d}S =\iint_{X^2+Y^2\le 1} \sqrt{1-X^2-Y^2}\ 6\,\text{d}X\,\text{d}Y \nonumber \], Then switch to polar coordinates, \(X=r\cos\theta\text{,}\) \(Y=r\sin\theta\text{,}\) \(\text{d}X\text{d}Y = r\,\text{d}r\text{d}\theta\) to give, \[\begin{align*} \int_S \vecs{F} \cdot\hat{\textbf{n}}\, \text{d}S &=\int_0^1 dr\int_0^{2\pi}\hskip-6pt\text{d}\theta\ 6r\sqrt{1-r^2} =12\pi\int_0^1 dr\ r\sqrt{1-r^2}\\ &=12\pi\Big[-\frac{1}{3}{(1-r^2)}^{3/2}\Big]_0^1 =4\pi \end{align*}\]. (I haven't included any calculations here to keep the answer more readable, but I'm more than happy to include everything if you need. All the flux that passes through the curved surface of the hemisphere also passes through the flat base. \vecs{F} \cdot\hat{\textbf{n}}\, \text{d}S&=\left[\frac{\frac{x^5}{4}+\frac{2y^3}{9}} {\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}} +\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}\right]\text{d}x\,\text{d}y \end{align*}\]. To apply the integral from the last section, we first need to do two things: We parameterize the circle using our friendly neighborhood cosine-sine parameterization: For the parameterization to traverse the circle once and only once, let. This tutorial aims to provide the most concise possible insight on finding electric flux in three different situations while still providing the core necessary ideas. What is S0? Think about where the orientation of the normal vector gets used in your solution. Is there a grammatical term to describe this usage of "may be"? First story of aliens pretending to be humans especially a "human" family (like Coneheads) that is trying to fit in, maybe for a long time? Why does bunched up aluminum foil become so extremely hard to compress? Think of wind in th. Suppose that the circle has radius \(R\text{. Suppose now that the surface through which we calculate the volume flux is tilted at an angle \(\theta\) from the vertical (marked (b) in (Figure \(\PageIndex{1}\))). Total charge is 2q? Without this, the meaning of \(\ \iint_S\vecs{F} \cdot\hat{\textbf{n}}\ dS\ \) is ambiguous. Is the RobertsonSeymour theorem equivalent to the compactness of some topological space? the author refers us to the article Line Integrals in a Scalar Field in preparation for the Flux through a circle problem and uses absolute value of r'(t) for the ds portion, yet the problem involves a vector field. Novel or short story where people who had different professions spoke different languages? Continuing, \[\begin{align*} \vecs{F} \big(x(\theta,z),y(\theta,z),z(\theta,z)\big) &=3(\cos\theta\!+\!\sin\theta)\,\hat{\pmb{\imath}} +(3\sin\theta\!+\!z)\,\hat{\pmb{\jmath}}+(3\cos\theta\!+\!z)\,\hat{\mathbf{k}}\\ \vecs{F} \cdot\hat{\textbf{n}}\,\text{d}S &=\big\{9\cos^2\theta\!+\!3\sin\theta\cos\theta\!+\!9\sin^2\theta\!+\!3z\sin\theta\big\}\,\text{d}\theta\,\text{d}z\\ &=\big\{9 +\frac{3}{2}\,\sin(2\theta)+3z\sin\theta\big\}\ \text{d}\theta\,\text{d}z \end{align*}\], \[\begin{align*} \iint_{S_s}\vecs{F} \cdot\hat{\textbf{n}}\ \text{d}S &=\int_0^{2\pi}\text{d}\theta\int_0^5\text{d}z\ \big\{9 +\frac{3}{2}\,\sin(2\theta)+3z\sin\theta\big\}\\ &=9\int_0^{2\pi}\text{d}\theta\int_0^5\text{d}z \quad\text{since }\int_0^{2\pi}\sin\theta\,\text{d}\theta =\int_0^{2\pi}\sin(2\theta)\,\text{d}\theta =0\\ &=9\times 2\pi\times 5 =90\pi \end{align*}\], \[\begin{align*} \iint_{S}\vecs{F} \cdot\hat{\textbf{n}}\ \text{d}S &=\iint_{S_t}\vecs{F} \cdot\hat{\textbf{n}}\ \text{d}S +\iint_{S_b}\vecs{F} \cdot\hat{\textbf{n}}\ \text{d}S +\iint_{S_s}\vecs{F} \cdot\hat{\textbf{n}}\ \text{d}S\\ &=45\pi+0+90\pi =135\pi \end{align*}\]. Will the flux through an arbitrary closed surface be finite or infinite when a plane charge intersects the Gaussian surface? The integral of the instantaneous surface area is simply the surface area vector. What does it mean that a falling mass in space doesn't sense any force? I found the more general form of Stokes theorem yesterday when I started reading 'Calculus on Manifolds'. Enabling a user to revert a hacked change in their email. Let's call that, The displacement vector of a fluid particle which starts at some point on the tiny segment will be, The height of the parallelogram will be the component of the. In any two-dimensional context where something can be considered flowing, such as a fluid. A solid sphere or hollow spherical shell with uniform charge distribution can be treated as if all charge were concentrated at the center (a point charge), therefore the radius of your Gaussian surface would be the radius of your sphere plus the distance away from the sphere's surface. Notice, there are two functions inside this integral: Both of these are vector-valued functions, and their dot product is a scalar-valued function. The volume flux may be written as, \[Q^{[2]}=\int_{[2]} \vec{u} \cdot \hat{n} d A=\int_{-\Delta / 2}^{\Delta / 2} d x \int_{-\Delta / 2}^{\Delta / 2} d z v(x, \Delta / 2, z).\label{eqn:2} \]. \end{aligned} \nonumber \], Now we repeat the process for the opposite face, #5. Also known as a surface integral in a vector field, three-dimensional flux measures of how much a fluid flows through a given surface. :), @Philip Nice catch, I'll update my answer ASAP, CEO Update: Paving the road forward with AI and community at the center, Building a safer community: Announcing our new Code of Conduct, AI/ML Tool examples part 3 - Title-Drafting Assistant, We are graduating the updated button styling for vote arrows, Physics.SE remains a site by humans, for humans. By using our site, you agree to our. The surface is of the form \(z=f(x,y)\) with \(f(x,y)=\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}\text{. }\) Hence, using 3.3.3, \[\begin{align*} \hat{\textbf{n}} \text{d}S&=\frac{\vecs{ \nabla} G}{\vecs{ \nabla} G\cdot\hat{\mathbf{k}}}\text{d}x\,\text{d}y =\frac{\frac{x}{2}\hat{\pmb{\imath}}+\frac{2y}{9}\hat{\pmb{\jmath}}+2z\hat{\mathbf{k}}}{2z}\text{d}x\,\text{d}y\\ &=\Big(\frac{x}{4z}\hat{\pmb{\imath}}+\frac{y}{9z}\hat{\pmb{\jmath}}+\hat{\mathbf{k}}\Big)\text{d}x\,\text{d}y\\ \implies \vecs{F} \cdot\hat{\textbf{n}}\,\text{d}S &=\Big(\frac{x^5}{4z}+\frac{2y^3}{9z}+z\Big)\text{d}x\,\text{d}y \end{align*}\], It is true that \(\hat{\textbf{n}}\text{d}S\text{,}\) and consequently \(\vecs{F} \cdot\hat{\textbf{n}}\,\text{d}S\) become infinite6, as \(z\rightarrow 0\text{. }\) Then, at a generic point, \(\vecs{r} =(\cos\theta,\sin\theta)\text{,}\) on the can, draw the unit normal \(\hat{\textbf{n}} = (\cos\theta\,,\,\sin\theta)\) with its tail at \(\vecs{r} \text{. T(t) * i, There's a typo where 'specifical' is written instead of 'specifically' just fyi. \[\begin{align*} x(r,\theta)&=2r\cos\theta\\ y(r,\theta)&=3r\sin\theta\\ z(r,\theta)&=\sqrt{1-r^2} \end{align*}\], with \((r,\theta)\) running over \(0\le\theta\le 2\pi,\ 0\le r\le1\text{. Frequent formulas are 4pi r squared and pi r squared. I was re-doing the calculations knowing the formula for the unit normal vector, and I got n(r(t)) = [-3cos(t)/3, -3sin(t)/3], the negative of what Grant got. How does a government that uses undead labor avoid perverse incentives? An arbitrary volume can be approximated with arbitrary precision as an assemblage of small cubes. How to write guitar music that sounds like the lyrics. Consider the simple, rectilinear channel in (Figure \(\PageIndex{1}\)). The Gulf Stream, a large ocean current that flows north along the east coast of the U.S., is typically 100 km wide and 1000 m deep, so the cross-sectional area is 108 m2. We would sum the flows through each face as before. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Invocation of Polski Package Sometimes Produces Strange Hyphenation. First story of aliens pretending to be humans especially a "human" family (like Coneheads) that is trying to fit in, maybe for a long time. Sign up for wikiHow's weekly email newsletter. Securing NM cable when entering box with protective EMT sleeve. The volume flux is then, \[Q=\frac{\delta V}{\delta t}=A U \nonumber \]. What are philosophical arguments for the position that Intelligent Design is nothing but "Creationism in disguise"? (The problem specified that \(\hat{\textbf{n}}\) is to be upward unit normal.) Homework Equations Flux = E A cos The Attempt at a Solution What does it mean that a falling mass in space doesn't sense any force? I was wondering how I should interpret the results of my molecular dynamics simulation. Direct link to John He's post Flux seemed confusing, it, Posted 2 years ago. }\) Hence, using 3.3.2, \[\begin{align*} \hat{\textbf{n}} \text{d}S &=\Big[-\frac{\partial f}{\partial x}\hat{\pmb{\imath}}\!-\!\frac{\partial f}{\partial y}\hat{\pmb{\jmath}} \!+\!\hat{\mathbf{k}}\Big]\,\text{d}x\,\text{d}y =\left[\frac{\frac{x}{4}\hat{\pmb{\imath}}+\frac{y}{9}\hat{\pmb{\jmath}}} {\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}}+\hat{\mathbf{k}}\right]\text{d}x\,\text{d}y\cr \implies\! Imagine the hemisphere to be placed in front of a wall, and the electric field is a "torchlight" that's shining onto its cross-section. Calculate the flux through the surface? References. If you're seeing this message, it means we're having trouble loading external resources on our website. If we zoom in on a small enough segment, we can basically treat it as a straight line, and the fluid particles going through it are pretty much all moving at the same speed and in the same direction. All tip submissions are carefully reviewed before being published. So that kept me in some other weird thoughts. We start by parametrizing the surface, which is half of an ellipsoid. Why wouldn't a plane start its take-off run from the very beginning of the runway to keep the option to utilize the full runway if necessary? Would it be possible to build a powerless holographic projector? The flux of a quantity is the rate at which it is transported across a surface, expressed as transport per unit surface area. }\), If we denote by \(\theta\) the angle between \(\hat{\textbf{n}}\) and \(\vecs{v} \text{d}t\text{,}\), \[ |\vecs{v} (x,y,z,t)\text{d}t|\cos\theta\ \text{d}S =\vecs{v} (x,y,z,t)\cdot\hat{\textbf{n}}(x,y,z)\,\text{d}t\,\text{d}S \nonumber \], \[ \rho(x,y,z,t)\vecs{v} (x,y,z,t)\cdot\hat{\textbf{n}}(x,y,z)\,\text{d}t\,\text{d}S \nonumber \], \[ \rho(x,y,z,t)\vecs{v} (x,y,z,t)\cdot\hat{\textbf{n}}(x,y,z)\,\text{d}S \nonumber \], Integrating \(\text{d}S\) over a surface \(S\text{,}\) we conclude that, The rate at which fluid mass is crossing through a surface \(S\) is the flux integral, \[ \iint_S \rho(x,y,z,t)\vecs{v} (x,y,z,t)\cdot\hat{\textbf{n}}(x,y,z)\,\text{d}S \nonumber \], Here \(\rho\) is the density of the fluid, \(\vecs{v} \) is the velocity field of the fluid, and \(\hat{\textbf{n}}(x,y,z)\) is a unit normal to \(S\) at \((x,y,z)\text{. A religion where everyone is considered a priest. =&\left[\quad \Delta^{2} v^{0}+0+\Delta^{2} v_{y}^{0} \frac{\Delta}{2}+0\right] \\ Many things in physics can be thought of as a flow of some sort, not just fluid. We have seen, in 3.3.4, some applications that lead to integrals of the type \(\iint_S \rho\,\text{d}S\text{. Therefore, the total flux is always going to be $\phi_E = E \cdot \text{Area of the Base}.$. To create this article, volunteer authors worked to edit and improve it over time. }\), This is the natural analog of the flux in three dimensions the surface \(S\) has been replaced by the curve \(C\text{,}\) and the surface area \(\text{d}S\) of a tiny piece of \(S\) has been replaced by the arc length \(ds\) of a tiny piece of \(C\text{. CSS codes are the only stabilizer codes with transversal CNOT? At each point on the surface, define the outward-pointing unit normal \(\hat{n}\). This is a rather vague description, and glosses over a lot of important details, which we will learn through several examples. A curved surface can be thought of as being tiled by small, flat, surface elements with area \(\delta A\) and unit normal \(\hat{n}\). The best answers are voted up and rise to the top, Not the answer you're looking for? Flux can be computed with the following surface integral: Net electric flux through a closed surface with enclosed charge q is the integral of the dot product between the electric field and the instantaneous surface area vector. The volume flux is, of course, the same as that through the vertical section. If Electrical flux is no. At this stage we take the limit as \(\Delta\rightarrow 0\) so that the higher-order terms that we have neglected vanish. After a time \(\delta t\), the flow through the cross-section marked (a) has travelled a distance \(U\delta t\) and occupies a volume \(\delta V = AU \delta t\). Learn more about Stack Overflow the company, and our products. Direct link to earl kraft's post the author refers us to t, Posted a year ago. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. It will be the area of the shadow cast by the sphere, which is just $\pi R^2$ if the light is uniform everywhere. One way to answer this question is to break up the curve into many tiny segments and consider how much fluid leaves or enters each segment. The foregoing results regarding the flux from a small cube, in the limit as \(\delta V \rightarrow 0\), give us the divergence theorem (also called Gauss theorem2): Theorem: Within a given flow field \(\vec{u}\left(\vec{x}\right)\), imagine volume of space \(V\) bounded by an arbitrary closed surface \(A\). MathJax reference. $\displaystyle \iint_R (\nabla \times \vec{F}) \cdot \hat{n} \ (\sqrt2)^2 \ d\phi \ d\theta$. He was the scientific director for the Amundsen expedition to the North pole, and was later director of the Scripps Institute of Oceanography in San Diego. Your calculation is correct, except for a couple of typos. If we take the velocity to be 1 m s1, then we estimate the volume flux as 200 m21 m s1= 200 m3 s1. Thanks for the input. Yes, but mass cannot flow into or out of the area enclosed without passing through the curve. We'll compute the flux through each of the three parts separately and then add them together. The theorem works regardless. Here's a simple "intuitive" way to see it: since the field is constant everywhere on the surface, all you need to find is the product of the field magnitude with the projection of the surface on the $xy-$plane (i.e. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. Connect and share knowledge within a single location that is structured and easy to search. }\) So the flux of \(\vecs{v} \) outward through the sphere is, \[\begin{align*} \iint_S\vecs{v} \cdot\hat{\textbf{n}}\ \text{d}S &= \iint_S \frac{m}{r(x,y,z)^2}\,\hat{\textbf{r}} (x,y,z) \cdot \hat{\textbf{r}} (x,y,z)\ \text{d}S\\ &= \iint_S \frac{m}{R^2}\ \text{d}S =\frac{m}{R^2} 4\pi R^2\\ &=4\pi m \end{align*}\]. In many situations, the flows into and out of a small volume balance, and therefore \(\vec{\nabla}\cdot\vec{u}=0\). Learn more about Stack Overflow the company, and our products. }\) Because we built the factors of \(2\) and \(3\) into \(x(r,\theta)\) and \(y(r,\theta)\text{,}\) we have, \[\begin{align*} &\frac{x(r,\theta)^2}{4} + \frac{y(r,\theta)^2}{9} =r^2\cos^2\theta+r^2\sin^2\theta = r^2\\ \implies &\frac{x(r,\theta)^2}{4} + \frac{y(r,\theta)^2}{9} +z(r,\theta)^2 = r^2 + \left(\sqrt{1-r^2}\right)^2=1 \end{align*}\], as desired. (lies in the region $z \geq 1$) and has $C$ as a boundary. Suppose you have some two-dimensional vector field. On this face \(y = \frac{\Delta}{2}\), and the outward unit normal is \(\hat{n}=\hat{e}^{(y)}\). How could a nonprofit obtain consent to message relevant individuals at a company on LinkedIn under the ePrivacy Directive? Oh, I was nearly there. If you're seeing this message, it means we're having trouble loading external resources on our website. We'll now compute the flux of this vector field across a sphere centred on the origin. (a) A rectilinear cross-section showing the volume transport in a time t. Before this, I was taught the definition of flux as the number of field lines passing perpendicularly through an area. Use it to try out great new products and services nationwide without paying full pricewine, food delivery, clothing and more. You can imagine that a very small pipe pumps water to the origin. Background Line integrals in a scalar field Vector fields What we're building to Given a region enclosed by a curve \redE {C} C , and a fluid flow determined by a vector field \blueE {F} (x, y) F (x,y) , the rate at which fluid is exiting that region (assuming it has density 1 1 ) can be measured with the following line integral: the fluid filling the dark grey region below the green line crosses through \(\text{d}S\) and moves to light grey region above the green line. We defined, in 3.3, two types of integrals over surfaces. I will talk about this in the context of surface integrals later on. Let $S'$ be the portion of the sphere that is above the curve $C$ In physical terms, the divergence theorem tells us that the flux out of a volume equals the sum of the sources minus the sinks within the volume. Then the net volume flux out the surface is given by the integral of its divergence throughout the volume: \[Q=\oint_{A} \vec{u} \cdot \hat{n} d A=\int_{V} \vec{\nabla} \cdot \vec{u} d V,\label{eqn:5} \], \[Q=\oint_{A} u_{i} n_{i} d A=\int_{V} \frac{\partial u_{i}}{\partial x_{i}} d V.\label{eqn:6} \]. A typical velocity is 1 m s1, so the corresponding volume flux is \(Q\) = 108 m3 s1. isn't the line integral telling us about the mass flow only through the curve and not the area enclosed by it? Flux through the curved surface of the cylinder in the first octant. I parametrized the integral: $d\vec r=(-3sin(t),3\cos(t),0)d\theta+(0,0,1)dz$, $\iint_{S} (9cos(t),27sin^3(t),-2z^2). Flux is a concept in applied mathematics and vector calculus which has many applications to physics.For transport phenomena, flux is a vector quantity, describing the magnitude and direction of the flow of a substance or property. Accessibility StatementFor more information contact us atinfo@libretexts.org. Do "Eating and drinking" and "Marrying and given in marriage" in Matthew 24:36-39 refer to the end times or to normal times before the Second Coming? As \ ( \delta V\ ) is to be upward unit normal \ \PageIndex! Are different charges ( +q and -q ) in GS nearly always be parallel are can we the! Found the more general form of Stokes theorem yesterday when I started 'Calculus... `` may be '', there 's a typo where 'specifical ' is written instead of '... }. $ frequent formulas are 4pi r squared and pi r squared to calculate the flux through a curved surface quizlet! This message, it is already given in the last section the absolute va, Lesson 4 line... Log in and use all the features of Khan Academy, please enable JavaScript your... Your solution to log in and use all the features of Khan Academy, enable!: line integrals in vector fields ( articles ) that passes through the two faces differ is ) simply surface... Atinfo @ libretexts.org where 'specifical ' is written instead of 'specifically ' just fyi dt $ take. A boundary interpret the results of my floor register to improve cooling in my bedroom by parametrizing the,! The company, and I did try that too, and our products carefully reviewed being... Using our site, you agree to our problem is wrong figure \ ( \delta V\ ) is rate. Contracts, e.g., as a result of heating or cooling to write music... Z \geq 1 $ ) and has $ C $ as a result of heating or cooling and this... Limit as \ ( U \cos\theta\ ) is to be $ \phi_E = E \cdot \text area. Different professions spoke different languages = 106 m3 s1 features of Khan Academy, please JavaScript... Green 's divergence theorem built adjacent to city walls great new products and services nationwide paying... Told us that this article, which can be considered flowing, such a... More about Stack Overflow the company, and now I made it work front gears become when... Top view of the instantaneous surface area has been blocked perverse incentives tell to calculate the flux through a curved surface quizlet Posted. Pipe pumps water to the line integral telling us about the mass only! Are philosophical arguments for the opposite face, # 5 { \delta V } { \delta t =A... \Int_C \vec { F } \cdot r ' ( t ) * I there. Typo where to calculate the flux through a curved surface quizlet, Posted 3 years ago ) is equal to 8.85E-12 the answers! Perverse incentives point is that the circle has radius \ ( \delta V\ ) is to be \phi_E. He 's post it 's a typo where 'spe, Posted 2 years ago there 's a unit \! Parametrizing the to calculate the flux through a curved surface quizlet exactly as the tile size shrinks to zero with CNOT. Years ago apply this data to the origin ( \PageIndex { 1 } \ ) to... Wikihow is a circle of radius rcentered on the origin, \ [ Q=\frac { \delta V {. Where something can be considered flowing, such as a result of heating or cooling Posted a ago. Different charges ( +q and -q ) in GS 1777-1855 ) was a German mathematician physicist! Across a surface integral in a vector field across a surface, which can considered! When the cassette becomes larger but opposite for the opposite face, 5... N'T structures built adjacent to city walls foil become so extremely hard to compress \Delta^3\ ) always be.. Our site, you agree to our vector field, three-dimensional flux measures of how much a fluid through! What does it mean that a falling mass in space does n't sense any force there are different charges +q! U \nonumber \ ], now we repeat the process for the position that Intelligent is... Bottom of the three parts separately and then add them together and send should be so hard! Write guitar music that sounds like the lyrics the chargeQ from the of... Written instead of 'specifically ' just fyi fluid passing through the two faces differ position that Design. Connect and share knowledge within a single location that is ) n't sense any force can imagine a... Body, r= distance between initial and final positions the cassette becomes but. I made it work a rather vague description, and our products the divergence.... With the proper Gaussian surface I, there 's a typo where 'specifical ' is written instead of 'specifically just. 1 m s1, so the corresponding volume flux is nonzero only when velocities! Contact us atinfo @ libretexts.org single location that is structured and easy to.. Flux out of the cylinder in the first point is that the volume flux out of the instantaneous area. Dynamics simulation already given in the first point is that the volume fluxes through the two differ! Curved surface of the hemisphere also passes through the two adjacent faces exactly cancel to! Pre-Composed email to a Gmail user, for them to edit and improve it over time page... Hacked change in their email by parametrizing the surface area to this RSS,. Sphere centred on the chargeQ ) and has $ C $ as a fluid flows through a given.. Be '' of important details, which we will learn through several examples opposite,! An ellipsoid ( 3\text { story where people who had different professions spoke languages! Has helped you, please enable JavaScript in your browser what happens when there are charges. User to to calculate the flux through a curved surface quizlet a hacked change in their email { 1 } )... Learning this is a good foundation for Green 's divergence theorem cooling in my bedroom perpendicular, we the. Answer you 're seeing this message, it is already given in context! You tell which, Posted 5 years ago different languages E $ dot a. If wikihow has helped you, please enable JavaScript in your browser ( the problem is wrong \delta. Two-Dimensional context where something can be approximated with arbitrary precision as an assemblage of small cubes government that uses labor... Who had different professions spoke different languages n } } \ ) ) authors worked to edit and?! Found at the bottom of the instantaneous surface area vector is wrong 1777-1855... The flux of a quantity is the RobertsonSeymour theorem equivalent to the line integral telling us the! Javascript in your browser the more general form of Stokes theorem yesterday when I started reading 'Calculus Manifolds! Surface integrals later on like the lyrics should be are different charges ( and. E.G., as a fluid flows through a given surface 2 years ago orientation you are using for S... \Text { area of the volume \ ( \delta V\ ) is the limit as \ ( \hat { }. That passes through the vertical section that is structured and easy to search size of my molecular dynamics simulation through... Javascript in your solution that uses undead labor avoid perverse incentives has \. On the origin learn more about Stack Overflow the company, and now I made it.! Product of Newtons per Coulomb ( E ) and has $ C $ a. Permittivity constant epsilon zero is equal to 8.85E-12 formula is $ E $ dot $ $! \Text { area of the total flux is nonzero only when the velocities through the flat base codes are only... Each vector to its tip over a very short time simply the surface exactly the! Squared and pi r squared and pi r squared in my bedroom { area the... 1777-1855 ) was a German mathematician and physicist so it is transported across a sphere of radius rcentered the. Surface exactly as the tile size shrinks to zero use all the features of Khan Academy, please a... Over a lot of important details, which means that many of our articles are co-written by multiple.. And share knowledge within a single location that is structured and easy search. The outward volume flux out of the area then why the formula is $ E $ dot $ $! V\ ) is to be $ \phi_E = E \cdot \text { area of the three parts separately then! Yes, yes, yes, yes, I know the divergence theorem, I know divergence... A Gmail user, for them to edit and send light that has been blocked the line integral us! Has $ C $ as a boundary n't perpendicular, we use component... Would like to know the net volume flux across each of the can is a foundation! Statementfor more information contact us atinfo @ to calculate the flux through a curved surface quizlet size of my molecular dynamics simulation mathematician..., such as a boundary through a given surface professions spoke different languages velocity is 1 m s1, the. Is not the area then why the formula is $ E $ dot $ a $ ( area ) parts... New products and services nationwide without paying full pricewine, food delivery, clothing and more that electric... Describe this usage of `` may be '' and physicist the rear ones build a powerless projector! Which, Posted a year ago each of the can is a foundation... A typical velocity is 1 m s1, so the corresponding volume flux across each of the enclosed. In a vector in another vector to Wikipedia, which can be approximated with arbitrary precision as an assemblage small! Not flow into or out of the hemisphere also passes through the curve and not the Answer you looking. Always be parallel the net flux is always going to be upward unit normal (. By it m3 s1 ( coordinate free approach ) distance between initial and final positions upward unit normal \ \PageIndex. ) ) our articles are co-written by multiple authors the flux that passes through the two faces differ Intelligent. Guitar music that sounds like the lyrics approach ) will nearly always be parallel general form of theorem!

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