surface integral calculator

By the definition of the line integral (Section 16.2), \[\begin{align*} m &= \iint_S x^2 yz \, dS \\[4pt] Find a parameterization r ( t) for the curve C for interval t. Find the tangent vector. Surface integral of vector field calculator For a vector function over a surface, the surface integral is given by Phi = int_SFda (3) = int_S(Fn^^)da (4) = int_Sf_xdydz+f_ydzdx+f_zdxdy Solve Now. Since the disk is formed where plane $z = 1$ intersects sphere $x^2 + y^2 + z^2 = 4$, we can substitute $z = 1$ into equation $x^2 + y^2 + z^2 = 4$: \[x^2 + y^2 + 1 = 4 \Rightarrow x^2 + y^2 = 3. Multiple Integrals Calculator - Symbolab Multiple Integrals Calculator Solve multiple integrals step-by-step full pad Examples Related Symbolab blog posts Advanced Math Solutions - Integral Calculator, trigonometric substitution In the previous posts we covered substitution, but standard substitution is not always enough. Find more Mathematics widgets in Wolfram|Alpha. Alternatively, you can view it as a way of generalizing double integrals to curved surfaces. Here is that work. In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero. Therefore, the flux of $\vecs{F}$ across $S$ is 340. By Equation \ref{scalar surface integrals}, \[\begin{align*} \iint_S f(x,y,z)dS &= \iint_D f (\vecs r(u,v)) ||\vecs t_u \times \vecs t_v|| \, dA \\ When you're done entering your function, click "Go! Again, this is set up to use the initial formula we gave in this section once we realize that the equation for the bottom is given by $g\left( {x,y} \right) = 0$ and $D$ is the disk of radius $\sqrt 3 $ centered at the origin. \end{align*}\], To calculate this integral, we need a parameterization of $S_2$. Remember that the plane is given by $z = 4 - y$. Thus, a surface integral is similar to a line integral but in one higher dimension. Then the heat flow is a vector field proportional to the negative temperature gradient in the object. Also, dont forget to plug in for $z$. There is a lot of information that we need to keep track of here. Then the curve traced out by the parameterization is $\langle \cos K, \, \sin K, \, v \rangle $, which gives a vertical line that goes through point $(\cos K, \sin K, v \rangle$ in the $xy$-plane. Each choice of $u$ and $v$ in the parameter domain gives a point on the surface, just as each choice of a parameter $t$ gives a point on a parameterized curve. If piece $S_{ij}$ is small enough, then the tangent plane at point $P_{ij}$ is a good approximation of piece $S_{ij}$. After putting the value of the function y and the lower and upper limits in the required blocks, the result appears as follows: \[S = \int_{1}^{2} 2 \pi x^2 \sqrt{1+ (\dfrac{d(x^2)}{dx})^2}\, dx \], \[S = \dfrac{1}{32} pi (-18\sqrt{5} + 132\sqrt{17} + sinh^{-1}(2) sinh^{-1}(4)) \]. and For example, spheres, cubes, and . Is the surface parameterization $\vecs r(u,v) = \langle u^{2v}, v + 1, \, \sin u \rangle, \, 0 \leq u \leq 2, \, 0 \leq v \leq 3$ smooth? If $v$ is held constant, then the resulting curve is a vertical parabola. This is the two-dimensional analog of line integrals. Direct link to Qasim Khan's post Wow thanks guys! Show that the surface area of cylinder $x^2 + y^2 = r^2, \, 0 \leq z \leq h$ is $2\pi rh$. If $v = 0$ or $v = \pi$, then the only choices for $u$ that make the $\mathbf{\hat{j}}$ component zero are $u = 0$ or $u = \pi$. Therefore, the choice of unit normal vector, \[\vecs N = \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \nonumber \]. The tangent vectors are $\vecs t_u = \langle - kv \, \sin u, \, kv \, \cos u, \, 0 \rangle$ and $\vecs t_v = \langle k \, \cos u, \, k \, \sin u, \, 1 \rangle$. Comment ( 11 votes) Upvote Downvote Flag more The tangent plane at $P_{ij}$ contains vectors $\vecs t_u(P_{ij})$ and $\vecs t_v(P_{ij})$ and therefore the parallelogram spanned by $\vecs t_u(P_{ij})$ and $\vecs t_v(P_{ij})$ is in the tangent plane. [2v^3u + v^2u - vu^2 - u^2]\right|_0^3 \, dv \\[4pt] &= \int_0^4 (6v^3 + 3v^2 - 9v - 9) \, dv \\[4pt] &= \left[ \dfrac{3v^4}{2} + v^3 - \dfrac{9v^2}{2} - 9v\right]_0^4\\[4pt] &= 340. While graphing, singularities (e.g. poles) are detected and treated specially. For now, assume the parameter domain $D$ is a rectangle, but we can extend the basic logic of how we proceed to any parameter domain (the choice of a rectangle is simply to make the notation more manageable). Computing surface integrals can often be tedious, especially when the formula for the outward unit normal vector at each point of  changes. Put the value of the function and the lower and upper limits in the required blocks on the calculator then press the submit button. Similarly, the average value of a function of two variables over the rectangular perform a surface integral. Flux = = S F n d . In the field of graphical representation to build three-dimensional models. &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \sqrt{\sin^2\phi + \cos^2\phi} \, d\phi \\ Calculate surface integral Scurl F d S, where S is the surface, oriented outward, in Figure 16.7.6 and F = z, 2xy, x + y . The surface in Figure $\PageIndex{8a}$ can be parameterized by, \[\vecs r(u,v) = \langle (2 + \cos v) \cos u, \, (2 + \cos v) \sin u, \, \sin v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v < 2\pi \nonumber \], (we can use technology to verify). The simplest parameterization of the graph of $f$ is $\vecs r(x,y) = \langle x,y,f(x,y) \rangle$, where $x$ and $y$ vary over the domain of $f$ (Figure $\PageIndex{6}$). For example, let's say you want to calculate the magnitude of the electric flux through a closed surface around a 10 n C 10\ \mathrm{nC} 10 nC electric charge. Let $\vecs r(u,v)$ be a parameterization of $S$ with parameter domain $D$. Given that the thermal conductivity of cast iron is 55, find the heat flow across the boundary of the solid if this boundary is oriented outward. \nonumber \]. (1) where the left side is a line integral and the right side is a surface integral. The Divergence Theorem can be also written in coordinate form as. Letting the vector field $\rho \vecs{v}$ be an arbitrary vector field $\vecs{F}$ leads to the following definition. Parallelogram Theorems: Quick Check-in ; Kite Construction Template C F d s. using Stokes' Theorem. At this point weve got a fairly simple double integral to do. A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (a two-dimensional object) rather than a curve (a one-dimensional object). Direct link to Aiman's post Why do you add a function, Posted 3 years ago. This approximation becomes arbitrarily close to $\displaystyle \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \Delta S_{ij}$ as we increase the number of pieces $S_{ij}$ by letting $m$ and $n$ go to infinity. If it is possible to choose a unit normal vector $\vecs N$ at every point $(x,y,z)$ on $S$ so that $\vecs N$ varies continuously over $S$, then $S$ is orientable. Such a choice of unit normal vector at each point gives the orientation of a surface $S$. Flux through a cylinder and sphere. Divide rectangle $D$ into subrectangles $D_{ij}$ with horizontal width $\Delta u$ and vertical length $\Delta v$. Then, $S$ can be parameterized with parameters $x$ and $\theta$ by, \[\vecs r(x, \theta) = \langle x, f(x) \, \cos \theta, \, f(x) \sin \theta \rangle, \, a \leq x \leq b, \, 0 \leq x \leq 2\pi. \nonumber \], Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. Learning Objectives. Symbolab is the best integral calculator solving indefinite integrals, definite integrals, improper integrals, double integrals, triple integrals, multiple integrals, antiderivatives, and more. If you think of the normal field as describing water flow, then the side of the surface that water flows toward is the negative side and the side of the surface at which the water flows away is the positive side. In the first grid line, the horizontal component is held constant, yielding a vertical line through $(u_i, v_j)$. If you have any questions or ideas for improvements to the Integral Calculator, don't hesitate to write me an e-mail. The tangent vectors are $ \vecs t_x = \langle 1, \, 2x \, \cos \theta, \, 2x \, \sin \theta \rangle$ and $\vecs t_{\theta} = \langle 0, \, -x^2 \sin \theta, \, -x^2 \cos \theta \rangle$. Recall that when we defined a scalar line integral, we did not need to worry about an orientation of the curve of integration. There are two moments, denoted by M x M x and M y M y. A surface integral over a vector field is also called a flux integral. The total surface area is calculated as follows: SA = 4r 2 + 2rh where r is the radius and h is the height Horatio is manufacturing a placebo that purports to hone a person's individuality, critical thinking, and ability to objectively and logically approach different situations. So I figure that in order to find the net mass outflow I compute the surface integral of the mass flow normal to each plane and add them all up. To be precise, consider the grid lines that go through point $(u_i, v_j)$. MathJax takes care of displaying it in the browser. We assume this cone is in $\mathbb{R}^3$ with its vertex at the origin (Figure $\PageIndex{12}$). &= - 55 \int_0^{2\pi} \int_0^1 (2v \, \cos^2 u + 2v \, \sin^2 u ) \, dv \,du \\[4pt] The Integral Calculator will show you a graphical version of your input while you type. With the standard parameterization of a cylinder, Equation \ref{equation1} shows that the surface area is $2 \pi rh$. David Scherfgen 2023 all rights reserved. \nonumber \] Notice that $S$ is not a smooth surface but is piecewise smooth, since $S$ is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). First, we calculate $\displaystyle \iint_{S_1} z^2 \,dS.$ To calculate this integral we need a parameterization of $S_1$. This is easy enough to do. Dont forget that we need to plug in for $x$, $y$ and/or $z$ in these as well, although in this case we just needed to plug in $z$. For each point $\vecs r(a,b)$ on the surface, vectors $\vecs t_u$ and $\vecs t_v$ lie in the tangent plane at that point. Therefore, the lateral surface area of the cone is $\pi r \sqrt{h^2 + r^2}$. However, before we can integrate over a surface, we need to consider the surface itself. If the density of the sheet is given by $\rho (x,y,z) = x^2 yz$, what is the mass of the sheet? In "Options", you can set the variable of integration and the integration bounds. By Equation, \[ \begin{align*} \iint_{S_3} -k \vecs \nabla T \cdot dS &= - 55 \int_0^{2\pi} \int_1^4 \vecs \nabla T(u,v) \cdot (\vecs t_u \times \vecs t_v) \, dv\, du \\[4pt] The surface area of $S$ is, \[\iint_D ||\vecs t_u \times \vecs t_v || \,dA, \label{equation1} \], where $\vecs t_u = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle$, \[\vecs t_v = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle. We can also find different types of surfaces given their parameterization, or we can find a parameterization when we are given a surface. For any point $(x,y,z)$ on $S$, we can identify two unit normal vectors $\vecs N$ and $-\vecs N$. For example, consider curve parameterization $\vecs r(t) = \langle 1,2\rangle, \, 0 \leq t \leq 5$. What does to integrate mean? Calculus: Integral with adjustable bounds. Posted 5 years ago. Sets up the integral, and finds the area of a surface of revolution. Wow what you're crazy smart how do you get this without any of that background? which leaves out the density. To create a Mbius strip, take a rectangular strip of paper, give the piece of paper a half-twist, and the glue the ends together (Figure $\PageIndex{20}$). Next, we need to determine just what $D$ is. Find the mass of the piece of metal. For a scalar function over a surface parameterized by and , the surface integral is given by. If $u$ is held constant, then we get vertical lines; if $v$ is held constant, then we get circles of radius 1 centered around the vertical line that goes through the origin. Similarly, when we define a surface integral of a vector field, we need the notion of an oriented surface. However, weve done most of the work for the first one in the previous example so lets start with that. Note as well that there are similar formulas for surfaces given by $y = g\left( {x,z} \right)$ (with $D$ in the $xz$-plane) and $x = g\left( {y,z} \right)$ (with $D$ in the $yz$-plane). GLAPS Model: Sea Surface and Ground Temperature, http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx. We gave the parameterization of a sphere in the previous section. Hold $u$ and $v$ constant, and see what kind of curves result. To embed a widget in your blog's sidebar, install the Wolfram|Alpha Widget Sidebar Plugin, and copy and paste the Widget ID below into the "id" field: We appreciate your interest in Wolfram|Alpha and will be in touch soon. Calculate surface integral \[\iint_S (x + y^2) \, dS, \nonumber \] where $S$ is cylinder $x^2 + y^2 = 4, \, 0 \leq z \leq 3$ (Figure $\PageIndex{15}$). The double integrals calculator displays the definite and indefinite double integral with steps against the given function with comprehensive calculations. \[\iint_S f(x,y,z) \,dS = \iint_D f (\vecs r(u,v)) ||\vecs t_u \times \vecs t_v||\,dA \nonumber \], \[\iint_S \vecs F \cdot \vecs N \, dS = \iint_S \vecs F \cdot dS = \iint_D \vecs F (\vecs r (u,v)) \cdot (\vecs t_u \times \vecs t_v) \, dA \nonumber \]. Otherwise, it tries different substitutions and transformations until either the integral is solved, time runs out or there is nothing left to try. The parameterization of the cylinder and $\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|$ is. Some surfaces cannot be oriented; such surfaces are called nonorientable. The arc length formula is derived from the methodology of approximating the length of a curve. A useful parameterization of a paraboloid was given in a previous example. It is used to calculate the area covered by an arc revolving in space. In "Examples", you can see which functions are supported by the Integral Calculator and how to use them. You can accept it (then it's input into the calculator) or generate a new one. How to calculate the surface integral of the vector field: $$\iint\limits_{S^+} \vec F\cdot \vec n {\rm d}S $$ Is it the same thing to: $$\iint\limits_{S^+}x^2{\rm d}y{\rm d}z+y^2{\rm d}x{\rm d}z+z^2{\rm d}x{\rm d}y$$ There is another post here with an answer by@MichaelE2 for the cases when the surface is easily described in parametric form . Like so many things in multivariable calculus, while the theory behind surface integrals is beautiful, actually computing one can be painfully labor intensive. Where L is the length of the function y = f (x) on the x interval [ a, b] and dy / dx is the derivative of the function y = f (x) with respect to x. The abstract notation for surface integrals looks very similar to that of a double integral: Computing a surface integral is almost identical to computing, You can find an example of working through one of these integrals in the. The temperature at point $(x,y,z)$ in a region containing the cylinder is $T(x,y,z) = (x^2 + y^2)z$. The definition of a scalar line integral can be extended to parameter domains that are not rectangles by using the same logic used earlier. Therefore, we have the following equation to calculate scalar surface integrals: \[\iint_S f(x,y,z)\,dS = \iint_D f(\vecs r(u,v)) ||\vecs t_u \times \vecs t_v||\,dA. To parameterize a sphere, it is easiest to use spherical coordinates. Since $S_{ij}$ is small, the dot product $\rho v \cdot N$ changes very little as we vary across $S_{ij}$ and therefore $\rho \vecs v \cdot \vecs N$ can be taken as approximately constant across $S_{ij}$. In fact the integral on the right is a standard double integral. The Surface Area Calculator uses a formula using the upper and lower limits of the function for the axis along which the arc revolves. With a parameterization in hand, we can calculate the surface area of the cone using Equation \ref{equation1}. Step 1: Chop up the surface into little pieces. Notice that this parameterization involves two parameters, $u$ and $v$, because a surface is two-dimensional, and therefore two variables are needed to trace out the surface. \nonumber \]. Did this calculator prove helpful to you? That is: To make the work easier I use the divergence theorem, to replace the surface integral with a . Describe the surface integral of a vector field. This surface has parameterization $\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 1 \leq v \leq 4$. \end{align*}\], By Equation \ref{equation1}, the surface area of the cone is, \[ \begin{align*}\iint_D ||\vecs t_u \times \vecs t_v|| \, dA &= \int_0^h \int_0^{2\pi} kv \sqrt{1 + k^2} \,du\, dv \\[4pt] &= 2\pi k \sqrt{1 + k^2} \int_0^h v \,dv \\[4pt] &= 2 \pi k \sqrt{1 + k^2} \left[\dfrac{v^2}{2}\right]_0^h \\[4pt] \\[4pt] &= \pi k h^2 \sqrt{1 + k^2}. If parameterization $\vec{r}$ is regular, then the image of $\vec{r}$ is a two-dimensional object, as a surface should be. What Is a Surface Area Calculator in Calculus? The mass of a sheet is given by Equation \ref{mass}. Direct link to Is Better Than 's post Well because surface inte, Posted 2 years ago. The domain of integration of a surface integral is a surface in a plane or space, rather than a curve in a plane or space. The component of the vector $\rho v$ at P in the direction of $\vecs{N}$ is $\rho \vecs v \cdot \vecs N$ at $P$. Since some surfaces are nonorientable, it is not possible to define a vector surface integral on all piecewise smooth surfaces. When the "Go!" This was to keep the sketch consistent with the sketch of the surface. \end{align*}\]. To approximate the mass of fluid per unit time flowing across $S_{ij}$ (and not just locally at point $P$), we need to multiply $(\rho \vecs v \cdot \vecs N) (P)$ by the area of $S_{ij}$. Then, the unit normal vector is given by $\vecs N = \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||}$ and, from Equation \ref{surfaceI}, we have, \[\begin{align*} \int_C \vecs F \cdot \vecs N\, dS &= \iint_S \vecs F \cdot \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \,dS \\[4pt] Their difference is computed and simplified as far as possible using Maxima. Introduction. To obtain a parameterization, let $\alpha$ be the angle that is swept out by starting at the positive z-axis and ending at the cone, and let $k = \tan \alpha$. Let S be a smooth surface. \nonumber \]. A surface integral is like a line integral in one higher dimension. Free online 3D grapher from GeoGebra: graph 3D functions, plot surfaces, construct solids and much more! Describe the surface integral of a scalar-valued function over a parametric surface. Double integrals also can compute volume, but if you let f(x,y)=1, then double integrals boil down to the capabilities of a plain single-variable definite integral (which can compute areas). x-axis. The surface element contains information on both the area and the orientation of the surface. Following are the steps required to use the, The first step is to enter the given function in the space given in front of the title. You can use this calculator by first entering the given function and then the variables you want to differentiate against. where $D$ is the range of the parameters that trace out the surface $S$. This is not an issue though, because Equation \ref{scalar surface integrals} does not place any restrictions on the shape of the parameter domain. Solution. Here are the two individual vectors. Direct link to benvessely's post Wow what you're crazy sma. 192. y = x 3 y = x 3 from x = 0 x = 0 to x = 1 x = 1. If you imagine placing a normal vector at a point on the strip and having the vector travel all the way around the band, then (because of the half-twist) the vector points in the opposite direction when it gets back to its original position. where $S$ is the surface with parameterization $\vecs r(u,v) = \langle u, \, u^2, \, v \rangle$ for $0 \leq u \leq 2$ and $0 \leq v \leq u$. Enter the function you want to integrate into the Integral Calculator. \[S = \int_{0}^{4} 2 \pi y^{\dfrac1{4}} \sqrt{1+ (\dfrac{d(y^{\dfrac1{4}})}{dy})^2}\, dy \]. We know the formula for volume of a sphere is ( 4 / 3) r 3, so the volume we have computed is ( 1 / 8) ( 4 / 3) 2 3 = ( 4 / 3) , in agreement with our answer. &= \dfrac{2560 \sqrt{6}}{9} \approx 696.74. The vendor states an area of 200 sq cm. The interactive function graphs are computed in the browser and displayed within a canvas element (HTML5). To get an idea of the shape of the surface, we first plot some points. The surface integral of the vector field over the oriented surface (or the flux of the vector field across the surface ) can be written in one of the following forms: Here is called the vector element of the surface. Direct link to Surya Raju's post What about surface integr, Posted 4 years ago. In this video we come up formulas for surface integrals, which are when we accumulate the values of a scalar function over a surface. That is, we need a working concept of a parameterized surface (or a parametric surface), in the same way that we already have a concept of a parameterized curve. Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. I'll go over the computation of a surface integral with an example in just a bit, but first, I think it's important for you to have a good grasp on what exactly a surface integral, The double integral provides a way to "add up" the values of, Multiply the area of each piece, thought of as, Image credit: By Kormoran (Self-published work by Kormoran). Therefore we use the orientation, $\vecs N = \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle $, \[\begin{align*} \iint_S \rho v \cdot \,dS &= 80 \int_0^{2\pi} \int_0^{\pi/2} v (r(\phi, \theta)) \cdot (t_{\phi} \times t_{\theta}) \, d\phi \, d\theta \\ Math Assignments. Describe the surface parameterized by $\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle, \, - \infty < u < \infty, \, 0 \leq v < 2\pi$. 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. Step #4: Fill in the lower bound value. This surface has parameterization $\vecs r(x, \theta) = \langle x, \, x^2 \cos \theta, \, x^2 \sin \theta \rangle, \, 0 \leq x \leq b, \, 0 \leq x < 2\pi.$. Therefore, a parameterization of this cone is, \[\vecs s(u,v) = \langle kv \, \cos u, \, kv \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq h. \nonumber \]. is given explicitly by, If the surface is surface parameterized using Find the flux of F = y z j ^ + z 2 k ^ outward through the surface S cut from the cylinder y 2 + z 2 = 1, z 0, by the planes x = 0 and x = 1. Since we are working on the upper half of the sphere here are the limits on the parameters. Calculate line integral $\displaystyle \iint_S (x - y) \, dS,$ where $S$ is cylinder $x^2 + y^2 = 1, \, 0 \leq z \leq 2$, including the circular top and bottom. It follows from Example $\PageIndex{1}$ that we can parameterize all cylinders of the form $x^2 + y^2 = R^2$. Stokes' theorem is the 3D version of Green's theorem. For example, the graph of $f(x,y) = x^2 y$ can be parameterized by $\vecs r(x,y) = \langle x,y,x^2y \rangle$, where the parameters $x$ and $y$ vary over the domain of $f$. In Vector Calculus, the surface integral is the generalization of multiple integrals to integration over the surfaces. Therefore, the mass flux is, \[\iint_s \rho \vecs v \cdot \vecs N \, dS = \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij}. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface. The way to tell them apart is by looking at the differentials. &= - 55 \int_0^{2\pi} \int_0^1 \langle 8v \, \cos u, \, 8v \, \sin u, \, v^2\rangle \cdot \langle 0, 0, -v \rangle\, \, dv \,du\\[4pt]

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