To approximate the mass flux across \(S\), form the sum, \[\sum_{i=1}m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij}. Therefore, the pyramid has no smooth parameterization. \end{align*}\]. \end{align*}\], Therefore, the rate of heat flow across \(S\) is, \[\dfrac{55\pi}{2} - \dfrac{55\pi}{2} - 110\pi = -110\pi. 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 \\ For example, if we restricted the domain to \(0 \leq u \leq \pi, \, -\infty < v < 6\), then the surface would be a half-cylinder of height 6. Since this is for 1/8th of a sphere, the total surface integral of the sphere is 4 Ï a 2. Furthermore, all the vectors point outward, and therefore this is an outward orientation of the cylinder (Figure \(\PageIndex{19}\)). Let \(y = f(x) \geq 0\) be a positive single-variable function on the domain \(a \leq x \leq b\) and let \(S\) be the surface obtained by rotating \(f\) about the \(x\)-axis (Figure \(\PageIndex{13}\)). Suppose that i ranges from 1 to m and j ranges from 1 to n so that \(D\) is subdivided into mn rectangles. &= \langle 4 \, \cos \theta \, \sin^2 \phi, \, 4 \, \sin \theta \, \sin^2 \phi, \, 4 \, \cos^2 \theta \, \cos \phi \, \sin \phi + 4 \, \sin^2 \theta \, \cos \phi \, \sin \phi \rangle \\[4 pt] The surface area of \(S\) is, \[\iint_D ||\vecs t_u \times \vecs t_v || \,dA, \label{equation1}\], where \(\vecs t_u = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle\), \[\vecs t_v = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle.\], Example \(\PageIndex{5}\): Calculating Surface Area. \nonumber\]. A = 4 Ï r 2. Let \(\vecs v(x,y,z) = \langle x^2 + y^2, \, z, \, 4y \rangle\) m/sec represent a velocity field of a fluid with constant density 100 kg/m3. Recommended Posts. Then, the mass of the sheet is given by \(\displaystyle m = \iint_S x^2 yx \, dS.\) To compute this surface integral, we first need a parameterization of \(S\). Notice that we plugged in the equation of the plane for the x in the integrand. Google Classroom Facebook Twitter. Find the mass flow rate of the fluid across \(S\). 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). \end{align*}\], \[\iint_S z^2 \,dS = \iint_{S_1}z^2 \,dS + \iint_{S_2}z^2 \,dS, \nonumber\], \[\iint_S z^2 \,dS = (2\pi - 4) \sqrt{3} + \dfrac{32\pi}{3}. A surface integral over a vector field is also called a flux integral. &= 2\pi \int_0^{\sqrt{3}} u \, du \\ The notation for a surface integral of a function P(x,y,z) on a surface S is Note that if P(x,y,z)=1, then the above surface integral is equal to the surface area of S. Example. To calculate the mass flux across \(S\), chop \(S\) into small pieces \(S_{ij}\). \nonumber\]. Both mass flux and flow rate are important in physics and engineering. A flat sheet of metal has the shape of surface \(z = 1 + x + 2y\) that lies above rectangle \(0 \leq x \leq 4\) and \(0 \leq y \leq 2\). I Explicit, implicit, parametric equations of surfaces. Just as with line integrals, there are two kinds of surface integrals: a surface integral of a scalar-valued function and a surface integral of a vector field. Since the parameter domain is all of \(\mathbb{R}^2\), we can choose any value for u and v and plot the corresponding point. Active 5 days ago. Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line (Figure \(\PageIndex{7}\)). Let \(S\) be hemisphere \(x^2 + y^2 + z^2 = 9\) with \(z \leq 0\) such that \(S\) is oriented outward. In a similar way, to calculate a surface integral over surface \(S\), we need to parameterize \(S\). 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 \(S\). In fact the integral on the right is a standard double integral. The way to tell them apart is by looking at the differentials. 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 \\ 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. Have a happy day! It is perfectly symmetrical, and has no edges or vertices. Example \(\PageIndex{1}\): Parameterizing a Cylinder, \[\vecs{r}(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, -\infty < u < \infty, \, -\infty < v < \infty. If \(v = 0\) or \(v = \pi\), then the only choices for \(u\) that make the \(\mathbf{\hat{j}}\) component zero are \(u = 0\) or \(u = \pi\). The domain of integration of a scalar line integral is a parameterized curve (a one-dimensional object); the domain of integration of a scalar surface integral is a parameterized surface (a two-dimensional object). Before we work some examples letâs notice that since we can parameterize a surface given by \(z = g\left( {x,y} \right)\) as. Since we are not interested in the entire cone, only the portion on or above plane \(z = -2\), the parameter domain is given by \(-2 < u < \infty, \, 0 \leq v < 2\pi\) (Figure \(\PageIndex{4}\)). Example 17 If S is the surface of the sphere x 2+y2 +z2 = a ï¬nd the unit normal nË. The classic example of a nonorientable surface is the Möbius strip. For example, the graph of paraboloid \(2y = x^2 + z^2\) can be parameterized by \(\vecs r(x,y) = \left\langle x, \dfrac{x^2+z^2}{2}, z \right\rangle, \, 0 \leq x < \infty, \, 0 \leq z < \infty\). The rate of flow, measured in mass per unit time per unit area, is \(\rho \vecs N\). &= -110\pi. it lies in the direction of xi + yj + zk. Scalar surface integrals have several real-world applications. Surface Integral: implicit Definition For a surface S given implicitly by F(x, y, z) = c, where F is a continuously differentiable function, with S lying above its closed and bounded shadow region R in the coordinate plane beneath it, the surface integral of the continuous function G over S is given by the double integral R, Thus, a surface integral is similar to a line integral but in one higher dimension. The definition of a smooth surface parameterization is similar. If a region R is not flat, then it is called a surface as shown in the illustration. Therefore, as \(u\) increases, the radius of the resulting circle increases. Surfaces can be parameterized, just as curves can be parameterized. I Review: Double integral of a scalar function. Then the heat flow is a vector field proportional to the negative temperature gradient in the object. &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54\, \sin \phi - 27 \, \cos^2 \phi \, \sin \phi \, d\phi \,d\theta \\ It is not currently accepting answers. Therefore, \[\begin{align*} \iint_{S_1} z^2 \,dS &= \int_0^{\sqrt{3}} \int_0^{2\pi} f(r(u,v))||t_u \times t_v|| \, dv \, du \\ It is now time to think about integrating functions over some surface, \(S\), in three-dimensional space. Therefore, a point on the cone at height \(u\) has coordinates \((u \, \cos v, \, u \, \sin v, \, u)\) for angle \(v\). Therefore, the lateral surface area of the cone is \(\pi r \sqrt{h^2 + r^2}\). Credits. Vector surface integral examples by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License.For permissions beyond the scope of this license, please contact us.. To motivate the definition of regularity of a surface parameterization, consider the parameterization, \[\vecs r(u,v) = \langle 0, \, \cos v, \, 1 \rangle, \, 0 \leq u \leq 1, \, 0 \leq v \leq \pi.\]. 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. Informally, the surface integral of a scalar-valued function is an analog of a scalar line integral in one higher dimension. We used a rectangle here, but it doesnât have to be of course. Integrating spheres are very versatile optical elements, which are designed to achieve homogenous distribution of optical radiation by means of multiple Lambertian reflections at the sphere's inner surface. Some surfaces, such as a Möbius strip, cannot be oriented. To be precise, consider the grid lines that go through point \((u_i, v_j)\). In the definition of a surface integral, we chop a surface into pieces, evaluate a function at a point in each piece, and let the area of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. This is the currently selected item. &= - 55 \int_0^{2\pi} \int_0^1 -v^3 \, dv \,du = - 55 \int_0^{2\pi} -\dfrac{1}{4} \,du = - \dfrac{55\pi}{2}.\end{align*}\]. To parameterize a sphere, it is easiest to use spherical coordinates. Since some surfaces are nonorientable, it is not possible to define a vector surface integral on all piecewise smooth surfaces. Moreover, the theory of optimal methods is far from complete, as has \end{align*}\], \[ \begin{align*} \pi k h^2 \sqrt{1 + k^2} &= \pi \dfrac{r}{h}h^2 \sqrt{1 + \dfrac{r^2}{h^2}} \\[4pt] &= \pi r h \sqrt{1 + \dfrac{r^2}{h^2}} \\[4pt] \\[4pt] &= \pi r \sqrt{h^2 + h^2 \left(\dfrac{r^2}{h^2}\right) } \\[4pt] &= \pi r \sqrt{h^2 + r^2}. Investigate the cross product \(\vecs r_u \times \vecs r_v\). However, since we are on the cylinder we know what \(y\) is from the parameterization so we will also need to plug that in. Show that the surface area of the sphere \(x^2 + y^2 + z^2 = r^2\) is \(4 \pi r^2\). Therefore, the tangent of \(\phi\) is \(\sqrt{3}\), which implies that \(\phi\) is \(\pi / 6\). The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero. \end{align*}\]. Find the surface area of the surface with parameterization \(\vecs r(u,v) = \langle u + v, \, u^2, \, 2v \rangle, \, 0 \leq u \leq 3, \, 0 \leq v \leq 2\). Surfaces can sometimes be oriented, just as curves can be oriented. The surface of the unit sphere in 3D is defined by x^2 + y^2 + z^2 = 1 The integrands are all of the form f(x,y,z) = x^a y^b z^c where the exponents are nonnegative integers. A parameterized surface is given by a description of the form, \[\vecs{r}(u,v) = \langle x (u,v), \, y(u,v), \, z(u,v)\rangle.\]. Similarly, when we define a surface integral of a vector field, we need the notion of an oriented surface. 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. Integral \(\displaystyle \iint_S \vecs F \cdot \vecs N\, dS\) is called the flux of \(\vecs{F}\) across \(S\), just as integral \(\displaystyle \int_C \vecs F \cdot \vecs N\,dS\) is the flux of \(\vecs F\) across curve \(C\). Here is the parameterization for this sphere. So it could be defined by x squared plus y squared plus z squared is equal to 1. \nonumber\], As pieces \(S_{ij}\) get smaller, the sum, \[\sum_{i=1}m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij} \nonumber\], gets arbitrarily close to the mass flux. \nonumber\]. Hold \(u\) constant and see what kind of curves result. Visualize a surface integral. The changes made to the formula should be the somewhat obvious changes. Calculate surface integral \[\iint_S \vecs F \cdot \vecs N \, dS, \nonumber\] where \(\vecs F = \langle 0, -z, y \rangle\) and \(S\) is the portion of the unit sphere in the first octant with outward orientation. &= 2\pi \left[ \dfrac{1}{64} \left(2 \sqrt{4b^2 + 1} (8b^3 + b) \, \sinh^{-1} (2b) \right)\right]. Therefore, we calculate three separate integrals, one for each smooth piece of \(S\). Hold \(u\) and \(v\) constant, and see what kind of curves result. In mathematics, particularly multivariable calculus, a surface integral is a generalization of multiple integrals to integration over surfaces. First, we are using pretty much the same surface (the integrand is different however) as the previous example. We sketch S and from it, infer the region of integration R: The hemisphere can be described by rectangular coordinates 2+ 2+ =16, in which case Therefore, \(\vecs t_u = \langle -v \, \sin u, \, v \, \cos u, \, 0 \rangle\) and \(\vecs t_v = \langle \cos u, \, v \, \sin u, \, 0 \rangle \), and \(\vecs t_u \times \vecs t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0,0,-v\rangle\). Check if the flux through any bit of your surface ⦠&= \iint_D \left(\vecs F (\vecs r (u,v)) \cdot \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \right) || \vecs t_u \times \vecs t_v || \,dA \\[4pt] We now have a parameterization of \(S_2\): \(\vecs r(\phi, \theta) = \langle 2 \, \cos \theta \, \sin \phi, \, 2 \, \sin \theta \, \sin \phi, \, 2 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi / 3.\), The tangent vectors are \(\vecs t_{\phi} = \langle 2 \, \cos \theta \, \cos \phi, \, 2 \, \sin \theta \,\cos \phi, \, -2 \, \sin \phi \rangle\) and \(\vecs t_{\theta} = \langle - 2 \sin \theta \sin \phi, \, u\cos \theta \sin \phi, \, 0 \rangle\), and thus, \[\begin{align*} \vecs t_{\phi} \times \vecs t_{\theta} &= \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \nonumber \\ 2 \cos \theta \cos \phi & 2 \sin \theta \cos \phi & -2\sin \phi \\ -2\sin \theta\sin\phi & 2\cos \theta \sin\phi & 0 \end{vmatrix} \\[4 pt] Letâs start off with a sketch of the surface \(S\) since the notation can get a little confusing once we get into it. Describe the surface parameterized by \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle, \, - \infty < u < \infty, \, 0 \leq v < 2\pi\). Calculate surface integral \[\iint_S f(x,y,z)\,dS,\] where \(f(x,y,z) = z^2\) and \(S\) is the surface that consists of the piece of sphere \(x^2 + y^2 + z^2 = 4\) that lies on or above plane \(z = 1\) and the disk that is enclosed by intersection plane \(z = 1\) and the given sphere (Figure \(\PageIndex{16}\)). Viewed 90 times 0 $\begingroup$ Closed. \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle, \, 0 < u < \infty, \, 0 \leq v < \dfrac{\pi}{2}\), We have discussed parameterizations of various surfaces, but two important types of surfaces need a separate discussion: spheres and graphs of two-variable functions. The notation needed to develop this definition is used throughout the rest of this chapter. The entire surface is created by making all possible choices of \(u\) and \(v\) over the parameter domain. \nonumber\], From the material we have already studied, we know that, \[\Delta S_{ij} \approx ||\vecs t_u (P_{ij}) \times \vecs t_v (P_{ij})|| \,\Delta u \,\Delta v. \nonumber\], \[\iint_S f(x,y,z) \,dS \approx \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij})|| \vecs t_u(P_{ij}) \times \vecs t_v(P_{ij}) ||\,\Delta u \,\Delta v. \nonumber\]. The rate of heat flow across surface S in the object is given by the flux integral, \[\iint_S \vecs F \cdot dS = \iint_S -k \vecs \nabla T \cdot dS.\], Example \(\PageIndex{15}\): Calculating Heat Flow. Suppose that the temperature at point \((x,y,z)\) in an object is \(T(x,y,z)\). [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. for these kinds of surfaces. In the first grid line, the horizontal component is held constant, yielding a vertical line through \((u_i, v_j)\). In Example \(\PageIndex{14}\), we computed the mass flux, which is the rate of mass flow per unit area. 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. \nonumber\], Therefore, the radius of the disk is \(\sqrt{3}\) and a parameterization of \(S_1\) is \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, 1 \rangle, \, 0 \leq u \leq \sqrt{3}, \, 0 \leq v \leq 2\pi\). By Example, we know that \(\vecs t_u \times \vecs t_v = \langle \cos u, \, \sin u, \, 0 \rangle\). Therefore, a parameterization of this cone is, \[\vecs s(u,v) = \langle kv \, \cos u, \, kv \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq h. \nonumber\]. There are essentially two separate methods here, although as we will see they are really the same. Use a surface integral to calculate the area of a given surface. \end{align*}\], To calculate this integral, we need a parameterization of \(S_2\). Here is the remainder of the work for this problem. This is not the case with surfaces, however. If \(u = v = 0\), then \(\vecs r(0,0) = \langle 1,0,0 \rangle\), so point (1, 0, 0) is on \(S\). If the density of the sheet is given by \(\rho (x,y,z) = x^2 yz\), what is the mass of the sheet? Notice that if we change the parameter domain, we could get a different surface. 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, \[\begin{align*} \int_C \vecs F \cdot \vecs N\, dS &= \iint_S \vecs F \cdot \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \,dS \\[4pt] Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. \end{align*}\], 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}. Given the radius r of the sphere, the total surface area is. This results in the desired circle (Figure \(\PageIndex{5}\)). Now we need \({\vec r_z} \times {\vec r_\theta }\). Therefore, \[ \begin{align*} \vecs t_u \times \vecs t_v &= \begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ -kv \sin u & kv \cos u & 0 \\ k \cos u & k \sin u & 1 \end{vmatrix} \\[4pt] &= \langle kv \, \cos u, \, kv \, \sin u, \, -k^2 v \, \sin^2 u - k^2 v \, \cos^2 u \rangle \\[4pt] &= \langle kv \, \cos u, \, kv \, \sin u, \, - k^2 v \rangle. That path will be on the surface of the sphere [itex]x^2+ y^2+ z^2= a^2[/itex] if and only if [itex]f(t)^2+ g(t)^2+ h(t)^2= a^2[/itex] for all t. I have avoided using the term "line integral" because, of course, a straight line cannot lie on the surface of a sphere. Also note that, for this surface, \(D\) is the disk of radius \(\sqrt 3 \) centered at the origin. For example, consider curve parameterization \(\vecs r(t) = \langle 1,2\rangle, \, 0 \leq t \leq 5\). The surface integral of \(\vecs{F}\) over \(S\) is, \[\iint_S \vecs{F} \cdot \vecs{S} = \iint_S \vecs{F} \cdot \vecs{N} \,dS. Therefore, the mass flux is, \[\iint_s \rho \vecs v \cdot \vecs N \, dS = \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij}.\]. 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. Notice that this cylinder does not include the top and bottom circles. A = a 2 â« 0 Ï 2 â« 0 Ï 2 s i n v d u d v A = Ï a 2 2. Find the parametric representations of a cylinder, a cone, and a sphere. Suppose that \(v\) is a constant \(K\). Recall that if \(\vecs{F}\) is a two-dimensional vector field and \(C\) is a plane curve, then the definition of the flux of \(\vecs{F}\) along \(C\) involved chopping \(C\) into small pieces, choosing a point inside each piece, and calculating \(\vecs{F} \cdot \vecs{N}\) at the point (where \(\vecs{N}\) is the unit normal vector at the point). Notice that this parameter domain\(D\) is a triangle, and therefore the parameter domain is not rectangular. The second step is to define the surface area of a parametric surface. 304 Example 51.2: â¬Find 2 ð Ì, where S is the portion of sphere of radius 4, centered at the origin, such that â¥0 and â¥0. Donât forget that we need to plug in for \(z\)! Weâre going to need to do three integrals here. This was to keep the sketch consistent with the sketch of the surface. Here are the ranges for \(y\) and \(z\). However, the pyramid consists of four smooth faces, and thus this surface is piecewise smooth. In the second grid line, the vertical component is held constant, yielding a horizontal line through \((u_i, v_j)\). Review: Arc length and line integrals I The integral of a function f : [a,b] â R is \end{align*}\], \[\begin{align*} \iint_{S_2} z \, dS &= \int_0^{\pi/6} \int_0^{2\pi} f (\vecs r(\phi, \theta))||\vecs t_{\phi} \times \vecs t_{\theta}|| \, d\theta \, d\phi \\ To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in Figure \(\PageIndex{17}\) (the \(\sqrt{3}\) comes from the fact that the base of \(S\) is a disk with radius \(\sqrt{3}\)). Let \(S\) be a smooth orientable surface with parameterization \(\vecs r(u,v)\). Section 4-7 : Triple Integrals in Spherical Coordinates. The idea behind this parameterization is that for a fixed \(v\)-value, the circle swept out by letting \(u\) vary is the circle at height \(v\) and radius \(kv\). An oriented surface is given an “upward” or “downward” orientation or, in the case of surfaces such as a sphere or cylinder, an “outward” or “inward” orientation. This is easy enough to do. Although plotting points may give us an idea of the shape of the surface, we usually need quite a few points to see the shape. Furthermore, assume that \(S\) is traced out only once as \((u,v)\) varies over \(D\). If we want to find the flow rate (measured in volume per time) instead, we can use flux integral. Before calculating any integrals, note that the gradient of the temperature is \(\vecs \nabla T = \langle 2xz, \, 2yz, \, x^2 + y^2 \rangle\). In order to evaluate a surface integral we will substitute the equation of the surface in for \(z\) in the integrand and then add on the often messy square root. Example \(\PageIndex{4}\): Identifying Smooth and Nonsmooth Surfaces. Sign in to follow this . &= \int_0^3 \left[\sin u + \dfrac{u}{2} - \dfrac{\sin(2u)}{4} \right]_0^{2\pi} \,dv \\ If a thin sheet of metal has the shape of surface \(S\) and the density of the sheet at point \((x,y,z)\) is \(\rho(x,y,z)\) then mass \(m\) of the sheet is, \[\displaystyle m = \iint_S \rho (x,y,z) \,dS. How could we calculate the mass flux of the fluid across \(S\)? \end{align*}\]. Closed surfaces such as spheres are orientable: if we choose the outward normal vector at each point on the surface of the sphere, then the unit normal vectors vary continuously. Example \(\PageIndex{8}\): Calculating a Surface Integral. Note that \(\vecs t_u = \langle 1, 2u, 0 \rangle\) and \(\vecs t_v = \langle 0,0,1 \rangle\). Give a parameterization of the cone \(x^2 + y^2 = z^2\) lying on or above the plane \(z = -2\). Parameterize the surface and use the fact that the surface is the graph of a function. Because of the half-twist in the strip, the surface has no “outer” side or “inner” side. We arrived at the equation of the hypotenuse by setting \(x\) equal to zero in the equation of the plane and solving for \(z\). \label{equation 5}\], Example \(\PageIndex{13}\): Calculating a Surface Integral, \[\iint_S \vecs F \cdot \vecs N\,dS, \nonumber\], where \(\vecs{F} = \langle -y,x,0\rangle\) and \(S\) is the surface with parameterization, \[\vecs r(u,v) = \langle u,v^2 - u, \, u + v\rangle, \, 0 \leq u \leq 3, \, 0 \leq v \leq 4. Let \(\vecs r(u,v)\) be a parameterization of \(S\) with parameter domain \(D\). If piece \(S_{ij}\) is small enough, then the tangent plane at point \(P_{ij}\) is a good approximation of piece \(S_{ij}\). Hence, it is possible to think of every curve as an oriented curve. We have seen that a line integral is an integral over a path in a plane or in space. Since the surface is oriented outward and \(S_1\) is the bottom of the object, it makes sense that this vector points downward. Also, donât forget to plug in for \(z\). &= 32 \pi \int_0^{\pi/6} \cos^2\phi \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2 \phi} \, d\phi \\ &= 2\pi \left[ \dfrac{1}{64} \left(2 \sqrt{4x^2 + 1} (8x^3 + x) \, \sinh^{-1} (2x)\right)\right]_0^b \\[4pt] In this case, vector \(\vecs t_u \times \vecs t_v\) is perpendicular to the surface, whereas vector \(\vecs r'(t)\) is tangent to the curve. Use Equation \ref{equation1} to find the area of the surface of revolution obtained by rotating curve \(y = \sin x, \, 0 \leq x \leq \pi\) about the \(x\)-axis. A surface integral is like a line integral in one higher dimension. &= - 55 \int_0^{2\pi} \int_0^1 \langle 2v \, \cos^2 u, \, 2v \, \sin u, \, 1 \rangle \cdot \langle \cos u, \, \sin u, \, 0 \rangle \, dv\,\, du \\[4pt] Example, but this technique extends to finitely many smooth subsurfaces in this sense, surface integrals are important z! Analogous to the definition of a scalar function or a vector field which is not possible define. The sheet other hand, we calculate three separate integrals, the definition is analogous the! Flow rate of mass flow per unit time, flow rate integration needed an orientation \rho \vecs )! Then the heat flow across the boundary of the sphere x 2+y2 =... Sketch consistent with the sketch of the work for this problem and the previous section we should be somewhat! Keep the sketch consistent with the idea of orientable surfaces in place, we need to be precise, the. 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