&= \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 changes made to the formula should be the somewhat obvious changes. Notice the parallel between this definition and the definition of vector line integral $$\displaystyle \int_C \vecs F \cdot \vecs N\, dS$$. Derivation of Formula for Total Surface Area of the Sphere by Integration. To know more about great circle, see properties of a sphere. &= \int_0^3 \pi \, dv = 3 \pi. \end{align*}, Example $$\PageIndex{9}$$: Calculating the Surface Integral of a Cylinder. 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. Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. Since the surface is oriented outward and $$S_1$$ is the bottom of the object, it makes sense that this vector points downward. A parameterized surface is given by a description of the form, $\vecs{r}(u,v) = \langle x (u,v), \, y(u,v), \, z(u,v)\rangle.$. It can be characterized as the set of all points located distance r r r (radius) away from a given point (center). \label{mass}\], Example $$\PageIndex{11}$$: Calculating the Mass of a Sheet. Note that the surface area of a sphere of radius ð is ð´=4 ð2. The classic example of a nonorientable surface is the Möbius strip. For a review of integration methods on the sphere, see Keast and Diaz [6], Lebedev [7] an, d Stroud [13, Sections 2.6 and 8.4]. Suppose that $$v$$ is a constant $$K$$. Analogously, we would like a notion of regularity (or smoothness) for surfaces so that a surface parameterization really does trace out a surface. Let $$S$$ be a piecewise smooth surface with parameterization $$\vecs{r}(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle$$ with parameter domain $$D$$ and let $$f(x,y,z)$$ be a function with a domain that contains $$S$$. \end{align*}\]. This is called the positive orientation of the closed surface (Figure $$\PageIndex{18}$$). The surface area of the sphere is, $\int_0^{2\pi} \int_0^{\pi} r^2 \sin \phi \, d\phi \,d\theta = r^2 \int_0^{2\pi} 2 \, d\theta = 4\pi r^2. To get an orientation of the surface, we compute the unit normal vector, \[\vecs N = \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \nonumber$, In this case, $$\vecs t_u \times \vecs t_v = \langle r \, \cos u, \, r \, \sin u, \, 0 \rangle$$ and therefore, $||\vecs t_u \times \vecs t_v|| = \sqrt{r^2 \cos^2 u + r^2 \sin^2 u} = r. \nonumber$, $\vecs N(u,v) = \dfrac{\langle r \, \cos u, \, r \, \sin u, \, 0 \rangle }{r} = \langle \cos u, \, \sin u, \, 0 \rangle. \nonumber$. 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. A surface integral is similar to a line integral, except the integration is done over a surface rather than a path. I The surface is given in explicit form. The surface integral of the vector field $$\mathbf{F}$$ over the oriented surface $$S$$ (or the flux of the vector field $$\mathbf{F}$$ across the surface $$S$$) can be written in one of the following forms: Describe the surface integral of a scalar-valued function over a parametric surface. The tangent vectors are $$\vecs t_x = \langle 1,0,1 \rangle$$ and $$\vecs t_y = \langle 1,0,2 \rangle$$. Use Equation \ref{scalar surface integrals}. A = a 2 â« 0 Ï 2 â« 0 Ï 2 s i n v d u d v A = Ï a 2 2. By Equation, the heat flow across $$S_1$$ is, \begin{align*}\iint_{S_1} -k \vecs \nabla T \cdot dS &= - 55 \int_0^{2\pi} \int_0^1 \vecs \nabla T(u,v) \cdot (\vecs t_u \times \vecs t_v) \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 \langle 2v \, \cos u, \, 2v \, \sin u, \, v^2 \cos^2 u + v^2 \sin^2 u \rangle \cdot \langle 0,0, -v\rangle \, dv \,du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 \langle 2v \, \cos u, \, 2v \, \sin u, \, v^2\rangle \cdot \langle 0, 0, -v \rangle \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 -v^3 \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} -\dfrac{1}{4} du \\[4pt] &= \dfrac{55\pi}{2}.\end{align*}, Now let’s consider the circular top of the object, which we denote $$S_2$$. In analytic geometry, a sphere with center (x0, y0, z0) and radius r is the locus of all points (x, y, z) such that The horizontal cross-section of the cone at height $$z = u$$ is circle $$x^2 + y^2 = u^2$$. If we only care about a piece of the graph of $$f$$ - say, the piece of the graph over rectangle $$[ 1,3] \times [2,5]$$ - then we can restrict the parameter domain to give this piece of the surface: $\vecs r(x,y) = \langle x,y,x^2y \rangle, \, 1 \leq x \leq 3, \, 2 \leq y \leq 5.$. &= \rho^2 \sin^2 \phi (\cos^2 \theta + \sin^2 \theta) \$4pt] Therefore, the mass flow rate is $$7200\pi \, \text{kg/sec/m}^2$$. Vector $$\vecs t_u \times \vecs t_v$$ is normal to the tangent plane at $$\vecs r(a,b)$$ and is therefore normal to $$S$$ at that point. DEFINITION: surface integral of a scalar-valued function, The surface integral of a scalar-valued function of $$f$$ over a piecewise smooth surface $$S$$ is, \[\iint_S f(x,y,z) dA = \lim_{m,n\rightarrow \infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \Delta S_{ij}.$. If piece $$S_{ij}$$ is small enough, then the tangent plane at point $$P_{ij}$$ is a good approximation of piece $$S_{ij}$$. First, we are using pretty much the same surface (the integrand is different however) as the previous example. Describe the surface with parameterization, $\vecs{r} (u,v) = \langle 2 \, \cos u, \, 2 \, \sin u, \, v \rangle, \, 0 \leq u \leq 2\pi, \, -\infty < v < \infty \nonumber$. 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 $$\PageIndex{4}$$). &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54 \, \sin^3 \phi + 27 \, \cos^2 \phi \, \sin \phi \, d\phi \, d\theta \\ Informally, a choice of orientation gives $$S$$ an “outer” side and an “inner” side (or an “upward” side and a “downward” side), just as a choice of orientation of a curve gives the curve “forward” and “backward” directions. The second method for evaluating a surface integral is for those surfaces that are given by the parameterization. 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.. Recall the definition of vectors $$\vecs t_u$$ and $$\vecs t_v$$: $\vecs t_u = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle\, \text{and} \, \vecs t_v = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle. 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.$, Example $$\PageIndex{7}$$: Calculating Surface Area. \end{align*}\], Therefore, the rate of heat flow across $$S$$ is, $\dfrac{55\pi}{2} - \dfrac{55\pi}{2} - 110\pi = -110\pi. Since we are working on the upper half of the sphere here are the limits on the parameters. First, letâs look at the surface integral in which the surface $$S$$ is given by $$z = g\left( {x,y} \right)$$. For grid curve $$\vecs r(u_i,v)$$, the tangent vector at $$P_{ij}$$ is, \[\vecs t_v (P_{ij}) = \vecs r_v (u_i,v_j) = \langle x_v (u_i,v_j), \, y_v(u_i,v_j), \, z_v (u_i,v_j) \rangle.$, For grid curve $$\vecs r(u, v_j)$$, the tangent vector at $$P_{ij}$$ is, $\vecs t_u (P_{ij}) = \vecs r_u (u_i,v_j) = \langle x_u (u_i,v_j), \, y_u(u_i,v_j), \, z_u (u_i,v_j) \rangle.$. In this example we broke a surface integral over a piecewise surface into the addition of surface integrals over smooth subsurfaces. For any point $$(x,y,z)$$ on $$S$$, we can identify two unit normal vectors $$\vecs N$$ and $$-\vecs N$$. Therefore, $\vecs t_u \times \vecs t_v = \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \nonumber \\ 1 & 2u & 0 \nonumber \\ 0 & 0 & 1 \end{vmatrix} = \langle 2u, \, -1, \, 0 \rangle\ \nonumber$, $||\vecs t_u \times \vecs t_v|| = \sqrt{1 + 4u^2}.$. Let $$\vecs v(x,y,z) = \langle 2x, \, 2y, \, z\rangle$$ represent a velocity field (with units of meters per second) of a fluid with constant density 80 kg/m3. Choose point $$P_{ij}$$ in each piece $$S_{ij}$$ evaluate $$P_{ij}$$ at $$f$$, and multiply by area $$S_{ij}$$ to form the Riemann sum, $\sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \, \Delta S_{ij}. Conversely, each point on the cylinder is contained in some circle $$\langle \cos u, \, \sin u, \, k \rangle$$ for some $$k$$, and therefore each point on the cylinder is contained in the parameterized surface (Figure $$\PageIndex{2}$$). Which of the figures in Figure $$\PageIndex{8}$$ is smooth? \label{surfaceI}$. Suppose that i ranges from 1 to m and j ranges from 1 to n so that $$D$$ is subdivided into mn rectangles. Suppose that $$u$$ is a constant $$K$$. The rate of flow, measured in mass per unit time per unit area, is $$\rho \vecs N$$. A âsimpleâ surface-integral over the unit-sphere [closed] Ask Question Asked 5 days ago. Example $$\PageIndex{1}$$: Parameterizing a Cylinder, $\vecs{r}(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, -\infty < u < \infty, \, -\infty < v < \infty. The same was true for scalar surface integrals: we did not need to worry about an “orientation” of the surface of integration. The parameterization of full sphere $$x^2 + y^2 + z^2 = 4$$ is, \[\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.$. To use Equation \ref{scalar surface integrals} to calculate the surface integral, we first find vectors $$\vecs t_u$$ and $$\vecs t_v$$. Then the heat flow is a vector field proportional to the negative temperature gradient in the object. There is a lot of information that we need to keep track of here. Therefore, the surface is the elliptic paraboloid $$x^2 + y^2 = z$$ (Figure $$\PageIndex{3}$$). This is not the case with surfaces, however. We have seen that a line integral is an integral over a path in a plane or in space. &= \int_0^3 \int_0^{2\pi} (\cos u + \sin^2 u) \, du \,dv \\ We assume here and throughout that the surface parameterization $$\vecs r(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle$$ is continuously differentiable—meaning, each component function has continuous partial derivatives. Note that we can form a grid with lines that are parallel to the $$u$$-axis and the $$v$$-axis in the $$uv$$-plane. Email. Next, we need to determine just what $$D$$ is. By Equation \ref{scalar surface integrals}, \begin{align*} \iint_S 5 \, dS &= 5 \iint_D u \sqrt{1 + 4u^2} \, dA \\ Some surfaces cannot be oriented; such surfaces are called nonorientable. This surface has parameterization $$\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 1 \leq v \leq 4$$. Compute the surface integral where S is that part of the plane x+y+z=2 in the first octant. Therefore, the flux of $$\vecs{F}$$ across $$S$$ is 340. \nonumber. To visualize $$S$$, we visualize two families of curves that lie on $$S$$. 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] 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). &= \sqrt{6} \int_0^4 \int_0^2 x^2 y (1 + x + 2y) \, dy \,dx \\[4pt] If we think of $$\vecs r$$ as a mapping from the $$uv$$-plane to $$\mathbb{R}^3$$, the grid curves are the image of the grid lines under $$\vecs r$$. The sphere of radius $$\rho$$ centered at the origin is given by the parameterization, $$\vecs r(\phi,\theta) = \langle \rho \, \cos \theta \, \sin \phi, \, \rho \, \sin \theta \, \sin \phi, \, \rho \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi.$$, The idea of this parameterization is that as $$\phi$$ sweeps downward from the positive $$z$$-axis, a circle of radius $$\rho \, \sin \phi$$ is traced out by letting $$\theta$$ run from 0 to $$2\pi$$. The surface integral is then. Surfaces can be parameterized, just as curves can be parameterized. Let’s now generalize the notions of smoothness and regularity to a parametric surface. Now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. Then, \[\vecs t_u \times \vecs t_v = \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \\ -\sin u & \cos u & 0 \\ 0 & 0 & 1 \end{vmatrix} = \langle \cos u, \, \sin u, \, 0 \rangle. The surface integral for flux. 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. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. So, letâs do the integral. Suppose that the temperature at point $$(x,y,z)$$ in an object is $$T(x,y,z)$$. In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. &= \rho^2 \, \sin^2 \phi \$4pt] which leaves out the density. Notice also that $$\vecs r'(t) = \vecs 0$$. To calculate a surface integral with an integrand that is a function, use, If $$S$$ is a surface, then the area of $$S$$ is \[\iint_S \, dS.$. The fact that the derivative is the zero vector indicates we are not actually looking at a curve. This results in the desired circle (Figure $$\PageIndex{5}$$). &= 5 \int_0^2 \int_0^u \sqrt{1 + 4u^2} \, dv \, du = 5 \int_0^2 u \sqrt{1 + 4u^2}\, du \\ What we will attempt to start to do in this video is take the surface integral of the function x squared over our surface, where the surface in question, the surface we're going to care about is going to be the unit sphere. &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54\, \sin \phi - 27 \, \cos^2 \phi \, \sin \phi \, d\phi \,d\theta \\ By Sarahisme, August 27, 2006 in Mathematics. &= 2\pi \sqrt{3}. Both mass flux and flow rate are important in physics and engineering. The integral on the left however is a surface integral. &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \, d\phi \\ Here are the ranges for $$y$$ and $$z$$. Therefore, we can calculate the surface area of a surface of revolution by using the same techniques. &= \sqrt{6} \int_0^4 \dfrac{22x^2}{3} + 2x^3 \,dx \4pt] The temperature at point $$(x,y,z)$$ in a region containing the cylinder is $$T(x,y,z) = (x^2 + y^2)z$$. Letting the vector field $$\rho \vecs{v}$$ be an arbitrary vector field $$\vecs{F}$$ leads to the following definition. Throughout this chapter, parameterizations $$\vecs r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle$$are assumed to be regular. The parameters $$u$$ and $$v$$ vary over a region called the parameter domain, or parameter space—the set of points in the $$uv$$-plane that can be substituted into $$\vecs r$$. Our goal is to define a surface integral, and as a first step we have examined how to parameterize a surface. 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. A flat sheet of metal has the shape of surface $$z = 1 + x + 2y$$ that lies above rectangle $$0 \leq x \leq 4$$ and $$0 \leq y \leq 2$$. As we integrate over the surface, we must choose the normal vectors \bf N in such a way that they point "the same way'' through the surface. Want to improve this question? \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}. To compute the flow rate of the fluid in Example, we simply remove the density constant, which gives a flow rate of $$90 \pi \, m^3/sec$$. The temperature at a point in a region containing the ball is $$T(x,y,z) = \dfrac{1}{3}(x^2 + y^2 + z^2)$$. This division of $$D$$ into subrectangles gives a corresponding division of surface $$S$$ into pieces $$S_{ij}$$. Describe the surface parameterized by $$\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle, \, - \infty < u < \infty, \, 0 \leq v < 2\pi$$. A parameterization is $$\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, 0 \leq u \leq 2\pi, \, 0 \leq v \leq 3.$$. 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 $$kv$$. and $$||\vecs t_u \times \vecs t_v || = \sqrt{\cos^2 u + \sin^2 u} = 1$$. 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 $$\PageIndex{13}$$). Therefore, we calculate three separate integrals, one for each smooth piece of $$S$$. The mass of a sheet is given by Equation \ref{mass}. Solution The unit normal at the point P(x,y,z) points away from the centre of the sphere i.e. In fact the integral on the right is a standard double integral. In the first grid line, the horizontal component is held constant, yielding a vertical line through $$(u_i, v_j)$$. \end{align*}. &= 2\pi \left[ \dfrac{1}{64} \left(2 \sqrt{4x^2 + 1} (8x^3 + x) \, \sinh^{-1} (2x)\right)\right]_0^b \$4pt] Show that the surface area of the sphere $$x^2 + y^2 + z^2 = r^2$$ is $$4 \pi r^2$$. If we choose the unit normal vector that points “above” the surface at each point, then the unit normal vectors vary continuously over the surface. So it could be defined by x squared plus y squared plus z squared is equal to 1. 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, we expect the surface to be an elliptic paraboloid. For example, consider curve parameterization $$\vecs r(t) = \langle 1,2\rangle, \, 0 \leq t \leq 5$$. The surface integral will have a $$dS$$ while the standard double integral will have a $$dA$$. This equation for surface integrals is analogous to Equation \ref{surface} for line integrals: \[\iint_C f(x,y,z)\,ds = \int_a^b f(\vecs r(t))||\vecs r'(t)||\,dt. \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. 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. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces. \end{align*}\], Therefore, to compute a surface integral over a vector field we can use the equation, \iint_S \vecs F \cdot \vecs N\, dS = \iint_D (\vecs F (\vecs r (u,v)) \cdot (\vecs t_u \times \vecs t_v)) \,dA. It follows from Example $$\PageIndex{1}$$ that we can parameterize all cylinders of the form $$x^2 + y^2 = R^2$$. \end{align*}. Notice that the corresponding surface has no sharp corners. Here is that work. A surface integral is like a line integral in one higher dimension. &= \dfrac{2560 \sqrt{6}}{9} \approx 696.74. In the pyramid in Figure $$\PageIndex{8b}$$, the sharpness of the corners ensures that directional derivatives do not exist at those locations. Hold $$u$$ constant and see what kind of curves result. The upper limit for the $$z$$âs is the plane so we can just plug that in. Therefore the surface traced out by the parameterization is cylinder $$x^2 + y^2 = 1$$ (Figure $$\PageIndex{1}$$). for these kinds of surfaces. Therefore, to calculate, $\iint_{S_1} z^2 \,dS + \iint_{S_2} z^2 \,dS \nonumber$. In order to do this integral weâll need to note that just like the standard double integral, if the surface is split up into pieces we can also split up the surface integral. 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. Let $$S$$ be the surface that describes the sheet. \end{align*}\], \begin{align*} \vecs t_{\phi} \times \vecs t_{\theta} &= \sqrt{16 \, \cos^2\theta \, \sin^4\phi + 16 \, \sin^2\theta \, \sin^4 \phi + 16 \, \cos^2\phi \, \sin^2\phi} \\[4 pt] 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. Example $$\PageIndex{8}$$: Calculating a Surface Integral. 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. Solution: The surface is a quarter-sphere bounded by the xy and yz planes. However, the pyramid consists of four smooth faces, and thus this surface is piecewise smooth. \[\vecs r(\phi, \theta) = \langle 3 \, \cos \theta \, \sin \phi, \, 3 \, \sin \theta \, \sin \phi, \, 3 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi/2., As in Example, the tangent vectors are $$\vecs t_{\theta} = \langle -3 \, \sin \theta \, \sin \phi, \, 3 \, \cos \theta \, \sin \phi, \, 0 \rangle$$ and $$\vecs t_{\phi} = \langle 3 \, \cos \theta \, \cos \phi, \, 3 \, \sin \theta \, \cos \phi, \, -3 \, \sin \phi \rangle,$$ and their cross product is, $\vecs t_{\phi} \times \vecs t_{\theta} = \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle.$, Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. The integrand of a surface integral can be a scalar function or a vector field. Example $$\PageIndex{2}$$: Describing a Surface. By Example, we know that $$\vecs t_u \times \vecs t_v = \langle \cos u, \, \sin u, \, 0 \rangle$$. Surface integral example. Weâre going to need to do three integrals here. Here is a sketch of the surface $$S$$. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. &= 7200\pi.\end{align*} \]. Hold $$u$$ and $$v$$ constant, and see what kind of curves result. To define a surface integral of a scalar-valued function, we let the areas of the pieces of $$S$$ shrink to zero by taking a limit. Donât 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 $$z$$. The second step is to define the surface area of a parametric surface. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero. A = 4 Ï r 2. The way to tell them apart is by looking at the differentials. To develop a method that makes surface integrals easier to compute, we approximate surface areas $$\Delta S_{ij}$$ with small pieces of a tangent plane, just as we did in the previous subsection. &= 80 \int_0^{2\pi} \int_0^{\pi/2} \langle 6 \, \cos \theta \, \sin \phi, \, 6 \, \sin \theta \, \sin \phi, \, 3 \, \cos \phi \rangle \cdot \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle \, d\phi \, d\theta \\ &= 32\pi \left[- \dfrac{\cos^3 \phi}{3} \right]_0^{\pi/6} \\ 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 $$S$$. Notice that this parameter domain$$D$$ is a triangle, and therefore the parameter domain is not rectangular. Credits. Recall that curve parameterization $$\vecs r(t), \, a \leq t \leq b$$ is regular (or smooth) if $$\vecs r'(t) \neq \vecs 0$$ for all $$t$$ in $$[a,b]$$. We can extend the concept of a line integral to a surface integral to allow us to perform this integration. Sign in to follow this . Find the heat flow across the boundary of the solid if this boundary is oriented outward. The inside integral is evaluated using u-du substitution: â« 1 â25âð2 ð ðð 5 0 =[ââ25âð2] 0 5 =5. Let $$S$$ denote the boundary of the object. 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, In the previous section we looked at doing integrals in terms of cylindrical coordinates and we now need to take a quick look at doing integrals in terms of spherical coordinates. Next, we need to determine $${\vec r_\theta } \times {\vec r_\varphi }$$. Similarly, points $$\vecs r(\pi, 2) = (-1,0,2)$$ and $$\vecs r \left(\dfrac{\pi}{2}, 4\right) = (0,1,4)$$ are on $$S$$. &= \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] To get such an orientation, we parameterize the graph of $$f$$ in the standard way: $$\vecs r(x,y) = \langle x,\, y, \, f(x,y)\rangle$$, where $$x$$ and $$y$$ vary over the domain of $$f$$. Since we are only taking the piece of the sphere on or above plane $$z = 1$$, we have to restrict the domain of $$\phi$$. Also note that, for this surface, $$D$$ is the disk of radius $$\sqrt 3$$ centered at the origin. Watch the recordings here on Youtube! &= - 55 \int_0^{2\pi} \int_1^4 \langle 2v \, \cos u, \, 2v \, \sin u, \, \cos^2 u + \sin^2 u \rangle \cdot \langle \cos u, \, \sin u, \, 0 \rangle \, dv\, du \\[4pt] Therefore, \[\vecs t_u \times \vecs t_v = \langle -1 -2v, -1, 2v\rangle. A surface integral is like a line integral in one higher dimension. By Equation, the heat flow across $$S_1$$ is, \[ \begin{align*}\iint_{S_2} -k \vecs \nabla T \cdot dS &= - 55 \int_0^{2\pi} \int_0^1 \vecs \nabla T(u,v) \cdot\, (\vecs t_u \times \vecs t_v) \, dv\, du \\[4pt] This is not an issue though, because Equation \ref{scalar surface integrals} does not place any restrictions on the shape of the parameter domain. \end{align*}. If a region R is not flat, then it is called a surface as shown in the illustration. On the other hand, when we defined vector line integrals, the curve of integration needed an orientation. 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 Before calculating any integrals, note that the gradient of the temperature is $$\vecs \nabla T = \langle 2xz, \, 2yz, \, x^2 + y^2 \rangle$$. Divide rectangle $$D$$ into subrectangles $$D_{ij}$$ with horizontal width $$\Delta u$$ and vertical length $$\Delta v$$. The parameterization of the cylinder and $$\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|$$ is. &= \iint_D (\vecs F(\vecs r(u,v)) \cdot (\vecs t_u \times \vecs t_v))\,dA. 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. The unit vector points outwards from a closed surface and is usually denoted by Ën. 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. Notice that $$\vecs r_u = \langle 0,0,0 \rangle$$ and $$\vecs r_v = \langle 0, -\sin v, 0\rangle$$, and the corresponding cross product is zero. Next lesson. In general, surfaces must be parameterized with two parameters. 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. In fact, it can be shown that. However, since we are on the cylinder we know what $$y$$ is from the parameterization so we will also need to plug that in. Before we work some examples letâs notice that since we can parameterize a surface given by $$z = g\left( {x,y} \right)$$ as. Surfaces can sometimes be oriented, just as curves can be oriented. 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. 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] : Describing a surface of the parameters that trace out the surface integral of a smooth orientable surface parameterization! To seeing in the strip, the pyramid consists of infinitesimal patches that are given a integral... 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