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Notice the parallel between this definition and the definition of vector line integral $$\displaystyle \int_C \vecs F \cdot \vecs N\, dS$$. Notice that we do not need to vary over the entire domain of $$y$$ because $$x$$ and $$z$$ are squared. The changes made to the formula should be the somewhat obvious changes. The definition is analogous to the definition of the flux of a vector field along a plane curve. Therefore, the definition of a surface integral follows the definition of a line integral quite closely. Calculate surface integral $\iint_S (x + y^2) \, dS,$ where $$S$$ is cylinder $$x^2 + y^2 = 4, \, 0 \leq z \leq 3$$ (Figure $$\PageIndex{15}$$). Now we need $${\vec r_z} \times {\vec r_\theta }$$. The surface integral will have a $$dS$$ while the standard double integral will have a $$dA$$. However, weâve done most of the work for the first one in the previous example so letâs start with that. Finally, the bottom of the cylinder (not shown here) is the disk of radius $$\sqrt 3$$ in the $$xy$$-plane and is denoted by $${S_3}$$. A sphere is a perfectly round geometrical 3-dimensional object. Notice that we plugged in the equation of the plane for the x in the integrand. \end{align*}\], \begin{align*}||\vecs t_{\phi} \times \vecs t_{\theta} || &= \sqrt{r^4\sin^4\phi \, \cos^2 \theta + r^4 \sin^4 \phi \, \sin^2 \theta + r^4 \sin^2 \phi \, \cos^2 \phi} \\[4pt] &= \sqrt{r^4 \sin^4 \phi + r^4 \sin^2 \phi \, \cos^2 \phi} \\[4pt] &= r^2 \sqrt{\sin^2 \phi} \\[4pt] &= r \, \sin \phi.\end{align*}, Notice that $$\sin \phi \geq 0$$ on the parameter domain because $$0 \leq \phi < \pi$$, and this justifies equation $$\sqrt{\sin^2 \phi} = \sin \phi$$. 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. Let $$\theta$$ be the angle of rotation. The tangent vectors are $$\vecs t_x = \langle 1,0,1 \rangle$$ and $$\vecs t_y = \langle 1,0,2 \rangle$$. The component of the vector $$\rho v$$ at P in the direction of $$\vecs{N}$$ is $$\rho \vecs v \cdot \vecs N$$ at $$P$$. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface. \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 \\ Varying point $$P_{ij}$$ over all pieces $$S_{ij}$$ and the previous approximation leads to the following definition of surface area of a parametric surface (Figure $$\PageIndex{11}$$). 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. Suppose that $$u$$ is a constant $$K$$. But, these choices of $$u$$ do not make the $$\mathbf{\hat{i}}$$ component zero. Thanks to William Sears for correcting errors. Finally, to parameterize the graph of a two-variable function, we first let $$z = f(x,y)$$ be a function of two variables. In the pyramid in Figure $$\PageIndex{8b}$$, the sharpness of the corners ensures that directional derivatives do not exist at those locations. Therefore, we calculate three separate integrals, one for each smooth piece of $$S$$. 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). To be precise, consider the grid lines that go through point $$(u_i, v_j)$$. It is perfectly symmetrical, and has no edges or vertices. and $$||\vecs t_u \times \vecs t_v || = \sqrt{\cos^2 u + \sin^2 u} = 1$$. Surface integrals are important for the same reasons that line integrals are important. Therefore, we have the following characterization of the flow rate of a fluid with velocity $$\vecs v$$ across a surface $$S$$: $\text{Flow rate of fluid across S} = \iint_S \vecs v \cdot dS. &= 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 \\ \[\vecs{N}(x,y) = \left\langle \dfrac{-y}{\sqrt{1+x^2+y^2}}, \, \dfrac{-x}{\sqrt{1+x^2+y^2}}, \, \dfrac{1}{\sqrt{1+x^2+y^2}} \right\rangle \nonumber$. This surface has parameterization $$\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 1 \leq v \leq 4$$. The surface integral for flux. \nonumber\]. A cast-iron solid cylinder is given by inequalities $$x^2 + y^2 \leq 1, \, 1 \leq z \leq 4$$. \end{align*}\], Therefore, the rate of heat flow across $$S$$ is, \dfrac{55\pi}{2} - \dfrac{55\pi}{2} - 110\pi = -110\pi. In the first family of curves we hold $$u$$ constant; in the second family of curves we hold $$v$$ constant. 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. The classic example of a nonorientable surface is the Möbius strip. &=80 \int_0^{2\pi} 45 \, d\theta \\ 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. &= \int_0^{\sqrt{3}} \int_0^{2\pi} u \, dv \, du \\ Let S be a smooth surface. 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. 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. 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. Similarly, points $$\vecs r(\pi, 2) = (-1,0,2)$$ and $$\vecs r \left(\dfrac{\pi}{2}, 4\right) = (0,1,4)$$ are on $$S$$. To compute the flow rate of the fluid in Example, we simply remove the density constant, which gives a flow rate of $$90 \pi \, m^3/sec$$. \end{align*}, To calculate this integral, we need a parameterization of $$S_2$$. A useful parameterization of a paraboloid was given in a previous example. 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 \\ $$r \, \cos \theta \, \sin \phi, \, r \, \sin \theta \, \sin \phi, \, r \, \cos \phi \rangle, \, 0 \leq \theta < 2\pi, \, 0 \leq \phi \leq \pi.$$, $$\vecs t_{\theta} = \langle -r \, \sin \theta \, \sin \phi, \, r \, \cos \theta \, \sin \phi, \, 0 \rangle$$, $$\vecs t_{\phi} = \langle r \, \cos \theta \, \cos \phi, \, r \, \sin \theta \, \cos \phi, \, -r \, \sin \phi \rangle.$$, \[ \begin{align*}\vecs t_{\phi} \times \vecs t_{\theta} &= \langle r^2 \cos \theta \, \sin^2 \phi, \, r^2 \sin \theta \, \sin^2 \phi, \, r^2 \sin^2 \theta \, \sin \phi \, \cos \phi + r^2 \cos^2 \theta \, \sin \phi \, \cos \phi \rangle \\[4pt] &= \langle r^2 \cos \theta \, \sin^2 \phi, \, r^2 \sin \theta \, \sin^2 \phi, \, r^2 \sin \phi \, \cos \phi \rangle. In fact the integral on the right is a standard double integral. Credits. A portion of the graph of any smooth function $$z = f(x,y)$$ is also orientable. The tangent vectors are $$\vecs t_u = \langle 1,-1,1\rangle$$ and $$\vecs t_v = \langle 0,2v,1\rangle$$. Now, we need to be careful here as both of these look like standard double integrals. Divide rectangle $$D$$ into subrectangles $$D_{ij}$$ with horizontal width $$\Delta u$$ and vertical length $$\Delta v$$. Because of the half-twist in the strip, the surface has no “outer” side or “inner” side. Since some surfaces are nonorientable, it is not possible to define a vector surface integral on all piecewise smooth surfaces. Let $$S$$ denote the boundary of the object. Since the parameter domain is all of $$\mathbb{R}^2$$, we can choose any value for u and v and plot the corresponding point. In a similar way, to calculate a surface integral over surface $$S$$, we need to parameterize $$S$$. Find the mass flow rate of the fluid across $$S$$. Now at this point we can proceed in one of two ways. In this case the surface integral is. 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. A âsimpleâ surface-integral over the unit-sphere [closed] Ask Question Asked 5 days ago. Break the integral into three separate surface integrals. Notice that if we change the parameter domain, we could get a different surface. The basic idea is to chop the parameter domain into small pieces, choose a sample point in each piece, and so on. Let $$\vecs{F}$$ be a continuous vector field with a domain that contains oriented surface $$S$$ with unit normal vector $$\vecs{N}$$. Therefore, the unit normal vector at $$P$$ can be used to approximate $$\vecs N(x,y,z)$$ across the entire piece $$S_{ij}$$ because the normal vector to a plane does not change as we move across the plane. For scalar surface integrals, we chop the domain region (no longer a curve) into tiny pieces and proceed in the same fashion. \nonumber. Then the curve traced out by the parameterization is $$\langle \cos u, \, \sin u, \, K \rangle$$, which gives a circle in plane $$z = K$$ with radius 1 and center $$(0, 0, K)$$. A surface may also be piecewise smooth if it has smooth faces but also has locations where the directional derivatives do not exist. &= - 55 \int_0^{2\pi} \int_0^1 2v \, dv \,du \$4pt] 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, $\vecs t_u (P_{ij}) = \vecs r_u (u_i,v_j) = \langle x_u (u_i,v_j), \, y_u(u_i,v_j), \, z_u (u_i,v_j) \rangle.$. Let $$S$$ be the surface that describes the sheet. This is in contrast to vector line integrals, which can be defined on any piecewise smooth curve. 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 $$\PageIndex{16}$$). The notation needed to develop this definition is used throughout the rest of this chapter. 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. \nonumber\]. Show that the surface area of the sphere $$x^2 + y^2 + z^2 = r^2$$ is $$4 \pi r^2$$. Next, we need to determine $${\vec r_\theta } \times {\vec r_\varphi }$$. &= \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] $$\vecs t_u = \langle -v \, \sin u, \, v \, \cos u, \, 0 \rangle$$ and $$\vecs t_v = \langle \cos u, \, v \, \sin u, \, 0 \rangle$$, and $$\vecs t_u \times \vecs t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0, \, 0, -v \rangle$$. Scalar surface integrals have several real-world applications. Which of the figures in Figure $$\PageIndex{8}$$ is smooth? Decide how youâre going to split up your surface (see the next section) 2. where $$D$$ is the range of the parameters that trace out the surface $$S$$. Example $$\PageIndex{6}$$: Calculating Surface Area. With the standard parameterization of a cylinder, Equation \ref{equation1} shows that the surface area is $$2 \pi rh$$. 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. Therefore, we expect the surface to be an elliptic paraboloid. The horizontal cross-section of the cone at height $$z = u$$ is circle $$x^2 + y^2 = u^2$$. \[\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. 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. First, let’s look at the surface integral of a scalar-valued function. 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. &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \, d\phi \\ However, since we are on the cylinder we know what $$y$$ is from the parameterization so we will also need to plug that in. You might want to verify this for the practice of computing these cross products. Before we work some examples letâs notice that since we can parameterize a surface given by $$z = g\left( {x,y} \right)$$ as. Describe the surface parameterized by $$\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle, \, - \infty < u < \infty, \, 0 \leq v < 2\pi$$. Weâre going to let $${S_1}$$ be the portion of the cylinder that goes from the $$xy$$-plane to the plane. Just as with vector line integrals, surface integral $$\displaystyle \iint_S \vecs F \cdot \vecs N\, dS$$ is easier to compute after surface $$S$$ has been parameterized. 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. Notice that if $$u$$ is held constant, then the resulting curve is a circle of radius $$u$$ in plane $$z = u$$. Given the radius r of the sphere, the total surface area is. \label{surfaceI} \]. We will see one of these formulas in the examples and weâll leave the other to you to write down. First, letâs look at the surface integral in which the surface $$S$$ is given by $$z = g\left( {x,y} \right)$$. 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 $$\PageIndex{13}$$). Consider the parameter domain for this surface. Example $$\PageIndex{10}$$: Calculating the Surface Integral of a Piece of a Sphere. For example, consider curve parameterization $$\vecs r(t) = \langle 1,2\rangle, \, 0 \leq t \leq 5$$. Some surfaces are twisted in such a fashion that there is no well-defined notion of an “inner” or “outer” side. 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 $$2\pi$$. A cast-iron solid ball is given by inequality $$x^2 + y^2 + z^2 \leq 1$$. We can now get the value of the integral that we are after. Then the heat flow is a vector field proportional to the negative temperature gradient in the object. 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). Recall that curve parameterization $$\vecs r(t), \, a \leq t \leq b$$ is smooth if $$\vecs r'(t)$$ is continuous and $$\vecs r'(t) \neq \vecs 0$$ for all $$t$$ in $$[a,b]$$. 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. The same was true for scalar surface integrals: we did not need to worry about an “orientation” of the surface of integration. If piece $$S_{ij}$$ is small enough, then the tangent plane at point $$P_{ij}$$ is a good approximation of piece $$S_{ij}$$. 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. &= (\rho \, \sin \phi)^2. Here is that work. Hence, a parameterization of the cone is $$\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle$$. If parameterization $$\vec{r}$$ is regular, then the image of $$\vec{r}$$ is a two-dimensional object, as a surface should be. The region $$S$$ will lie above (in this case) some region $$D$$ that lies in the $$xy$$-plane. However, before we can integrate over a surface, we need to consider the surface itself. &= 7200\pi.\end{align*} \]. The definition of a scalar line integral can be extended to parameter domains that are not rectangles by using the same logic used earlier. The surface consists of infinitesimal patches that are approximately flat. 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. Next lesson. Let $$S$$ be a surface with parameterization $$\vecs r(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle$$ over some parameter domain $$D$$. 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. Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line (Figure $$\PageIndex{7}$$). &= - 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] 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, Therefore, the tangent of $$\phi$$ is $$\sqrt{3}$$, which implies that $$\phi$$ is $$\pi / 6$$. \nonumber$, \begin{align*} \iint_S \vecs F \cdot dS &= \int_0^4 \int_0^3 F (\vecs r(u,v)) \cdot (\vecs t_u \times \vecs t_v) \, du \,dv \\[4pt] &= \int_0^4 \int_0^3 \langle u - v^2, \, u, \, 0\rangle \cdot \langle -1 -2v, \, -1, \, 2v\rangle \, du\,dv \\[4pt] &= \int_0^4 \int_0^3 [(u - v^2)(-1-2v) - u] \, du\,dv \\[4pt] &= \int_0^4 \int_0^3 (2v^3 + v^2 - 2uv - 2u) \, du\,dv \\[4pt] &= \int_0^4 \left. If $$S_{ij}$$ is small enough, then it can be approximated by a tangent plane at some point $$P$$ in $$S_{ij}$$. In addition to modeling fluid flow, surface integrals can be used to model heat flow. By Example, we know that $$\vecs t_u \times \vecs t_v = \langle \cos u, \, \sin u, \, 0 \rangle$$. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. Suppose that i ranges from 1 to m and j ranges from 1 to n so that $$D$$ is subdivided into mn rectangles. Thus, a surface integral is similar to a line integral but in one higher dimension. &= \dfrac{5(17^{3/2}-1)}{3} \approx 115.15. A surface parameterization $$\vecs r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle$$ is smooth if vector $$\vecs r_u \times \vecs r_v$$ is not zero for any choice of $$u$$ and $$v$$ in the parameter domain. SPHERE_INTEGRALS, a C library which returns the exact value of the integral of any monomial over the surface of the unit sphere in 3D. Similarly, if $$S$$ is a surface given by equation $$x = g(y,z)$$ or equation $$y = h(x,z)$$, then a parameterization of $$S$$ is $$\vecs r(y,z) = \langle g(y,z), \, y,z\rangle$$ or $$\vecs r(x,z) = \langle x,h(x,z), z\rangle$$, respectively. Imagine what happens as $$u$$ increases or decreases. Surfaces can be parameterized, just as curves can be parameterized. Viewed 90 times 0 \begingroup Closed. These grid lines correspond to a set of grid curves on surface $$S$$ that is parameterized by $$\vecs r(u,v)$$. \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}. Essentially, a surface can be oriented if the surface has an “inner” side and an “outer” side, or an “upward” side and a “downward” side. Example $$\PageIndex{3}$$: Finding a Parameterization. Now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. &= 5 \left[\dfrac{(1+4u^2)^{3/2}}{3} \right]_0^2 \\ This is not the case with surfaces, however. Since we are only taking the piece of the sphere on or above plane $$z = 1$$, we have to restrict the domain of $$\phi$$. &= - 55 \int_0^{2\pi} \int_0^1 \langle 8v \, \cos u, \, 8v \, \sin u, \, v^2 \cos^2 u + v^2 \sin^2 u \rangle \cdot \langle 0,0, -v\rangle \, dv\,du \\[4pt] 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$. Find the surface area of the surface with parameterization $$\vecs r(u,v) = \langle u + v, \, u^2, \, 2v \rangle, \, 0 \leq u \leq 3, \, 0 \leq v \leq 2$$. How can surface area lie inside the cylinder given that the radius of sphere is greater than radius of cylinder? 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 like a line integral in one higher dimension. The mass flux is measured in mass per unit time per unit area. 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. \end{align*}\], Example $$\PageIndex{9}$$: Calculating the Surface Integral of a Cylinder. Note that we can form a grid with lines that are parallel to the $$u$$-axis and the $$v$$-axis in the $$uv$$-plane. 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. \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. Find the parametric representations of a cylinder, a cone, and a sphere. Since $$S$$ is given by the function $$f(x,y) = 1 + x + 2y$$, a parameterization of $$S$$ is $$\vecs r(x,y) = \langle x, \, y, \, 1 + x + 2y \rangle, \, 0 \leq x \leq 4, \, 0 \leq y \leq 2$$. Example $$\PageIndex{2}$$: Describing a Surface. \end{align*}\]. \nonumber\]. Surface integral example. In particular, surface integrals allow us to generalize Green’s theorem to higher dimensions, and they appear in some important theorems we discuss in later sections. In the definition of a line integral we chop a curve into pieces, evaluate a function at a point in each piece, and let the length of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. In other words, the top of the cylinder will be at an angle. \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. Since the surface is oriented outward and $$S_1$$ is the bottom of the object, it makes sense that this vector points downward. &= \rho^2 \, \sin^2 \phi \\[4pt] &= \int_0^3 \pi \, dv = 3 \pi. \nonumber$, Therefore, the radius of the disk is $$\sqrt{3}$$ and a parameterization of $$S_1$$ is $$\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, 1 \rangle, \, 0 \leq u \leq \sqrt{3}, \, 0 \leq v \leq 2\pi$$. Visualize a surface integral. Review: Arc length and line integrals I The integral of a function f : [a,b] â R is Therefore, the strip really only has one side. 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. Therefore, the lateral surface area of the cone is $$\pi r \sqrt{h^2 + r^2}$$. At this point weâve got a fairly simple double integral to do. Show that the surface area of cylinder $$x^2 + y^2 = r^2, \, 0 \leq z \leq h$$ is $$2\pi rh$$. Watch the recordings here on Youtube! &= 80 \int_0^{2\pi} \Big[-54 \, \cos \phi + 9 \, \cos^3 \phi \Big]_{\phi=0}^{\phi=2\pi} \, d\theta \\ For a curve, this condition ensures that the image of $$\vecs r$$ really is a curve, and not just a point. Divide rectangle $$D$$ into subrectangles $$D_{ij}$$ with horizontal width $$\Delta u$$ and vertical length $$\Delta v$$. How could we avoid parameterizations such as this? &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54 (1 - \cos^2\phi) \, \sin \phi + 27 \cos^2\phi \, \sin \phi \, d\phi \, d\theta \\ Therefore, to calculate, $\iint_{S_1} z^2 \,dS + \iint_{S_2} z^2 \,dS \nonumber$. I The surface is given in explicit form. Use the parameterization of surfaces of revolution given before Example $$\PageIndex{7}$$. 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]. This surface has parameterization $$\vecs r(u,v) = \langle v \, \cos u, \, v \, \sin u, \, 4 \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq 1.$$. &= \int_0^{\pi/6} \int_0^{2\pi} 16 \, \cos^2\phi \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2\phi} \, d\theta \, d\phi \\ Use Equation \ref{scalar surface integrals}. Here is the parameterization for this sphere. We can extend the concept of a line integral to a surface integral to allow us to perform this integration. A piece of metal has a shape that is modeled by paraboloid $$z = x^2 + y^2, \, 0 \leq z \leq 4,$$ and the density of the metal is given by $$\rho (x,y,z) = z + 1$$. Of points, we can calculate their surface areas, surface integral sphere need to plug in for \ ( v\ increases... Bounded by the xy and yz planes of curves result first one in the strip really only has one.... 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