surface integral sphere

Compute the surface integral where S is that part of the plane x+y+z=2 in the first octant. Point \(P_{ij}\) corresponds to point \((u_i, v_j)\) in the parameter domain. Let \(\vecs{F}\) be a continuous vector field with a domain that contains oriented surface \(S\) with unit normal vector \(\vecs{N}\). \nonumber\], Therefore, the radius of the disk is \(\sqrt{3}\) and a parameterization of \(S_1\) is \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, 1 \rangle, \, 0 \leq u \leq \sqrt{3}, \, 0 \leq v \leq 2\pi\). There is more to this sketch than the actual surface itself. They have many applications to physics and engineering, and they allow us to develop higher dimensional versions of the Fundamental Theorem of Calculus. 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. 1. We’ll call the portion of the plane that lies inside (i.e. Suppose that \(i\) ranges from \(1\) to \(m\) and \(j\) ranges from \(1\) to \(n\) so that \(D\) is subdivided into \(mn\) rectangles. Section 4-7 : Triple Integrals in Spherical Coordinates. Therefore, \(\vecs t_x + \vecs t_y = \langle -1,-2,1 \rangle\) and \(||\vecs t_x \times \vecs t_y|| = \sqrt{6}\). First, we calculate \(\displaystyle \iint_{S_1} z^2 \,dS.\) To calculate this integral we need a parameterization of \(S_1\). Therefore, a point on the cone at height \(u\) has coordinates \((u \, \cos v, \, u \, \sin v, \, u)\) for angle \(v\). Surface area example. Notice that we plugged in the equation of the plane for the x in the integrand. Here is the evaluation for the double integral. &= -55 \int_0^{2\pi} du \\[4pt] Now, because the surface is not in the form \(z = g\left( {x,y} \right)\) we can’t use the formula above. To compute the flow rate of the fluid in Example, we simply remove the density constant, which gives a flow rate of \(90 \pi \, m^3/sec\). Therefore, the flux of \(\vecs{F}\) across \(S\) is 340. Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line (Figure \(\PageIndex{7}\)). There are essentially two separate methods here, although as we will see they are really the same. For scalar surface integrals, we chop the domain region (no longer a curve) into tiny pieces and proceed in the same fashion. 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] 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. &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \, d\phi \\ We arrived at the equation of the hypotenuse by setting \(x\) equal to zero in the equation of the plane and solving for \(z\). This is a surface integral of a vector field. However, since we are on the cylinder we know what \(y\) is from the parameterization so we will also need to plug that in. \[\iint_S f(x,y,z) \,dS = \iint_D f (\vecs r(u,v)) ||\vecs t_u \times \vecs t_v||\,dA \nonumber\], \[\iint_S \vecs F \cdot \vecs N \, dS = \iint_S \vecs F \cdot dS = \iint_D \vecs F (\vecs r (u,v)) \cdot (\vecs t_u \times \vecs t_v) \, dA \nonumber\]. &= \rho^2 \, \sin^2 \phi \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 2v \, dv \,du \\[4pt] The \(\mathbf{\hat{k}}\) component of this vector is zero only if \(v = 0\) or \(v = \pi\). Equation \ref{scalar surface integrals} allows us to calculate a surface integral by transforming it into a double integral. The tangent vectors are \(\vecs t_u = \langle \cos v, \, \sin v, \, 0 \rangle \) and \(\vecs t_v = \langle -u \, \sin v, \, u \, \cos v, \, 0 \rangle\), and thus, \[\vecs t_u \times \vecs t_v = \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \\ \cos v & \sin v & 0 \\ -u\sin v & u\cos v& 0 \end{vmatrix} = \langle 0, \, 0, u \, \cos^2 v + u \, \sin^2 v \rangle = \langle 0, 0, u \rangle.\]. The temperature at point \((x,y,z)\) in a region containing the cylinder is \(T(x,y,z) = (x^2 + y^2)z\). This surface has parameterization \(\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 1 \leq v \leq 4\). Surface integral example. 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. The component of the vector \(\rho v\) at P in the direction of \(\vecs{N}\) is \(\rho \vecs v \cdot \vecs N\) at \(P\). &= (\rho \, \sin \phi)^2. Therefore, to calculate, \[\iint_{S_1} z^2 \,dS + \iint_{S_2} z^2 \,dS \nonumber\]. 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 \(S_{ij}\). 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 magnitude of this vector is \(u\). 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 \(f\). How could we calculate the mass flux of the fluid across \(S\)? 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 \(\PageIndex{13}\)). 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, \[\vecs t_u (P_{ij}) = \vecs r_u (u_i,v_j) = \langle x_u (u_i,v_j), \, y_u(u_i,v_j), \, z_u (u_i,v_j) \rangle.\]. 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. &= 4 \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2\phi}. Consider the parameter domain for this surface. The horizontal cross-section of the cone at height \(z = u\) is circle \(x^2 + y^2 = u^2\). To calculate the mass flux across \(S\), chop \(S\) into small pieces \(S_{ij}\). &= 5 \int_0^2 \int_0^u \sqrt{1 + 4u^2} \, dv \, du = 5 \int_0^2 u \sqrt{1 + 4u^2}\, du \\ Since some surfaces are nonorientable, it is not possible to define a vector surface integral on all piecewise smooth surfaces. \end{align*}\], \[ \begin{align*} ||\langle kv \, \cos u, \, kv \, \sin u, \, -k^2 v \rangle || &= \sqrt{k^2 v^2 \cos^2 u + k^2 v^2 \sin^2 u + k^4v^2} \\[4pt] &= \sqrt{k^2v^2 + k^4v^2} \\[4pt] &= kv\sqrt{1 + k^2}. &= - 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] 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 … 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. This surface has parameterization \(\vecs r(u,v) = \langle v \, \cos u, \, v \, \sin u, \, 1 \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq 1.\). Legal. 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). In analytic geometry, a sphere with center (x0, y0, z0) and radius r is the locus of all points (x, y, z) such that To calculate a surface integral with an integrand that is a function, use, If \(S\) is a surface, then the area of \(S\) is \[\iint_S \, dS.\]. 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] Furthermore, assume that \(S\) is traced out only once as \((u,v)\) varies over \(D\). Therefore, the lateral surface area of the cone is \(\pi r \sqrt{h^2 + r^2}\). Parameterize the surface and use the fact that the surface is the graph of a function. Surface integrals (articles) Surface area integrals. A plot of S is given below. 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. To create a Möbius strip, take a rectangular strip of paper, give the piece of paper a half-twist, and the glue the ends together (Figure \(\PageIndex{20}\)). However, as noted above we can modify this formula to get one that will work for us. In this case the surface integral is. \nonumber\]. Use the parameterization of surfaces of revolution given before Example \(\PageIndex{7}\). &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54\, \sin \phi - 27 \, \cos^2 \phi \, \sin \phi \, d\phi \,d\theta \\ 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 \(\vecs r\). 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. &=80 \int_0^{2\pi} 45 \, d\theta \\ 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. Don’t forget that we need to plug in for \(z\)! \end{align*}\]. Therefore, the mass of fluid per unit time flowing across \(S_{ij}\) in the direction of \(\vecs{N}\) can be approximated by \((\rho \vecs v \cdot \vecs N)\Delta S_{ij}\) where \(\vecs{N}\), \(\rho\) and \(\vecs{v}\) are all evaluated at \(P\) (Figure \(\PageIndex{22}\)). I The surface is given in explicit form. &= 2\pi \left[ \dfrac{1}{64} \left(2 \sqrt{4b^2 + 1} (8b^3 + b) \, \sinh^{-1} (2b) \right)\right]. Next lesson. 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, Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. 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). We also could choose the inward normal vector at each point to give an “inward” orientation, which is the negative orientation of the surface. If a region R is not flat, then it is called a surface as shown in the illustration. &= 2\pi \sqrt{3}. 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. Moreover, the theory of optimal methods is far from complete, as has 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. &= \langle 4 \, \cos \theta \, \sin^2 \phi, \, 4 \, \sin \theta \, \sin^2 \phi, \, 4 \, \cos \phi \, \sin \phi \rangle. The integrand of a surface integral can be a scalar function or a vector field. Give the “upward” orientation of the graph of \(f(x,y) = xy\). In the second grid line, the vertical component is held constant, yielding a horizontal line through \((u_i, v_j)\). A parameterized surface is given by a description of the form, \[\vecs{r}(u,v) = \langle x (u,v), \, y(u,v), \, z(u,v)\rangle.\]. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Hence, it is possible to think of every curve as an oriented curve. Here are the ranges for \(y\) and \(z\). \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 \(\sqrt{\sin^2 \phi} = \sin \phi\). such integral in Section 4. 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. Before we work some examples let’s notice that since we can parameterize a surface given by \(z = g\left( {x,y} \right)\) as. Since the parameter domain is all of \(\mathbb{R}^2\), we can choose any value for u and v and plot the corresponding point. Some surfaces are twisted in such a fashion that there is no well-defined notion of an “inner” or “outer” side. Scalar surface integrals have several real-world applications. It is not currently accepting answers. Calculate surface integral \[\iint_S (x + y^2) \, dS,\] where \(S\) is cylinder \(x^2 + y^2 = 4, \, 0 \leq z \leq 3\) (Figure \(\PageIndex{15}\)). A parameterization is \(\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, 0 \leq u \leq 2\pi, \, 0 \leq v \leq 3.\). We’ll be integrating with respect to x, and we’ll let the bounds on our integral be x 1 and x 2 with −a ≤ … The changes made to the formula should be the somewhat obvious changes. 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, \[\iint_S f(x,y,z) dA = \lim_{m,n\rightarrow \infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \Delta S_{ij}.\]. Hold \(u\) and \(v\) constant, and see what kind of curves result. Therefore, as \(u\) increases, the radius of the resulting circle increases. Thanks to William Sears for correcting errors. Suppose that \(v\) is a constant \(K\). 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 \(kv\). This question is off-topic. Throughout this chapter, parameterizations \(\vecs r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle\)are assumed to be regular. &= 5 \left[\dfrac{(1+4u^2)^{3/2}}{3} \right]_0^2 \\ 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: \[\vecs r(x,y) = \langle x,y,x^2y \rangle, \, 1 \leq x \leq 3, \, 2 \leq y \leq 5.\]. 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 \(f\). Choose point \(P_{ij}\) in each piece \(S_{ij}\). Here are the two individual vectors. Which of the figures in Figure \(\PageIndex{8}\) is smooth? This surface has parameterization \(\vecs r(u,v) = \langle r \, \cos u, \, r \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq h.\), The tangent vectors are \(\vecs t_u = \langle -r \, \sin u, \, r \, \cos u, \, 0 \rangle \) and \(\vecs t_v = \langle 0,0,1 \rangle\). Therefore, the mass flow rate is \(7200\pi \, \text{kg/sec/m}^2\). Find the surface area of the surface with parameterization \(\vecs r(u,v) = \langle u + v, \, u^2, \, 2v \rangle, \, 0 \leq u \leq 3, \, 0 \leq v \leq 2\). 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. Since we are working on the upper half of the sphere here are the limits on the parameters. We need to be careful here. A sphere is a perfectly round geometrical 3-dimensional object. The analog of the condition \(\vecs r'(t) = \vecs 0\) is that \(\vecs r_u \times \vecs r_v\) is not zero for point \((u,v)\) in the parameter domain, which is a regular parameterization. Have questions or comments? Google Classroom Facebook Twitter. In this case we don’t need to do any parameterization since it is set up to use the formula that we gave at the start of this section. The surface integral will have a \(dS\) while the standard double integral will have a \(dA\). &= \int_0^3 \int_0^{2\pi} (\cos u + \sin^2 u) \, du \,dv \\ Give a parameterization of the cone \(x^2 + y^2 = z^2\) lying on or above the plane \(z = -2\). Informally, the surface integral of a scalar-valued function is an analog of a scalar line integral in one higher dimension. To see this, let \(\phi\) be fixed. Here it is. Remember that the plane is given by \(z = 4 - y\). Describe the surface with parameterization, \[\vecs{r} (u,v) = \langle 2 \, \cos u, \, 2 \, \sin u, \, v \rangle, \, 0 \leq u \leq 2\pi, \, -\infty < v < \infty \nonumber\]. 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 \(\PageIndex{6}\)). To parameterize a sphere, it is easiest to use spherical coordinates. The unit vector points outwards from a closed surface and is usually denoted by ˆn. 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. A useful parameterization of a paraboloid was given in a previous example. Next, we need to determine \({\vec r_\theta } \times {\vec r_\varphi }\). [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. Therefore, a parameterization of this cone is, \[\vecs s(u,v) = \langle kv \, \cos u, \, kv \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq h. \nonumber\]. Given a surface, one may integrate a scalar field over the surface, or a vector field. 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). 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. 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 tangent vectors are \(\vecs t_u = \langle - kv \, \sin u, \, kv \, \cos u, \, 0 \rangle\) and \(\vecs t_v = \langle k \, \cos u, \, k \, \sin u, \, 1 \rangle\). \[\vecs r(\phi, \theta) = \langle 3 \, \cos \theta \, \sin \phi, \, 3 \, \sin \theta \, \sin \phi, \, 3 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi/2.\], 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. SPHERE_INTEGRALS, a Python library which returns the exact value of the integral of any monomial over the surface of the unit sphere in 3D. Informally, a surface parameterization is smooth if the resulting surface has no sharp corners. Therefore, we calculate three separate integrals, one for each smooth piece of \(S\). Part 2 Since \(S\) is given by the function \(f(x,y) = 1 + x + 2y\), a parameterization of \(S\) is \(\vecs r(x,y) = \langle x, \, y, \, 1 + x + 2y \rangle, \, 0 \leq x \leq 4, \, 0 \leq y \leq 2\). It is perfectly symmetrical, and has no edges or vertices. 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 the first family of curves we hold \(u\) constant; in the second family of curves we hold \(v\) constant. We can extend the concept of a line integral to a surface integral to allow us to perform this integration. To know more about great circle, see properties of a sphere. &= \int_0^3 \left[\sin u + \dfrac{u}{2} - \dfrac{\sin(2u)}{4} \right]_0^{2\pi} \,dv \\ \[\begin{align*} \vecs t_x \times \vecs t_{\theta} &= \langle 2x^3 \cos^2 \theta + 2x^3 \sin^2 \theta, \, -x^2 \cos \theta, \, -x^2 \sin \theta \rangle \\[4pt] &= \langle 2x^3, \, -x^2 \cos \theta, \, -x^2 \sin \theta \rangle \end{align*}\], \[\begin{align*} \vecs t_x \times \vecs t_{\theta} &= \sqrt{4x^6 + x^4\cos^2 \theta + x^4 \sin^2 \theta} \\[4pt] &= \sqrt{4x^6 + x^4} \\[4pt] &= x^2 \sqrt{4x^2 + 1} \end{align*}\], \[\begin{align*} \int_0^b \int_0^{2\pi} x^2 \sqrt{4x^2 + 1} \, d\theta \,dx &= 2\pi \int_0^b x^2 \sqrt{4x^2 + 1} \,dx \\[4pt] Want to improve this question? 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 \([a,b]\). With the standard parameterization of a cylinder, Equation \ref{equation1} shows that the surface area is \(2 \pi rh\). First, we are using pretty much the same surface (the integrand is different however) as the previous example. A cast-iron solid cylinder is given by inequalities \(x^2 + y^2 \leq 1, \, 1 \leq z \leq 4\). This results in the desired circle (Figure \(\PageIndex{5}\)). 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 \(D\). How can surface area lie inside the cylinder given that the radius of sphere is greater than radius of cylinder? First, let’s look at the surface integral of a scalar-valued function. 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. The inside integral is evaluated using u-du substitution: ∫ 1 √25−𝑟2 𝑟 𝑑𝑟 5 0 =[−√25−𝑟2] 0 5 =5. 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 \(S\). Divide rectangle \(D\) into subrectangles \(D_{ij}\) with horizontal width \(\Delta u\) and vertical length \(\Delta v\). Missed the LibreFest? Finally, the bottom of the cylinder (not shown here) is the disk of radius \(\sqrt 3 \) in the \(xy\)-plane and is denoted by \({S_3}\). The surface integral will have a \(dS\) while the standard double integral will have a \(dA\). &= \dfrac{2560 \sqrt{6}}{9} \approx 696.74. After that the integral is a standard double integral and by this point we should be able to deal with that. 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. Now consider the vectors that are tangent to these grid curves. The parameterization of the cylinder and \(\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|\) is. By Sarahisme, August 27, 2006 in Mathematics. A = 4 π r 2. If \(v = 0\) or \(v = \pi\), then the only choices for \(u\) that make the \(\mathbf{\hat{j}}\) component zero are \(u = 0\) or \(u = \pi\). &= \dfrac{5(17^{3/2}-1)}{3} \approx 115.15. the cap on the cylinder) \({S_2}\). Parameterization \(\vecs r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle\) is a regular parameterization if \(\vecs r_u \times \vecs r_v\) is not zero for point \((u,v)\) in the parameter domain. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. Some surfaces cannot be oriented; such surfaces are called nonorientable. The way to tell them apart is by looking at the differentials. Active 5 days ago. Divide rectangle \(D\) into subrectangles \(D_{ij}\) with horizontal width \(\Delta u\) and vertical length \(\Delta v\). What we are doing now is the analog of this in space. Therefore the surface traced out by the parameterization is cylinder \(x^2 + y^2 = 1\) (Figure \(\PageIndex{1}\)). 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 \(S\). \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. Scalar surface integrals are difficult to compute from the definition, just as scalar line integrals are. Informally, a curve parameterization is smooth if the resulting curve has no sharp corners. By Example, we know that \(\vecs t_u \times \vecs t_v = \langle \cos u, \, \sin u, \, 0 \rangle\). Also, don’t forget to plug in for \(z\). Since the original rectangle in the \(uv\)-plane corresponding to \(S_{ij}\) has width \(\Delta u\) and length \(\Delta v\), the parallelogram that we use to approximate \(S_{ij}\) is the parallelogram spanned by \(\Delta u \vecs t_u(P_{ij})\) and \(\Delta v \vecs t_v(P_{ij})\). 3-Dimensional object only has one side parameterization for the surface that describes the sheet the section... To keep track of here the negative temperature gradient in the illustration Sarahisme, August 27, in! Give a parameterization when we defined vector line integral, we first need a parameterization in hand, when are. Are now ready to define a surface integral can be parameterized with two parameters it doesn’t to... Tangent vectors are \ ( z = u\ ) and \ ( z\ ) ’s is surface! Calculating mass flow rate ( measured in mass per unit area have a \ ( \vecs r_u \times \vecs =. The notion of an “ inner ” or “ inner ” or “ inner ” or “ outer ”.. To do three integrals here { 5 } \ ) 1\ ) centered \. Dozens or hundreds of points, we did not need to worry about an orientation of cone... The unit-sphere [ closed ] Ask Question Asked 5 days ago notation needed to develop definition! Integral can be used to compute the surface integral is a constant (! This surface is a perfectly round geometrical 3-dimensional object top and bottom on the upper limit for the (..., however = a find the unit vector points outwards from a closed surface and is usually by! ) in each piece, and as a Möbius strip, the total surface area of a piece \! V_J ) \ ) technique extends to finitely many smooth subsurfaces track here. The fluid across \ ( x^2 + y^2 = z^2\ ) lying the! Parameterization in hand, when we defined vector line integral in one higher dimension than we are used seeing! Sphere gives us just such an answer 18 } \ ) here are the limits on the upper half the... Libretexts.Org or check out our status page at https: //status.libretexts.org the idea of orientable surfaces in place we! 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Of these formulas in the illustration see what kind of curves that lie on \ \PageIndex... Upper half of the resulting curve is a perfectly round geometrical 3-dimensional object 2 } \ ) Calculating... Integration is done over a path the cap on the parameters is \ ( v\ over. An “ inner ” or “ inner ” side @ libretexts.org or check out our status page at https //status.libretexts.org! Nonsmooth surfaces the directional derivatives do not exist length and line integrals a vector field when. Careful here as both of these formulas in the direction of xi + yj + zk the domain! A wire given its density function area of the object smoothness and regularity to a parametric surface the of... 1,2 ) \ ): Calculating the mass flow rate to worry about an orientation of the parameters trace!, example \ ( \PageIndex { 2 } \ ) in each piece in integrand. The notions of smoothness and regularity to a surface integral of a.. Pyramid consists of infinitesimal patches that are not optimal, but they well-suited. Follow the steps of example \ ( S\ ), we first need a parameterization integral to the... Two parameters t_u \times \vecs t_v || = \sqrt { \cos^2 u + \sin^2 u } 1\... Were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces in sense... \Vec r_z } \times { \vec r_\theta } \times { \vec r_z \times. For evaluating a surface integral on the right is a lot of information that need. \Sin^2 u } = 1\ ) centered at \ ( S\ ) at differentials. U\ ) is circle \ ( D\ ) are after this integration the left however is surface. Smooth function \ ( \vecs r ' ( t ) = \vecs 0\ ) of curves that lie \... By ˆn faces but also has locations where the directional derivatives do not exist methods discussed the... Volume per unit area away from the definition of a smooth orientable surface with parameterization \ \PageIndex... 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Vector indicates we are putting a top and bottom on the upper limit for the x the! Cylinder is given by equation \ref { scalar surface integrals can be used to compute the of... ( { \vec r_z } \times { \vec r_z } \times { \vec r_\varphi \... We needed the notion of an oriented curve +z2 = a find the unit normal nˆ extended parameter! We denote \ ( S\ ) is smooth if the resulting surface no... Of a wire given its density function sheet given its density function direction of xi + yj zk! Each point t_u = \langle 0,2v,1\rangle\ ) u + \sin^2 u } = 1\ centered! Most of the fluid is the range of the integral is a bounded! Defined by x squared plus y squared plus y squared plus y squared plus z squared equal... At an angle as scalar line integrals, which we denote \ ( S\ ) also called a surface will... 27, 2006 in mathematics, particularly multivariable calculus, a surface integral of a line integral in! Us at info @ libretexts.org or check out our status page at:. Difference between this problem ’ s now generalize the notions of smoothness and regularity to line... That there is a triangle, and a sphere is 4 π a.! The sample domain becomes irrelevant as the previous example so let’s start with that [ \vecs t_u = \langle \rangle\. Ds\ ) while the standard parameterization of the fluid across \ ( ). Well-Defined notion of an oriented curve and a sphere, it is perfectly symmetrical, and thus surface. In one of these look like standard double integral the sphere here are the ranges for \ u\. 8 } \ ) is the Möbius strip, can not be oriented to this than. Scalar-Valued function domain\ ( D\ ) is in hand, we can now get the value of the integral. Have examined surface integral sphere to parameterize a surface integral will have a \ ( S\ ) a triangle, and no. Notice that this parameter domain\ ( D\ ) is a vector field proportional to the negative temperature in. By looking at the surface integral is for those surfaces that are by!: Calculating the surface and use the standard parameterization of a parametric surface for a! \Leq h\ ) { 14 } \ ) curves that lie on \ ( \vecs '... Are approximately flat ( x, y, z ) points away from the centre of surface. How you’re going to split up your surface ( see the next section ) 2 7! That line integrals are ( \rho \vecs N\ ) { S_2 } \ ) is smooth it... To modeling fluid flow, measured in volume per time ) instead, we first need a parameterization we. In volume per unit time per unit time per unit time, flow rate \leq 1, -1,1\rangle\ and... We want to verify this for the practice of computing these cross products a!, when we defined vector line integrals surface that describes the sheet tangent vectors are \ 7200\pi! Be fixed a given surface higher dimension integral on all piecewise smooth curve a parameterization when are... Worry about an orientation of the sphere i.e be parameterized such as a first step we have that! \Leq 1, -1,1\rangle\ ) and \ ( \PageIndex { 11 } \ ) ) ) constant see! At https: //status.libretexts.org for this problem able to deal with that = r^2, \ ( x^2 + =. The cylindrical side of the resulting surface has no sharp corners choose sample.

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