surface integral calculator

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. In Example \(\PageIndex{14}\), we computed the mass flux, which is the rate of mass flow per unit area. Stokes' theorem is the 3D version of Green's theorem. This is not the case with surfaces, however. 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. \nonumber \], 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, \[\vecs t_{\phi} \times \vecs t_{\theta} = \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle. \nonumber \], Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. An approximate answer of the surface area of the revolution is displayed. This is analogous to a . In addition to parameterizing surfaces given by equations or standard geometric shapes such as cones and spheres, we can also parameterize surfaces of revolution. To use Equation \ref{scalar surface integrals} to calculate the surface integral, we first find vectors \(\vecs t_u\) and \(\vecs t_v\). In fact the integral on the right is a standard double integral. Our goal is to define a surface integral, and as a first step we have examined how to parameterize a surface. In the field of graphical representation to build three-dimensional models. Volume and Surface Integrals Used in Physics. then, Weisstein, Eric W. "Surface Integral." &= 4 \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2\phi}. 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. 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. Note that all four surfaces of this solid are included in S S. Solution. surface integral Natural Language Math Input Use Math Input Mode to directly enter textbook math notation. Example 1. Also note that we could just as easily looked at a surface \(S\) that was in front of some region \(D\) in the \(yz\)-plane or the \(xz\)-plane. If vector \(\vecs N = \vecs t_u (P_{ij}) \times \vecs t_v (P_{ij})\) exists and is not zero, then the tangent plane at \(P_{ij}\) exists (Figure \(\PageIndex{10}\)). In order to evaluate a surface integral we will substitute the equation of the surface in for z z in the integrand and then add on the often messy square root. Our integral solver also displays anti-derivative calculations to users who might be interested in the mathematical concept and steps involved in integration. If piece \(S_{ij}\) is small enough, then the tangent plane at point \(P_{ij}\) is a good approximation of piece \(S_{ij}\). Yes, as he explained explained earlier in the intro to surface integral video, when you do coordinate substitution for dS then the Jacobian is the cross-product of the two differential vectors r_u and r_v. Some surfaces cannot be oriented; such surfaces are called nonorientable. Direct link to benvessely's post Wow what you're crazy sma. At the center point of the long dimension, it appears that the area below the line is about twice that above. \nonumber \], Therefore, the radius of the disk is \(\sqrt{3}\) and a parameterization of \(S_1\) is \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, 1 \rangle, \, 0 \leq u \leq \sqrt{3}, \, 0 \leq v \leq 2\pi\). 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 \([a,b]\). Following are the steps required to use the Surface Area Calculator: The first step is to enter the given function in the space given in front of the title Function. Notice the parallel between this definition and the definition of vector line integral \(\displaystyle \int_C \vecs F \cdot \vecs N\, dS\). For each function to be graphed, the calculator creates a JavaScript function, which is then evaluated in small steps in order to draw the graph. How could we avoid parameterizations such as this? We gave the parameterization of a sphere in the previous section. 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. &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54 \, \sin^3 \phi + 27 \, \cos^2 \phi \, \sin \phi \, d\phi \, d\theta \\ 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. However, the pyramid consists of four smooth faces, and thus this surface is piecewise smooth. This is in contrast to vector line integrals, which can be defined on any piecewise smooth curve. 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. A surface integral of a vector field. d S, where F = z, x, y F = z, x, y and S is the surface as shown in the following figure. 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. Explain the meaning of an oriented surface, giving an example. In the pyramid in Figure \(\PageIndex{8b}\), the sharpness of the corners ensures that directional derivatives do not exist at those locations. &= 80 \int_0^{2\pi} \Big[-54 \, \cos \phi + 9 \, \cos^3 \phi \Big]_{\phi=0}^{\phi=2\pi} \, d\theta \\ 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. 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 \(kv\). The formula for calculating the length of a curve is given as: L = a b 1 + ( d y d x) 2 d x. The surface element contains information on both the area and the orientation of the surface. Learning Objectives. This allows us to build a skeleton of the surface, thereby getting an idea of its shape. After that the integral is a standard double integral and by this point we should be able to deal with that. The vendor states an area of 200 sq cm. Describe the surface with parameterization, \[\vecs{r} (u,v) = \langle 2 \, \cos u, \, 2 \, \sin u, \, v \rangle, \, 0 \leq u \leq 2\pi, \, -\infty < v < \infty \nonumber \]. &= 2\pi \left[ \dfrac{1}{64} \left(2 \sqrt{4b^2 + 1} (8b^3 + b) \, \sinh^{-1} (2b) \right)\right]. &= \int_0^{\sqrt{3}} \int_0^{2\pi} u \, dv \, du \\ &= \dfrac{2560 \sqrt{6}}{9} \approx 696.74. 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 \(\sqrt{3}\)). 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 \(2\pi\). If , &= 5 \int_0^2 \int_0^u \sqrt{1 + 4u^2} \, dv \, du = 5 \int_0^2 u \sqrt{1 + 4u^2}\, du \\ Area of Surface of Revolution Calculator. However, before we can integrate over a surface, we need to consider the surface itself. The Integral Calculator has to detect these cases and insert the multiplication sign. Here they are. One line is given by \(x = u_i, \, y = v\); the other is given by \(x = u, \, y = v_j\). Calculate the surface integral where is the portion of the plane lying in the first octant Solution. Added Aug 1, 2010 by Michael_3545 in Mathematics. Recall that to calculate a scalar or vector line integral over curve \(C\), we first need to parameterize \(C\). Now, how we evaluate the surface integral will depend upon how the surface is given to us. Describe the surface integral of a scalar-valued function over a parametric surface. https://mathworld.wolfram.com/SurfaceIntegral.html. to denote the surface integral, as in (3). The step by step antiderivatives are often much shorter and more elegant than those found by Maxima. 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}. After studying line integrals, double integrals and triple integrals, you may recognize this idea of chopping something up and adding all its pieces as a more general pattern in how integration can be used to solve problems. 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. For more on surface area check my online book "Flipped Classroom Calculus of Single Variable" https://versal.com/learn/vh45au/ We can see that \(S_1\) is a circle of radius 1 centered at point \((0,0,1)\) sitting in plane \(z = 1\). ; 6.6.5 Describe the surface integral of a vector field. With the standard parameterization of a cylinder, Equation \ref{equation1} shows that the surface area is \(2 \pi rh\). Notice that if \(u\) is held constant, then the resulting curve is a circle of radius \(u\) in plane \(z = u\). Let C be the closed curve illustrated below. Therefore, the strip really only has one side. How does one calculate the surface integral of a vector field on a surface? where \(S\) is the surface with parameterization \(\vecs r(u,v) = \langle u, \, u^2, \, v \rangle\) for \(0 \leq u \leq 2\) and \(0 \leq v \leq u\). That is: To make the work easier I use the divergence theorem, to replace the surface integral with a . In the next block, the lower limit of the given function is entered. Let \(\theta\) be the angle of rotation. However, when now dealing with the surface integral, I'm not sure on how to start as I have that ( 1 + 4 z) 3 . Here it is. Integrals can be a little daunting for students, but they are essential to calculus and understanding more advanced mathematics. 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 \(\PageIndex{11}\)). Find the mass flow rate of the fluid across \(S\). Some surfaces, such as a Mbius strip, cannot be oriented. Like so many things in multivariable calculus, while the theory behind surface integrals is beautiful, actually computing one can be painfully labor intensive. Improve your academic performance SOLVING . Surface Area Calculator Author: Ravinder Kumar Topic: Area, Surface The present GeoGebra applet shows surface area generated by rotating an arc. Direct link to Andras Elrandsson's post I almost went crazy over , Posted 3 years ago. I'm not sure on how to start this problem. Now, we need to be careful here as both of these look like standard double integrals. The definition is analogous to the definition of the flux of a vector field along a plane curve. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces. &= - 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] Maxima takes care of actually computing the integral of the mathematical function. When the "Go!" eMathHelp: free math calculator - solves algebra, geometry, calculus, statistics, linear algebra, and linear programming problems step by step To compute the flow rate of the fluid in Example, we simply remove the density constant, which gives a flow rate of \(90 \pi \, m^3/sec\). Calculus: Fundamental Theorem of Calculus In principle, the idea of a surface integral is the same as that of a double integral, except that instead of "adding up" points in a flat two-dimensional region, you are adding up points on a surface in space, which is potentially curved. where \(D\) is the range of the parameters that trace out the surface \(S\). \nonumber \] Notice that \(S\) is not a smooth surface but is piecewise smooth, since \(S\) is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). Skip the "f(x) =" part and the differential "dx"! A cast-iron solid ball is given by inequality \(x^2 + y^2 + z^2 \leq 1\). Notice that the axes are labeled differently than we are used to seeing in the sketch of \(D\). If \(S_{ij}\) is small enough, then it can be approximated by a tangent plane at some point \(P\) in \(S_{ij}\). All common integration techniques and even special functions are supported. Put the value of the function and the lower and upper limits in the required blocks on the calculator then press the submit button. That's why showing the steps of calculation is very challenging for integrals. Let \(S\) be hemisphere \(x^2 + y^2 + z^2 = 9\) with \(z \leq 0\) such that \(S\) is oriented outward. 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 \(S\). Step #4: Fill in the lower bound value. It transforms it into a form that is better understandable by a computer, namely a tree (see figure below). Find the parametric representations of a cylinder, a cone, and a sphere. You appear to be on a device with a "narrow" screen width (, \[\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {x,y,g\left( {x,y} \right)} \right)\sqrt {{{\left( {\frac{{\partial g}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial g}}{{\partial y}}} \right)}^2} + 1} \,dA}}\], \[\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {\vec r\left( {u,v} \right)} \right)\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\|\,dA}}\], 2.4 Equations With More Than One Variable, 2.9 Equations Reducible to Quadratic in Form, 4.1 Lines, Circles and Piecewise Functions, 1.5 Trig Equations with Calculators, Part I, 1.6 Trig Equations with Calculators, Part II, 3.6 Derivatives of Exponential and Logarithm Functions, 3.7 Derivatives of Inverse Trig Functions, 4.10 L'Hospital's Rule and Indeterminate Forms, 5.3 Substitution Rule for Indefinite Integrals, 5.8 Substitution Rule for Definite Integrals, 6.3 Volumes of Solids of Revolution / Method of Rings, 6.4 Volumes of Solids of Revolution/Method of Cylinders, A.2 Proof of Various Derivative Properties, A.4 Proofs of Derivative Applications Facts, 7.9 Comparison Test for Improper Integrals, 9. What about surface integrals over a vector field? Surface integrals are used anytime you get the sensation of wanting to add a bunch of values associated with points on a surface. The parameterization of the cylinder and \(\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|\) is. A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (a two-dimensional object) rather than a curve (a one-dimensional object). 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 \(\PageIndex{22}\)). The definition of a smooth surface parameterization is similar. We rewrite the equation of the plane in the form Find the partial derivatives: Applying the formula we can express the surface integral in terms of the double integral: The region of integration is the triangle shown in Figure Figure 2. \end{align*}\]. In this article, we will discuss line, surface and volume integrals.We will start with line integrals, which are the simplest type of integral.Then we will move on to surface integrals, and finally volume integrals. Informally, the surface integral of a scalar-valued function is an analog of a scalar line integral in one higher dimension. Symbolab is the best integral calculator solving indefinite integrals, definite integrals, improper integrals, double integrals, triple integrals, multiple integrals, antiderivatives, and more. By the definition of the line integral (Section 16.2), \[\begin{align*} m &= \iint_S x^2 yz \, dS \\[4pt] In the second grid line, the vertical component is held constant, yielding a horizontal line through \((u_i, v_j)\). Here are the two individual vectors. Do my homework for me. \nonumber \]. A surface integral over a vector field is also called a flux integral. is the divergence of the vector field (it's also denoted ) and the surface integral is taken over a closed surface. The corresponding grid curves are \(\vecs r(u_i, v)\) and \((u, v_j)\) and these curves intersect at point \(P_{ij}\). 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 \(\vecs t_u \times \vecs t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0,0,-v\rangle\). \end{align*}\], To calculate this integral, we need a parameterization of \(S_2\). How To Use a Surface Area Calculator in Calculus? S curl F d S, where S is a surface with boundary C. In general, surfaces must be parameterized with two parameters. \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] Let's now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the x-axis. Dot means the scalar product of the appropriate vectors. If \(u = v = 0\), then \(\vecs r(0,0) = \langle 1,0,0 \rangle\), so point (1, 0, 0) is on \(S\). It could be described as a flattened ellipse. I'll go over the computation of a surface integral with an example in just a bit, but first, I think it's important for you to have a good grasp on what exactly a surface integral, The double integral provides a way to "add up" the values of, Multiply the area of each piece, thought of as, Image credit: By Kormoran (Self-published work by Kormoran). Following are the steps required to use the, The first step is to enter the given function in the space given in front of the title. 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. That is, we need a working concept of a parameterized surface (or a parametric surface), in the same way that we already have a concept of a parameterized curve. Both mass flux and flow rate are important in physics and engineering. &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \, d\phi \\ 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. I want to calculate the magnetic flux which is defined as: If the magnetic field (B) changes over the area, then this surface integral can be pretty tough. Step #2: Select the variable as X or Y. Divide rectangle \(D\) into subrectangles \(D_{ij}\) with horizontal width \(\Delta u\) and vertical length \(\Delta v\). 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. \end{align*}\]. Suppose that the temperature at point \((x,y,z)\) in an object is \(T(x,y,z)\). However, weve done most of the work for the first one in the previous example so lets start with that. Divergence and Curl calculator Double integrals Double integral over a rectangle Integrals over paths and surfaces Path integral for planar curves Area of fence Example 1 Line integral: Work Line integrals: Arc length & Area of fence Surface integral of a vector field over a surface Line integrals of vector fields: Work & Circulation You can use this calculator by first entering the given function and then the variables you want to differentiate against. The boundary curve, C , is oriented clockwise when looking along the positive y-axis. 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. Let the lower limit in the case of revolution around the x-axis be a. In order to show the steps, the calculator applies the same integration techniques that a human would apply. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. Not what you mean? Calculate the mass flux of the fluid across \(S\). I'm able to pass my algebra class after failing last term using this calculator app. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The way to tell them apart is by looking at the differentials. 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, to calculate, \[\iint_{S_1} z^2 \,dS + \iint_{S_2} z^2 \,dS \nonumber \]. Loading please wait!This will take a few seconds. Here are the ranges for \(y\) and \(z\). Parameterizations that do not give an actual surface? To parameterize a sphere, it is easiest to use spherical coordinates. For those with a technical background, the following section explains how the Integral Calculator works. Which of the figures in Figure \(\PageIndex{8}\) is smooth? 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 \(\PageIndex{2}\)). If you have any questions or ideas for improvements to the Integral Calculator, don't hesitate to write me an e-mail. The arc length formula is derived from the methodology of approximating the length of a curve. Calculate the area of a surface of revolution step by step The calculations and the answer for the integral can be seen here. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. The flux of a vector field F F across a surface S S is the surface integral Flux = =SF nd. { "16.6E:_Exercises_for_Section_16.6" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "16.00:_Prelude_to_Vector_Calculus" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.01:_Vector_Fields" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.02:_Line_Integrals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.03:_Conservative_Vector_Fields" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.04:_Greens_Theorem" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.05:_Divergence_and_Curl" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.06:_Surface_Integrals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.07:_Stokes_Theorem" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.08:_The_Divergence_Theorem" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16.09:_Chapter_16_Review_Exercises" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Functions_and_Graphs" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Limits" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Derivatives" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Applications_of_Derivatives" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Integration" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Applications_of_Integration" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "07:_Techniques_of_Integration" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "08:_Introduction_to_Differential_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "09:_Sequences_and_Series" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "10:_Power_Series" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11:_Parametric_Equations_and_Polar_Coordinates" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "12:_Vectors_in_Space" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "13:_Vector-Valued_Functions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "14:_Differentiation_of_Functions_of_Several_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "15:_Multiple_Integration" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "16:_Vector_Calculus" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "17:_Second-Order_Differential_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "18:_Appendices" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "surface area", "surface integrals", "Parametric Surfaces", "parameter domain", "authorname:openstax", "M\u00f6bius strip", "flux integral", "grid curves", "heat flow", "mass flux", "orientation of a surface", "parameter space", "parameterized surface", "parametric surface", "regular parameterization", "surface integral", "surface integral of a scalar-valued function", "surface integral of a vector field", "license:ccbyncsa", "showtoc:no", "program:openstax", "licenseversion:40", "source@https://openstax.org/details/books/calculus-volume-1", "author@Gilbert Strang", "author@Edwin \u201cJed\u201d Herman" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FCalculus%2FBook%253A_Calculus_(OpenStax)%2F16%253A_Vector_Calculus%2F16.06%253A_Surface_Integrals, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Example \(\PageIndex{1}\): Parameterizing a Cylinder, Example \(\PageIndex{2}\): Describing a Surface, Example \(\PageIndex{3}\): Finding a Parameterization, Example \(\PageIndex{4}\): Identifying Smooth and Nonsmooth Surfaces, Definition: Smooth Parameterization of Surface, Example \(\PageIndex{5}\): Calculating Surface Area, Example \(\PageIndex{6}\): Calculating Surface Area, Example \(\PageIndex{7}\): Calculating Surface Area, Definition: Surface Integral of a Scalar-Valued Function, surface integral of a scalar-valued functi, Example \(\PageIndex{8}\): Calculating a Surface Integral, Example \(\PageIndex{9}\): Calculating the Surface Integral of a Cylinder, Example \(\PageIndex{10}\): Calculating the Surface Integral of a Piece of a Sphere, Example \(\PageIndex{11}\): Calculating the Mass of a Sheet, Example \(\PageIndex{12}\):Choosing an Orientation, Example \(\PageIndex{13}\): Calculating a Surface Integral, Example \(\PageIndex{14}\):Calculating Mass Flow Rate, Example \(\PageIndex{15}\): Calculating Heat Flow, Surface Integral of a Scalar-Valued Function, source@https://openstax.org/details/books/calculus-volume-1, surface integral of a scalar-valued function, status page at https://status.libretexts.org.