surface integral calculator

This book makes you realize that Calculus isn't that tough after all. To get an idea of the shape of the surface, we first plot some points. By the definition of the line integral (Section 16.2), \[\begin{align*} m &= \iint_S x^2 yz \, dS \\[4pt] then Multiply the area of each tiny piece by the value of the function, Abstract notation and visions of chopping up airplane wings are all well and good, but how do you actually, Specifically, the way you tend to represent a surface mathematically is with a, The trick for surface integrals, then, is to find a way of integrating over the flat region, Almost all of the work for this was done in the article on, For our surface integral desires, this means you expand. By Example, we know that $\vecs t_u \times \vecs t_v = \langle \cos u, \, \sin u, \, 0 \rangle$. A surface integral of a vector field. Put the value of the function and the lower and upper limits in the required blocks on the calculator then press the submit button. To calculate the mass flux across $S$, chop $S$ into small pieces $S_{ij}$. A cast-iron solid cylinder is given by inequalities $x^2 + y^2 \leq 1, \, 1 \leq z \leq 4$. As $v$ increases, the parameterization sweeps out a stack of circles, resulting in the desired cone. Each choice of $u$ and $v$ in the parameter domain gives a point on the surface, just as each choice of a parameter $t$ gives a point on a parameterized curve. Throughout this chapter, parameterizations $\vecs r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle$are assumed to be regular. A portion of the graph of any smooth function $z = f(x,y)$ is also orientable. \nonumber \], From the material we have already studied, we know that, \[\Delta S_{ij} \approx ||\vecs t_u (P_{ij}) \times \vecs t_v (P_{ij})|| \,\Delta u \,\Delta v. \nonumber \], \[\iint_S f(x,y,z) \,dS \approx \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij})|| \vecs t_u(P_{ij}) \times \vecs t_v(P_{ij}) ||\,\Delta u \,\Delta v. \nonumber \]. 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. We have derived the familiar formula for the surface area of a sphere using surface integrals. Direct link to Qasim Khan's post Wow thanks guys! Evaluate S x zdS S x z d S where S S is the surface of the solid bounded by x2 . The same was true for scalar surface integrals: we did not need to worry about an orientation of the surface of integration. &= 5 \left[\dfrac{(1+4u^2)^{3/2}}{3} \right]_0^2 \\ The tangent vectors are $\vecs t_x = \langle 1,0,1 \rangle$ and $\vecs t_y = \langle 1,0,2 \rangle$. This surface has parameterization $\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 1 \leq v \leq 4$. Sets up the integral, and finds the area of a surface of revolution. \nonumber \], 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. Parameterize the surface and use the fact that the surface is the graph of a function. Hence, it is possible to think of every curve as an oriented curve. In this case we dont need to do any parameterization since it is set up to use the formula that we gave at the start of this section. Use parentheses! 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. 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. &= 32\pi \left[- \dfrac{\cos^3 \phi}{3} \right]_0^{\pi/6} \\ The parameterization of full sphere $x^2 + y^2 + z^2 = 4$ is, \[\vecs r(\phi, \theta) = \langle 2 \, \cos \theta \, \sin \phi, \, 2 \, \sin \theta \, \sin \phi, \, 2 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi. Then, the mass of the sheet is given by $\displaystyle m = \iint_S x^2 yx \, dS.$ To compute this surface integral, we first need a parameterization of $S$. It is now time to think about integrating functions over some surface, $S$, in three-dimensional space. &= 80 \int_0^{2\pi} \int_0^{\pi/2} \langle 6 \, \cos \theta \, \sin \phi, \, 6 \, \sin \theta \, \sin \phi, \, 3 \, \cos \phi \rangle \cdot \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle \, d\phi \, d\theta \\ Suppose that $u$ is a constant $K$. which leaves out the density. The reason for this is that the circular base is included as part of the cone, and therefore the area of the base $\pi r^2$ is added to the lateral surface area $\pi r \sqrt{h^2 + r^2}$ that we found. The parameterization of the cylinder and $\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|$ is. Find more Mathematics widgets in Wolfram|Alpha. This is in contrast to vector line integrals, which can be defined on any piecewise smooth curve. Legal. \end{align*}\], \[ \begin{align*}||\vecs t_{\phi} \times \vecs t_{\theta} || &= \sqrt{r^4\sin^4\phi \, \cos^2 \theta + r^4 \sin^4 \phi \, \sin^2 \theta + r^4 \sin^2 \phi \, \cos^2 \phi} \\[4pt] &= \sqrt{r^4 \sin^4 \phi + r^4 \sin^2 \phi \, \cos^2 \phi} \\[4pt] &= r^2 \sqrt{\sin^2 \phi} \\[4pt] &= r \, \sin \phi.\end{align*}\], Notice that $\sin \phi \geq 0$ on the parameter domain because $0 \leq \phi < \pi$, and this justifies equation $\sqrt{\sin^2 \phi} = \sin \phi$. 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. A specialty in mathematical expressions is that the multiplication sign can be left out sometimes, for example we write "5x" instead of "5*x". where Imagine what happens as $u$ increases or decreases. Hold $u$ and $v$ constant, and see what kind of curves result. The following theorem provides an easier way in the case when  is a closed surface, that is, when  encloses a bounded solid in $\mathbb{R}^ 3$. Describe the surface integral of a scalar-valued function over a parametric surface. Otherwise, a probabilistic algorithm is applied that evaluates and compares both functions at randomly chosen places. Furthermore, all the vectors point outward, and therefore this is an outward orientation of the cylinder (Figure $\PageIndex{19}$). It is the axis around which the curve revolves. \end{align*}\]. 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. First, we calculate $\displaystyle \iint_{S_1} z^2 \,dS.$ To calculate this integral we need a parameterization of $S_1$. x-axis. \nonumber \], Therefore, the radius of the disk is $\sqrt{3}$ and a parameterization of $S_1$ is $\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, 1 \rangle, \, 0 \leq u \leq \sqrt{3}, \, 0 \leq v \leq 2\pi$. In the field of graphical representation to build three-dimensional models. The result is displayed in the form of the variables entered into the formula used to calculate the. Both mass flux and flow rate are important in physics and engineering. The mass is, M =(Area of plate) = b a f (x) g(x) dx M = ( Area of plate) = a b f ( x) g ( x) d x Next, we'll need the moments of the region. &= -110\pi. What if you are considering the surface of a curved airplane wing with variable density, and you want to find its total mass? 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. Divide rectangle $D$ into subrectangles $D_{ij}$ with horizontal width $\Delta u$ and vertical length $\Delta v$. \nonumber \]. Therefore, the choice of unit normal vector, \[\vecs N = \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \nonumber \]. Learning Objectives. However, weve done most of the work for the first one in the previous example so lets start with that. Let $\vecs r(u,v)$ be a parameterization of $S$ with parameter domain $D$. We know the formula for volume of a sphere is ( 4 / 3) r 3, so the volume we have computed is ( 1 / 8) ( 4 / 3) 2 3 = ( 4 / 3) , in agreement with our answer. button is clicked, the Integral Calculator sends the mathematical function and the settings (variable of integration and integration bounds) to the server, where it is analyzed again. Solution. Set integration variable and bounds in "Options". If you don't specify the bounds, only the antiderivative will be computed. is a dot product and is a unit normal vector. Alternatively, you can view it as a way of generalizing double integrals to curved surfaces. 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. Interactive graphs/plots help visualize and better understand the functions. 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. The rotation is considered along the y-axis. In this section we introduce the idea of a surface integral. ; 6.6.4 Explain the meaning of an oriented surface, giving an example. , for which the given function is differentiated. Therefore, the mass flux is, \[\iint_s \rho \vecs v \cdot \vecs N \, dS = \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij}. &= - 55 \int_0^{2\pi} \int_0^1 -v^3 \, dv \,du = - 55 \int_0^{2\pi} -\dfrac{1}{4} \,du = - \dfrac{55\pi}{2}.\end{align*}\]. This makes a=23.7/2=11.85 and b=11.8/2=5.9, if it were symmetrical. \nonumber \]. Calculate the mass flux of the fluid across $S$. . How to calculate the surface integral of the vector field: $$\iint\limits_{S^+} \vec F\cdot \vec n {\rm d}S $$ Is it the same thing to: $$\iint\limits_{S^+}x^2{\rm d}y{\rm d}z+y^2{\rm d}x{\rm d}z+z^2{\rm d}x{\rm d}y$$ There is another post here with an answer by@MichaelE2 for the cases when the surface is easily described in parametric form . This can also be written compactly in vector form as (2) If the region is on the left when traveling around , then area of can be computed using the elegant formula (3) Which of the figures in Figure $\PageIndex{8}$ is smooth? 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. It transforms it into a form that is better understandable by a computer, namely a tree (see figure below). For each point $\vecs r(a,b)$ on the surface, vectors $\vecs t_u$ and $\vecs t_v$ lie in the tangent plane at that point. Calculate the lateral surface area (the area of the side, not including the base) of the right circular cone with height h and radius r. Before calculating the surface area of this cone using Equation \ref{equation1}, we need a parameterization. The surface integral of the vector field over the oriented surface (or the flux of the vector field across First we calculate the partial derivatives:. It is used to find the area under a curve by slicing it to small rectangles and summing up thier areas. \[\begin{align*} \vecs t_x \times \vecs t_{\theta} &= \langle 2x^3 \cos^2 \theta + 2x^3 \sin^2 \theta, \, -x^2 \cos \theta, \, -x^2 \sin \theta \rangle \\[4pt] &= \langle 2x^3, \, -x^2 \cos \theta, \, -x^2 \sin \theta \rangle \end{align*}\], \[\begin{align*} \vecs t_x \times \vecs t_{\theta} &= \sqrt{4x^6 + x^4\cos^2 \theta + x^4 \sin^2 \theta} \\[4pt] &= \sqrt{4x^6 + x^4} \\[4pt] &= x^2 \sqrt{4x^2 + 1} \end{align*}\], \[\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] They have many applications to physics and engineering, and they allow us to develop higher dimensional versions of the Fundamental Theorem of Calculus. Scalar surface integrals have several real-world applications. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero. [2v^3u + v^2u - vu^2 - u^2]\right|_0^3 \, dv \\[4pt] &= \int_0^4 (6v^3 + 3v^2 - 9v - 9) \, dv \\[4pt] &= \left[ \dfrac{3v^4}{2} + v^3 - \dfrac{9v^2}{2} - 9v\right]_0^4\\[4pt] &= 340. For a height value $v$ with $0 \leq v \leq h$, the radius of the circle formed by intersecting the cone with plane $z = v$ is $kv$. In the first family of curves we hold $u$ constant; in the second family of curves we hold $v$ constant. &= - 55 \int_0^{2\pi} \int_0^1 (2v \, \cos^2 u + 2v \, \sin^2 u ) \, dv \,du \\[4pt] The second step is to define the surface area of a parametric surface. Lets start off with a sketch of the surface $S$ since the notation can get a little confusing once we get into it. Let $S$ be a smooth orientable surface with parameterization $\vecs r(u,v)$. You can do so using our Gauss law calculator with two very simple steps: Enter the value 10 n C 10\ \mathrm{nC} 10 nC ** in the field "Electric charge Q". Choose point $P_{ij}$ in each piece $S_{ij}$. Informally, a surface parameterization is smooth if the resulting surface has no sharp corners. perform a surface integral. The surface element contains information on both the area and the orientation of the surface. &= - 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] &= \int_0^3 \pi \, dv = 3 \pi. In the next block, the lower limit of the given function is entered. We gave the parameterization of a sphere in the previous section. Make sure that it shows exactly what you want. Clicking an example enters it into the Integral Calculator. Describe the surface integral of a vector field. The rate of flow, measured in mass per unit time per unit area, is $\rho \vecs N$. In general, surfaces must be parameterized with two parameters. Moving the mouse over it shows the text. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. &= \int_0^3 \left[\sin u + \dfrac{u}{2} - \dfrac{\sin(2u)}{4} \right]_0^{2\pi} \,dv \\ With the standard parameterization of a cylinder, Equation \ref{equation1} shows that the surface area is $2 \pi rh$. However, as noted above we can modify this formula to get one that will work for us. 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$. Therefore, the tangent of $\phi$ is $\sqrt{3}$, which implies that $\phi$ is $\pi / 6$. \end{align*}\], Therefore, the rate of heat flow across $S$ is, \[\dfrac{55\pi}{2} - \dfrac{55\pi}{2} - 110\pi = -110\pi. How to compute the surface integral of a vector field.Join me on Coursera: https://www.coursera.org/learn/vector-calculus-engineersLecture notes at http://ww. The gesture control is implemented using Hammer.js. How To Use a Surface Area Calculator in Calculus? Try it Extended Keyboard Examples Assuming "surface integral" is referring to a mathematical definition | Use as a character instead Input interpretation Definition More details More information Related terms Subject classifications Embed this widget . In "Examples", you can see which functions are supported by the Integral Calculator and how to use them. \nonumber \]. Very useful and convenient. 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. When you're done entering your function, click "Go! The difference between this problem and the previous one is the limits on the parameters. &= 32 \pi \left[ \dfrac{1}{3} - \dfrac{\sqrt{3}}{8} \right] = \dfrac{32\pi}{3} - 4\sqrt{3}. Notice that we do not need to vary over the entire domain of $y$ because $x$ and $z$ are squared. To visualize $S$, we visualize two families of curves that lie on $S$. The parser is implemented in JavaScript, based on the Shunting-yard algorithm, and can run directly in the browser. Therefore, as $u$ increases, the radius of the resulting circle increases. Direct link to Andras Elrandsson's post I almost went crazy over , Posted 3 years ago. 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 $\PageIndex{6}$). Describe the surface parameterized by $\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle, \, - \infty < u < \infty, \, 0 \leq v < 2\pi$. In order to show the steps, the calculator applies the same integration techniques that a human would apply. 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 $[a,b]$. Otherwise, it tries different substitutions and transformations until either the integral is solved, time runs out or there is nothing left to try. This allows for quick feedback while typing by transforming the tree into LaTeX code. Direct link to Is Better Than 's post Well because surface inte, Posted 2 years ago. You can use this calculator by first entering the given function and then the variables you want to differentiate against. The result is displayed after putting all the values in the related formula. is given explicitly by, If the surface is surface parameterized using The classic example of a nonorientable surface is the Mbius strip. Since the surface is oriented outward and $S_1$ is the top of the object, we instead take vector $\vecs t_v \times \vecs t_u = \langle 0,0,v\rangle$. Use a surface integral to calculate the area of a given surface. This is analogous to the flux of two-dimensional vector field $\vecs{F}$ across plane curve $C$, in which we approximated flux across a small piece of $C$ with the expression $(\vecs{F} \cdot \vecs{N}) \,\Delta s$. \label{surfaceI} \]. What does to integrate mean? The Integral Calculator lets you calculate integrals and antiderivatives of functions online for free! When the "Go!" Since the parameter domain is all of $\mathbb{R}^2$, we can choose any value for u and v and plot the corresponding point. for these kinds of surfaces. The mass of a sheet is given by Equation \ref{mass}. While the line integral depends on a curve defined by one parameter, a two-dimensional surface depends on two parameters. 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. In order to do this integral well 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. Then, \[\begin{align*} x^2 + y^2 &= (\rho \, \cos \theta \, \sin \phi)^2 + (\rho \, \sin \theta \, \sin \phi)^2 \\[4pt] Here are the two vectors. 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] In addition to parameterizing surfaces given by equations or standard geometric shapes such as cones and spheres, we can also parameterize surfaces of revolution. The surface integral of $\vecs{F}$ over $S$ is, \[\iint_S \vecs{F} \cdot \vecs{S} = \iint_S \vecs{F} \cdot \vecs{N} \,dS. \label{scalar surface integrals} \]. In addition to modeling fluid flow, surface integrals can be used to model heat flow. 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. &= - 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] then, Weisstein, Eric W. "Surface Integral." Solution Note that to calculate Scurl F d S without using Stokes' theorem, we would need the equation for scalar surface integrals. We can start with the surface integral of a scalar-valued function. To obtain a parameterization, let $\alpha$ be the angle that is swept out by starting at the positive z-axis and ending at the cone, and let $k = \tan \alpha$. By double integration, we can find the area of the rectangular region. Find the parametric representations of a cylinder, a cone, and a sphere. Solution : Since we are given a line integral and told to use Stokes' theorem, we need to compute a surface integral. This can be used to solve problems in a wide range of fields, including physics, engineering, and economics. Remember that the plane is given by $z = 4 - y$. . Use parentheses, if necessary, e.g. "a/(b+c)". \nonumber \]. In the case of antiderivatives, the entire procedure is repeated with each function's derivative, since antiderivatives are allowed to differ by a constant. Surface integral of a vector field over a surface. 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"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.
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