&= \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 \\ A Surface Area Calculator is an online calculator that can be easily used to determine the surface area of an object in the x-y plane. Remember that the plane is given by \(z = 4 - y\). It is the axis around which the curve revolves. This book makes you realize that Calculus isn't that tough after all. &= \dfrac{2560 \sqrt{6}}{9} \approx 696.74. Stokes' theorem is the 3D version of Green's theorem. Therefore, the calculated surface area is: Find the surface area of the following function: where 0y4 and the rotation are along the y-axis. Notice that if \(u\) is held constant, then the resulting curve is a circle of radius \(u\) in plane \(z = u\). Since every curve has a forward and backward direction (or, in the case of a closed curve, a clockwise and counterclockwise direction), it is possible to give an orientation to any curve. \label{surfaceI} \]. 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:. In this section we introduce the idea of a surface integral. Find the parametric representations of a cylinder, a cone, and a sphere. We have seen that a line integral is an integral over a path in a plane or in space. &= -110\pi. If \(v\) is held constant, then the resulting curve is a vertical parabola. Therefore, the surface is the elliptic paraboloid \(x^2 + y^2 = z\) (Figure \(\PageIndex{3}\)). Explain the meaning of an oriented surface, giving an example. You can use this calculator by first entering the given function and then the variables you want to differentiate against. Now, how we evaluate the surface integral will depend upon how the surface is given to us. Choose point \(P_{ij}\) in each piece \(S_{ij}\). &= \int_0^{\sqrt{3}} \int_0^{2\pi} u \, dv \, du \\ Use Equation \ref{scalar surface integrals}. 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 \(\Delta v \vecs t_v(P_{ij})\). Were going to let \({S_1}\) be the portion of the cylinder that goes from the \(xy\)-plane to the plane. It helps you practice by showing you the full working (step by step integration). 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 \(f\). However, why stay so flat? Lets now generalize the notions of smoothness and regularity to a parametric surface. Since we are working on the upper half of the sphere here are the limits on the parameters. Alternatively, you can view it as a way of generalizing double integrals to curved surfaces. Figure-1 Surface Area of Different Shapes. \end{align*}\]. If \(v = 0\) or \(v = \pi\), then the only choices for \(u\) that make the \(\mathbf{\hat{j}}\) component zero are \(u = 0\) or \(u = \pi\). Notice that if we change the parameter domain, we could get a different surface. Surface Integral -- from Wolfram MathWorld Calculus and Analysis Differential Geometry Differential Geometry of Surfaces Algebra Vector Algebra Calculus and Analysis Integrals Definite Integrals Surface Integral For a scalar function over a surface parameterized by and , the surface integral is given by (1) (2) Let \(\theta\) be the angle of rotation. It helps you practice by showing you the full working (step by step integration). By Equation \ref{scalar surface integrals}, \[\begin{align*} \iint_S f(x,y,z)dS &= \iint_D f (\vecs r(u,v)) ||\vecs t_u \times \vecs t_v|| \, dA \\ In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. Here is the parameterization of this cylinder. The surface area of the sphere is, \[\int_0^{2\pi} \int_0^{\pi} r^2 \sin \phi \, d\phi \,d\theta = r^2 \int_0^{2\pi} 2 \, d\theta = 4\pi r^2. Parallelogram Theorems: Quick Check-in ; Kite Construction Template Assume that f is a scalar, vector, or tensor field defined on a surface S.To find an explicit formula for the surface integral of f over S, we need to parameterize S by defining a system of curvilinear coordinates on S, like the latitude and longitude on a sphere.Let such a parameterization be r(s, t), where (s, t) varies in some region T in the plane. and , To calculate the surface integral, we first need a parameterization of the cylinder. mass of a shell; center of mass and moments of inertia of a shell; gravitational force and pressure force; fluid flow and mass flow across a surface; electric charge distributed over a surface; electric fields (Gauss' Law . Here they are. Analogously, we would like a notion of regularity (or smoothness) for surfaces so that a surface parameterization really does trace out a surface. Since the disk is formed where plane \(z = 1\) intersects sphere \(x^2 + y^2 + z^2 = 4\), we can substitute \(z = 1\) into equation \(x^2 + y^2 + z^2 = 4\): \[x^2 + y^2 + 1 = 4 \Rightarrow x^2 + y^2 = 3. Legal. 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 \(S\). The surface is a portion of the sphere of radius 2 centered at the origin, in fact exactly one-eighth of the sphere. Here is the parameterization for this sphere. Having an integrand allows for more possibilities with what the integral can do for you. If the density of the sheet is given by \(\rho (x,y,z) = x^2 yz\), what is the mass of the sheet? In Physics to find the centre of gravity. 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". &=80 \int_0^{2\pi} 45 \, d\theta \\ Find the mass flow rate of the fluid across \(S\). Let's take a closer look at each form . Interactive graphs/plots help visualize and better understand the functions. Notice also that \(\vecs r'(t) = \vecs 0\). Maxima takes care of actually computing the integral of the mathematical function. ; 6.6.5 Describe the surface integral of a vector field. Let \(S\) be the half-cylinder \(\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, 0 \leq u \leq \pi, \, 0 \leq v \leq 2\) oriented outward. The notation needed to develop this definition is used throughout the rest of this chapter. For a curve, this condition ensures that the image of \(\vecs r\) really is a curve, and not just a point. A cast-iron solid cylinder is given by inequalities \(x^2 + y^2 \leq 1, \, 1 \leq z \leq 4\). Step #3: Fill in the upper bound value. \[\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. Because of the half-twist in the strip, the surface has no outer side or inner side. This surface has parameterization \(\vecs r(u,v) = \langle v \, \cos u, \, v \, \sin u, \, 4 \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq 1.\). Suppose that \(v\) is a constant \(K\). If you have any questions or ideas for improvements to the Integral Calculator, don't hesitate to write me an e-mail. At this point weve got a fairly simple double integral to do. 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. 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 \(\PageIndex{11}\)). Direct link to Andras Elrandsson's post I almost went crazy over , Posted 3 years ago. While the line integral depends on a curve defined by one parameter, a two-dimensional surface depends on two parameters. \end{align*}\]. Note that \(\vecs t_u = \langle 1, 2u, 0 \rangle\) and \(\vecs t_v = \langle 0,0,1 \rangle\). Therefore the surface traced out by the parameterization is cylinder \(x^2 + y^2 = 1\) (Figure \(\PageIndex{1}\)). Use a surface integral to calculate the area of a given surface. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Area of Surface of Revolution Calculator. In Example \(\PageIndex{14}\), we computed the mass flux, which is the rate of mass flow per unit area. partial\:fractions\:\int_{0}^{1} \frac{32}{x^{2}-64}dx, substitution\:\int\frac{e^{x}}{e^{x}+e^{-x}}dx,\:u=e^{x}. If parameterization \(\vec{r}\) is regular, then the image of \(\vec{r}\) is a two-dimensional object, as a surface should be. Explain the meaning of an oriented surface, giving an example. 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. 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. The calculator lacks the mathematical intuition that is very useful for finding an antiderivative, but on the other hand it can try a large number of possibilities within a short amount of time. &= \sqrt{6} \int_0^4 \int_0^2 x^2 y (1 + x + 2y) \, dy \,dx \\[4pt] To embed this widget in a post on your WordPress blog, copy and paste the shortcode below into the HTML source: To add a widget to a MediaWiki site, the wiki must have the. The surface integral is then. Dot means the scalar product of the appropriate vectors. 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. To approximate the mass flux across \(S\), form the sum, \[\sum_{i=1}m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij}. &= \int_0^3 \left[\sin u + \dfrac{u}{2} - \dfrac{\sin(2u)}{4} \right]_0^{2\pi} \,dv \\ The Integral Calculator has to detect these cases and insert the multiplication sign. Okay, since we are looking for the portion of the plane that lies in front of the \(yz\)-plane we are going to need to write the equation of the surface in the form \(x = g\left( {y,z} \right)\). For F ( x, y, z) = ( y, z, x), compute. Give a parameterization of the cone \(x^2 + y^2 = z^2\) lying on or above the plane \(z = -2\). Some surfaces, such as a Mbius strip, cannot be oriented. 193. \nonumber \]. This is in contrast to vector line integrals, which can be defined on any piecewise smooth curve. Figure-1 Surface Area of Different Shapes It calculates the surface area of a revolution when a curve completes a rotation along the x-axis or y-axis. An extremely well-written book for students taking Calculus for the first time as well as those who need a refresher. We arrived at the equation of the hypotenuse by setting \(x\) equal to zero in the equation of the plane and solving for \(z\). Finally, the bottom of the cylinder (not shown here) is the disk of radius \(\sqrt 3 \) in the \(xy\)-plane and is denoted by \({S_3}\). 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] 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) Surface Area Calculator Author: Ravinder Kumar Topic: Area, Surface The present GeoGebra applet shows surface area generated by rotating an arc. While graphing, singularities (e.g. poles) are detected and treated specially. Since \(S\) is given by the function \(f(x,y) = 1 + x + 2y\), a parameterization of \(S\) is \(\vecs r(x,y) = \langle x, \, y, \, 1 + x + 2y \rangle, \, 0 \leq x \leq 4, \, 0 \leq y \leq 2\). 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 \(\vecs t_u \times \vecs t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0,0,-v\rangle\). \nonumber \]. Use surface integrals to solve applied problems. For example, the graph of paraboloid \(2y = x^2 + z^2\) can be parameterized by \(\vecs r(x,y) = \left\langle x, \dfrac{x^2+z^2}{2}, z \right\rangle, \, 0 \leq x < \infty, \, 0 \leq z < \infty\). &= \langle 4 \, \cos \theta \, \sin^2 \phi, \, 4 \, \sin \theta \, \sin^2 \phi, \, 4 \, \cos \phi \, \sin \phi \rangle. Learning Objectives. Find step by step results, graphs & plot using multiple integrals, Step 1: Enter the function and the limits in the input field Step 2: Now click the button Calculate to get the value Step 3: Finally, the, For a scalar function f over a surface parameterized by u and v, the surface integral is given by Phi = int_Sfda (1) = int_Sf(u,v)|T_uxT_v|dudv. We can now get the value of the integral that we are after. \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. 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. So, we want to find the center of mass of the region below. Surface integrals of scalar fields. $\operatorname{f}(x) \operatorname{f}'(x)$. 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. Here is that work. Calculate the Surface Area using the calculator. tothebook. Now that we are able to parameterize surfaces and calculate their surface areas, we are ready to define surface integrals. The mass flux of the fluid is the rate of mass flow per unit area. Notice that this cylinder does not include the top and bottom circles. Sets up the integral, and finds the area of a surface of revolution. Clicking an example enters it into the Integral Calculator. 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. Similarly, the average value of a function of two variables over the rectangular then, Weisstein, Eric W. "Surface Integral." Before we work some examples lets notice that since we can parameterize a surface given by \(z = g\left( {x,y} \right)\) as. Did this calculator prove helpful to you? The definition is analogous to the definition of the flux of a vector field along a plane curve. A surface integral is like a line integral in one higher dimension. Step #4: Fill in the lower bound value. &= -55 \int_0^{2\pi} du \\[4pt] Therefore, we calculate three separate integrals, one for each smooth piece of \(S\). The \(\mathbf{\hat{k}}\) component of this vector is zero only if \(v = 0\) or \(v = \pi\). This surface has parameterization \(\vecs r(x, \theta) = \langle x, \, x^2 \cos \theta, \, x^2 \sin \theta \rangle, \, 0 \leq x \leq b, \, 0 \leq x < 2\pi.\). To be precise, consider the grid lines that go through point \((u_i, v_j)\). Furthermore, assume that \(S\) is traced out only once as \((u,v)\) varies over \(D\). For a vector function over a surface, the surface integral is given by Phi = int_SFda (3) = int_S(Fn^^)da (4) = int_Sf_xdydz+f . It could be described as a flattened ellipse. Surface integral calculator with steps Calculate the area of a surface of revolution step by step The calculations and the answer for the integral can be seen here. \end{align*}\], By Equation \ref{equation1}, the surface area of the cone is, \[ \begin{align*}\iint_D ||\vecs t_u \times \vecs t_v|| \, dA &= \int_0^h \int_0^{2\pi} kv \sqrt{1 + k^2} \,du\, dv \\[4pt] &= 2\pi k \sqrt{1 + k^2} \int_0^h v \,dv \\[4pt] &= 2 \pi k \sqrt{1 + k^2} \left[\dfrac{v^2}{2}\right]_0^h \\[4pt] \\[4pt] &= \pi k h^2 \sqrt{1 + k^2}. If we want to find the flow rate (measured in volume per time) instead, we can use flux integral, \[\iint_S \vecs v \cdot \vecs N \, dS, \nonumber \]. Then, \(\vecs t_x = \langle 1,0,f_x \rangle\) and \(\vecs t_y = \langle 0,1,f_y \rangle \), and therefore the cross product \(\vecs t_x \times \vecs t_y\) (which is normal to the surface at any point on the surface) is \(\langle -f_x, \, -f_y, \, 1 \rangle \)Since the \(z\)-component of this vector is one, the corresponding unit normal vector points upward, and the upward side of the surface is chosen to be the positive side. for these kinds of surfaces. For example, this involves writing trigonometric/hyperbolic functions in their exponential forms. We gave the parameterization of a sphere in the previous section. Otherwise, a probabilistic algorithm is applied that evaluates and compares both functions at randomly chosen places. Surface integrals are important for the same reasons that line integrals are important. A specialty in mathematical expressions is that the multiplication sign can be left out sometimes, for example we write "5x" instead of "5*x". To parameterize a sphere, it is easiest to use spherical coordinates. Surface integrals are used anytime you get the sensation of wanting to add a bunch of values associated with points on a surface. Therefore, the flux of \(\vecs{F}\) across \(S\) is 340. Therefore, the lateral surface area of the cone is \(\pi r \sqrt{h^2 + r^2}\). 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). 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. We have derived the familiar formula for the surface area of a sphere using surface integrals. Why do you add a function to the integral of surface integrals? With the standard parameterization of a cylinder, Equation \ref{equation1} shows that the surface area is \(2 \pi rh\). 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 \(D\). Use the parameterization of surfaces of revolution given before Example \(\PageIndex{7}\). 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. The rotation is considered along the y-axis. Computing a surface integral is almost identical to computing surface area using a double integral, except that you stick a function inside the integral. \nonumber \]. 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}. Also, dont forget to plug in for \(z\). \label{equation 5} \], \[\iint_S \vecs F \cdot \vecs N\,dS, \nonumber \], where \(\vecs{F} = \langle -y,x,0\rangle\) and \(S\) is the surface with parameterization, \[\vecs r(u,v) = \langle u,v^2 - u, \, u + v\rangle, \, 0 \leq u \leq 3, \, 0 \leq v \leq 4. 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. Math Assignments. In this sense, surface integrals expand on our study of line integrals. The magnitude of this vector is \(u\). We also could choose the inward normal vector at each point to give an inward orientation, which is the negative orientation of the surface. Calculate surface integral Scurl F d S, where S is the surface, oriented outward, in Figure 16.7.6 and F = z, 2xy, x + y . The partial derivatives in the formulas are calculated in the following way: 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 \(kv\). Give the upward orientation of the graph of \(f(x,y) = xy\). Our integral solver also displays anti-derivative calculations to users who might be interested in the mathematical concept and steps involved in integration. Hence, it is possible to think of every curve as an oriented curve. Integration is a way to sum up parts to find the whole. Multiple Integrals Calculator - Symbolab Multiple Integrals Calculator Solve multiple integrals step-by-step full pad Examples Related Symbolab blog posts Advanced Math Solutions - Integral Calculator, trigonometric substitution In the previous posts we covered substitution, but standard substitution is not always enough. Use surface integrals to solve applied problems. A parameterization is \(\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, 0 \leq u \leq 2\pi, \, 0 \leq v \leq 3.\). The idea behind this parameterization is that for a fixed \(v\)-value, the circle swept out by letting \(u\) vary is the circle at height \(v\) and radius \(kv\). Example 1. Describe surface \(S\) parameterized by \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u^2 \rangle, \, 0 \leq u < \infty, \, 0 \leq v < 2\pi\). There is Surface integral calculator with steps that can make the process much easier. New Resources. The classic example of a nonorientable surface is the Mbius strip. 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, \[\begin{align*} \vecs t_{\phi} \times \vecs t_{\theta} &= \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \nonumber \\ 2 \cos \theta \cos \phi & 2 \sin \theta \cos \phi & -2\sin \phi \\ -2\sin \theta\sin\phi & 2\cos \theta \sin\phi & 0 \end{vmatrix} \\[4 pt] Notice that we plugged in the equation of the plane for the x in the integrand. The Integral Calculator will show you a graphical version of your input while you type. \nonumber \] 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). \label{mass} \]. 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. \nonumber \]. Is the surface parameterization \(\vecs r(u,v) = \langle u^{2v}, v + 1, \, \sin u \rangle, \, 0 \leq u \leq 2, \, 0 \leq v \leq 3\) smooth? But, these choices of \(u\) do not make the \(\mathbf{\hat{i}}\) component zero. Find a parameterization r ( t) for the curve C for interval t. Find the tangent vector. So, lets do the integral. This approximation becomes arbitrarily close to \(\displaystyle \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \Delta S_{ij}\) as we increase the number of pieces \(S_{ij}\) by letting \(m\) and \(n\) go to infinity. Solve Now. Calculate surface integral \[\iint_S \vecs F \cdot \vecs N \, dS, \nonumber \] where \(\vecs F = \langle 0, -z, y \rangle\) and \(S\) is the portion of the unit sphere in the first octant with outward orientation. So I figure that in order to find the net mass outflow I compute the surface integral of the mass flow normal to each plane and add them all up. \(r \, \cos \theta \, \sin \phi, \, r \, \sin \theta \, \sin \phi, \, r \, \cos \phi \rangle, \, 0 \leq \theta < 2\pi, \, 0 \leq \phi \leq \pi.\), \(\vecs t_{\theta} = \langle -r \, \sin \theta \, \sin \phi, \, r \, \cos \theta \, \sin \phi, \, 0 \rangle\), \(\vecs t_{\phi} = \langle r \, \cos \theta \, \cos \phi, \, r \, \sin \theta \, \cos \phi, \, -r \, \sin \phi \rangle.\), \[ \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. We will see one of these formulas in the examples and well leave the other to you to write down. I have already found the area of the paraboloid which is: A = ( 5 5 1) 6. 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. First, a parser analyzes the mathematical function. 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.