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Here is the evaluation for the double integral. Sets up the integral, and finds the area of a surface of revolution. Step #5: Click on "CALCULATE" button. Give a parameterization of the cone \(x^2 + y^2 = z^2\) lying on or above the plane \(z = -2\). 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}. Explain the meaning of an oriented surface, giving an example. First, we are using pretty much the same surface (the integrand is different however) as the previous example. 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 \(k = \tan \alpha\). This helps me sooo much, im in seventh grade and this helps A LOT, i was able to pass and ixl in 3 minutes, and it was a word problems one. Calculate the area of a surface of revolution step by step The calculations and the answer for the integral can be seen here. The tangent vectors are \(\vecs t_x = \langle 1,0,1 \rangle\) and \(\vecs t_y = \langle 1,0,2 \rangle\). Choose point \(P_{ij}\) in each piece \(S_{ij}\). \nonumber \]. 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. Furthermore, all the vectors point outward, and therefore this is an outward orientation of the cylinder (Figure \(\PageIndex{19}\)). Integrations is used in various fields such as engineering to determine the shape and size of strcutures. 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. where \(D\) is the range of the parameters that trace out the surface \(S\). Let \(S\) denote the boundary of the object. They have many applications to physics and engineering, and they allow us to develop higher dimensional versions of the Fundamental Theorem of Calculus. By the definition of the line integral (Section 16.2), \[\begin{align*} m &= \iint_S x^2 yz \, dS \\[4pt] However, weve done most of the work for the first one in the previous example so lets start with that. To get an idea of the shape of the surface, we first plot some points. In the field of graphical representation to build three-dimensional models. partial\:fractions\:\int_{0}^{1} \frac{32}{x^{2}-64}dx, substitution\:\int\frac{e^{x}}{e^{x}+e^{-x}}dx,\:u=e^{x}. With surface integrals we will be integrating over the surface of a solid. Finally, the bottom of the cylinder (not shown here) is the disk of radius \(\sqrt 3 \) in the \(xy\)-plane and is denoted by \({S_3}\). Posted 5 years ago. If \(u = v = 0\), then \(\vecs r(0,0) = \langle 1,0,0 \rangle\), so point (1, 0, 0) is on \(S\). The Integral Calculator has to detect these cases and insert the multiplication sign. Example 1. Solutions Graphing Practice; New Geometry; Calculators; Notebook . It is the axis around which the curve revolves. If \(S_{ij}\) is small enough, then it can be approximated by a tangent plane at some point \(P\) in \(S_{ij}\). then If \(v = 0\) or \(v = \pi\), then the only choices for \(u\) that make the \(\mathbf{\hat{j}}\) component zero are \(u = 0\) or \(u = \pi\). On the other hand, when we defined vector line integrals, the curve of integration needed an orientation. \nonumber \]. 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". 193. \nonumber \]. Let the lower limit in the case of revolution around the x-axis be a. , the upper limit of the given function is entered. By Example, we know that \(\vecs t_u \times \vecs t_v = \langle \cos u, \, \sin u, \, 0 \rangle\). 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. 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 \(\vecs t_u \times \vecs t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0,0,-v\rangle\). Send feedback | Visit Wolfram|Alpha. The following theorem provides an easier way in the case when \(\) is a closed surface, that is, when \(\) encloses a bounded solid in \(\mathbb{R}^ 3\). 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 \(kv\). In fact, it can be shown that. The rotation is considered along the y-axis. Since we are working on the upper half of the sphere here are the limits on the parameters. Equation \ref{scalar surface integrals} allows us to calculate a surface integral by transforming it into a double integral. Again, this is set up to use the initial formula we gave in this section once we realize that the equation for the bottom is given by \(g\left( {x,y} \right) = 0\) and \(D\) is the disk of radius \(\sqrt 3 \) centered at the origin. Therefore, we expect the surface to be an elliptic paraboloid. Lets now generalize the notions of smoothness and regularity to a parametric surface. In the first family of curves we hold \(u\) constant; in the second family of curves we hold \(v\) constant. The surface element contains information on both the area and the orientation of the surface. \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] Finally, to parameterize the graph of a two-variable function, we first let \(z = f(x,y)\) be a function of two variables. Use the standard parameterization of a cylinder and follow the previous example. Step 3: Add up these areas. First, we calculate \(\displaystyle \iint_{S_1} z^2 \,dS.\) To calculate this integral we need a parameterization of \(S_1\). 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. Computing a surface integral is almost identical to computing surface area using a double integral, except that you stick a function inside the integral. With the standard parameterization of a cylinder, Equation \ref{equation1} shows that the surface area is \(2 \pi rh\). \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). 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. Note how the equation for a surface integral is similar to the equation for the line integral of a vector field C F d s = a b F ( c ( t)) c ( t) d t. For line integrals, we integrate the component of the vector field in the tangent direction given by c ( t). \end{align*}\]. Dont forget that we need to plug in for \(z\)! 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. Use a surface integral to calculate the area of a given surface. Here they are. It's just a matter of smooshing the two intuitions together. Describe the surface integral of a vector field. However, why stay so flat? I tried and tried multiple times, it helps me to understand the process. Solution. 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. &= 32\pi \left[- \dfrac{\cos^3 \phi}{3} \right]_0^{\pi/6} \\ Did this calculator prove helpful to you? I have been tasked with solving surface integral of ${\bf V} = x^2{\bf e_x}+ y^2{\bf e_y}+ z^2 {\bf e_z}$ on the surface of a cube bounding the region $0\le x,y,z \le 1$. The partial derivatives in the formulas are calculated in the following way: \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle, \, 0 < u < \infty, \, 0 \leq v < \dfrac{\pi}{2}\), We have discussed parameterizations of various surfaces, but two important types of surfaces need a separate discussion: spheres and graphs of two-variable functions. Last, lets consider the cylindrical side of the object. Step 2: Compute the area of each piece. &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \sqrt{\sin^2\phi + \cos^2\phi} \, d\phi \\ 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 \(\vecs r\). Make sure that it shows exactly what you want. We used a rectangle here, but it doesnt have to be of course. An extremely well-written book for students taking Calculus for the first time as well as those who need a refresher. Find the heat flow across the boundary of the solid if this boundary is oriented outward. We will see one of these formulas in the examples and well leave the other to you to write down. First, lets look at the surface integral of a scalar-valued function. Now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. 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. There are two moments, denoted by M x M x and M y M y. Then enter the variable, i.e., xor y, for which the given function is differentiated. 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 \([a,b]\). &= 2\pi \int_0^{\sqrt{3}} u \, du \\ eMathHelp: free math calculator - solves algebra, geometry, calculus, statistics, linear algebra, and linear programming problems step by step \nonumber \]. The upper limit for the \(z\)s is the plane so we can just plug that in. \end{align*}\], \[\begin{align*} \iint_{S_2} z \, dS &= \int_0^{\pi/6} \int_0^{2\pi} f (\vecs r(\phi, \theta))||\vecs t_{\phi} \times \vecs t_{\theta}|| \, d\theta \, d\phi \\ This idea of adding up values over a continuous two-dimensional region can be useful for curved surfaces as well. Taking a normal double integral is just taking a surface integral where your surface is some 2D area on the s-t plane. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces. If you think of the normal field as describing water flow, then the side of the surface that water flows toward is the negative side and the side of the surface at which the water flows away is the positive side. 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 \(x = g\left( {y,z} \right)\). Find the parametric representations of a cylinder, a cone, and a sphere. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero. Direct link to Aiman's post Why do you add a function, Posted 3 years ago. 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. Hold \(u\) constant and see what kind of curves result. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier . Now consider the vectors that are tangent to these grid curves. It is used to find the area under a curve by slicing it to small rectangles and summing up thier areas. 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 Since we are only taking the piece of the sphere on or above plane \(z = 1\), we have to restrict the domain of \(\phi\). If you imagine placing a normal vector at a point on the strip and having the vector travel all the way around the band, then (because of the half-twist) the vector points in the opposite direction when it gets back to its original position. Let \(S\) be a smooth orientable surface with parameterization \(\vecs r(u,v)\). Notice that if we change the parameter domain, we could get a different surface. Before we work some examples lets notice that since we can parameterize a surface given by \(z = g\left( {x,y} \right)\) as. However, as noted above we can modify this formula to get one that will work for us. 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 \(S\). Find the area of the surface of revolution obtained by rotating \(y = x^2, \, 0 \leq x \leq b\) about the x-axis (Figure \(\PageIndex{14}\)). Like so many things in multivariable calculus, while the theory behind surface integrals is beautiful, actually computing one can be painfully labor intensive. The practice problem generator allows you to generate as many random exercises as you want. Why do you add a function to the integral of surface integrals? The surface area of a right circular cone with radius \(r\) and height \(h\) is usually given as \(\pi r^2 + \pi r \sqrt{h^2 + r^2}\). This surface has parameterization \(\vecs r(u,v) = \langle v \, \cos u, \, v \, \sin u, \, 4 \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq 1.\). In Example \(\PageIndex{14}\), we computed the mass flux, which is the rate of mass flow per unit area. The result is displayed after putting all the values in the related formula. The Surface Area calculator displays these values in the surface area formula and presents them in the form of a numerical value for the surface area bounded inside the rotation of the arc. With the idea of orientable surfaces in place, we are now ready to define a surface integral of a vector field. The formula for calculating the length of a curve is given as: L = a b 1 + ( d y d x) 2 d x. Then the curve traced out by the parameterization is \(\langle \cos u, \, \sin u, \, K \rangle \), which gives a circle in plane \(z = K\) with radius 1 and center \((0, 0, K)\). 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. 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 . Use the Surface area calculator to find the surface area of a given curve. Notice that we plugged in the equation of the plane for the x in the integrand. Take the dot product of the force and the tangent vector. Symbolab is the best integral calculator solving indefinite integrals, definite integrals, improper integrals, double integrals, triple integrals, multiple integrals, antiderivatives, and more. Lets first start out with a sketch of the surface. 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 general, surfaces must be parameterized with two parameters. Let \(S\) be hemisphere \(x^2 + y^2 + z^2 = 9\) with \(z \leq 0\) such that \(S\) is oriented outward. When the "Go!" Since the surface is oriented outward and \(S_1\) is the bottom of the object, it makes sense that this vector points downward. 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. 2. 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 is an easy surface integral to calculate using the Divergence Theorem: $$ \iiint_E {\rm div} (F)\ dV = \iint_ {S=\partial E} \vec {F}\cdot d {\bf S}$$ However, to confirm the divergence theorem by the direct calculation of the surface integral, how should the bounds on the double integral for a unit ball be chosen? We discuss how Surface integral of vector field calculator can help students learn Algebra in this blog post. 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] The temperature at a point in a region containing the ball is \(T(x,y,z) = \dfrac{1}{3}(x^2 + y^2 + z^2)\). We also could choose the inward normal vector at each point to give an inward orientation, which is the negative orientation of the surface. Compute the net mass outflow through the cube formed by the planes x=0, x=1, y=0, y=1, z=0, z=1. The magnitude of this vector is \(u\). The rate of flow, measured in mass per unit time per unit area, is \(\rho \vecs N\). When the integrand matches a known form, it applies fixed rules to solve the integral (e.g. partial fraction decomposition for rational functions, trigonometric substitution for integrands involving the square roots of a quadratic polynomial or integration by parts for products of certain functions). I understood this even though I'm just a senior at high school and I haven't read the background material on double integrals or even Calc II. A specialty in mathematical expressions is that the multiplication sign can be left out sometimes, for example we write "5x" instead of "5*x". Maxima takes care of actually computing the integral of the mathematical function. Area of Surface of Revolution Calculator. 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. How To Use a Surface Area Calculator in Calculus? 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). 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. surface integral Natural Language Math Input Use Math Input Mode to directly enter textbook math notation. { "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 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"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. \nonumber \]. Set integration variable and bounds in "Options". The flux of a vector field F F across a surface S S is the surface integral Flux = =SF nd. In the first grid line, the horizontal component is held constant, yielding a vertical line through \((u_i, v_j)\). \nonumber \]. Boundary Value Problems & Fourier Series, 8.3 Periodic Functions & Orthogonal Functions, 9.6 Heat Equation with Non-Zero Temperature Boundaries, 1.14 Absolute Value Equations and Inequalities. For example, if we restricted the domain to \(0 \leq u \leq \pi, \, -\infty < v < 6\), then the surface would be a half-cylinder of height 6. If the density of the sheet is given by \(\rho (x,y,z) = x^2 yz\), what is the mass of the sheet? I almost went crazy over this but note that when you are looking for the SURFACE AREA (not surface integral) over some scalar field (z = f(x, y)), meaning that the vector V(x, y) of which you take the cross-product of becomes V(x, y) = (x, y, f(x, y)). 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. Since \(S_{ij}\) is small, the dot product \(\rho v \cdot N\) changes very little as we vary across \(S_{ij}\) and therefore \(\rho \vecs v \cdot \vecs N\) can be taken as approximately constant across \(S_{ij}\). &= 32 \pi \left[ \dfrac{1}{3} - \dfrac{\sqrt{3}}{8} \right] = \dfrac{32\pi}{3} - 4\sqrt{3}. Otherwise, a probabilistic algorithm is applied that evaluates and compares both functions at randomly chosen places.