Break the integral into three separate surface integrals. 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 \(D\). \nonumber \]. Put the value of the function and the lower and upper limits in the required blocks on the calculator t, Surface Area Calculator Calculus + Online Solver With Free Steps. Paid link. To parameterize this disk, we need to know its radius. Here is a sketch of the surface \(S\). \end{align*}\], \[ \begin{align*} \pi k h^2 \sqrt{1 + k^2} &= \pi \dfrac{r}{h}h^2 \sqrt{1 + \dfrac{r^2}{h^2}} \\[4pt] &= \pi r h \sqrt{1 + \dfrac{r^2}{h^2}} \\[4pt] \\[4pt] &= \pi r \sqrt{h^2 + h^2 \left(\dfrac{r^2}{h^2}\right) } \\[4pt] &= \pi r \sqrt{h^2 + r^2}. Use parentheses! \nonumber \]. Suppose that the temperature at point \((x,y,z)\) in an object is \(T(x,y,z)\). If \(S_{ij}\) is small enough, then it can be approximated by a tangent plane at some point \(P\) in \(S_{ij}\). 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 \(\vecs t_u \times \vecs t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0,0,-v\rangle\). The \(\mathbf{\hat{k}}\) component of this vector is zero only if \(v = 0\) or \(v = \pi\). Step #3: Fill in the upper bound value. Loading please wait!This will take a few seconds. Notice that \(S\) is not smooth but is piecewise smooth; \(S\) can be written as the union of its base \(S_1\) and its spherical top \(S_2\), and both \(S_1\) and \(S_2\) are smooth. The rotation is considered along the y-axis. Find the mass of the piece of metal. A specialty in mathematical expressions is that the multiplication sign can be left out sometimes, for example we write "5x" instead of "5*x". If it is possible to choose a unit normal vector \(\vecs N\) at every point \((x,y,z)\) on \(S\) so that \(\vecs N\) varies continuously over \(S\), then \(S\) is orientable. Such a choice of unit normal vector at each point gives the orientation of a surface \(S\). Let the upper limit in the case of revolution around the x-axis be b, and in the case of the y-axis, it is d. Press the Submit button to get the required surface area value. \nonumber \], Therefore, the radius of the disk is \(\sqrt{3}\) and a parameterization of \(S_1\) is \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, 1 \rangle, \, 0 \leq u \leq \sqrt{3}, \, 0 \leq v \leq 2\pi\). \nonumber \]. We can see that \(S_1\) is a circle of radius 1 centered at point \((0,0,1)\) sitting in plane \(z = 1\). &= 2\pi \left[ \dfrac{1}{64} \left(2 \sqrt{4b^2 + 1} (8b^3 + b) \, \sinh^{-1} (2b) \right)\right]. This is not an issue though, because Equation \ref{scalar surface integrals} does not place any restrictions on the shape of the parameter domain. Informally, a curve parameterization is smooth if the resulting curve has no sharp corners. Compute the net mass outflow through the cube formed by the planes x=0, x=1, y=0, y=1, z=0, z=1. \nonumber \]. 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. Notice that this cylinder does not include the top and bottom circles. Integration is a way to sum up parts to find the whole. Now, we need to be careful here as both of these look like standard double integrals. Dont forget that we need to plug in for \(z\)! The notation needed to develop this definition is used throughout the rest of this chapter. Here is the evaluation for the double integral. Step 2: Compute the area of each piece. It is now time to think about integrating functions over some surface, \(S\), in three-dimensional space. For any given surface, we can integrate over surface either in the scalar field or the vector field. 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? The horizontal cross-section of the cone at height \(z = u\) is circle \(x^2 + y^2 = u^2\). and , Since we are only taking the piece of the sphere on or above plane \(z = 1\), we have to restrict the domain of \(\phi\). Step #2: Select the variable as X or Y. &= - 55 \int_0^{2\pi} \int_0^1 \langle 8v \, \cos u, \, 8v \, \sin u, \, v^2\rangle \cdot \langle 0, 0, -v \rangle\, \, dv \,du\\[4pt] \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. If piece \(S_{ij}\) is small enough, then the tangent plane at point \(P_{ij}\) is a good approximation of piece \(S_{ij}\). Double integrals also can compute volume, but if you let f(x,y)=1, then double integrals boil down to the capabilities of a plain single-variable definite integral (which can compute areas). At its simplest, a surface integral can be thought of as the quantity of a vector field that penetrates through a given surface, as shown in Figure 5.1. \[\vecs{N}(x,y) = \left\langle \dfrac{-y}{\sqrt{1+x^2+y^2}}, \, \dfrac{-x}{\sqrt{1+x^2+y^2}}, \, \dfrac{1}{\sqrt{1+x^2+y^2}} \right\rangle \nonumber \]. To visualize \(S\), we visualize two families of curves that lie on \(S\). Find more Mathematics widgets in Wolfram|Alpha. However, why stay so flat? Our integral solver also displays anti-derivative calculations to users who might be interested in the mathematical concept and steps involved in integration. Having an integrand allows for more possibilities with what the integral can do for you. How do you add up infinitely many infinitely small quantities associated with points on a surface? Remember that the plane is given by \(z = 4 - y\). Do not get so locked into the \(xy\)-plane that you cant do problems that have regions in the other two planes. First, lets look at the surface integral of a scalar-valued function. You can use this calculator by first entering the given function and then the variables you want to differentiate against. Hold \(u\) and \(v\) constant, and see what kind of curves result. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. But, these choices of \(u\) do not make the \(\mathbf{\hat{i}}\) component zero. are tangent vectors and is the cross product. Again, notice the similarities between this definition and the definition of a scalar line integral. The rate of flow, measured in mass per unit time per unit area, is \(\rho \vecs N\). The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero. 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 surface traced out by the parameterization is cylinder \(x^2 + y^2 = 1\) (Figure \(\PageIndex{1}\)). \[\vecs r(\phi, \theta) = \langle 3 \, \cos \theta \, \sin \phi, \, 3 \, \sin \theta \, \sin \phi, \, 3 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi/2. \nonumber \], 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. I'm able to pass my algebra class after failing last term using this calculator app. Surfaces can be parameterized, just as curves can be parameterized. 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 useful parameterization of a paraboloid was given in a previous example. Hold \(u\) constant and see what kind of curves result. The result is displayed in the form of the variables entered into the formula used to calculate the. Since we are not interested in the entire cone, only the portion on or above plane \(z = -2\), the parameter domain is given by \(-2 < u < \infty, \, 0 \leq v < 2\pi\) (Figure \(\PageIndex{4}\)). Make sure that it shows exactly what you want. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. The definition of a smooth surface parameterization is similar. \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle, \, 0 < u < \infty, \, 0 \leq v < \dfrac{\pi}{2}\), We have discussed parameterizations of various surfaces, but two important types of surfaces need a separate discussion: spheres and graphs of two-variable functions. where \(S\) is the surface with parameterization \(\vecs r(u,v) = \langle u, \, u^2, \, v \rangle\) for \(0 \leq u \leq 2\) and \(0 \leq v \leq u\). 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 . This makes a=23.7/2=11.85 and b=11.8/2=5.9, if it were symmetrical. The mass of a sheet is given by Equation \ref{mass}. 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. The classic example of a nonorientable surface is the Mbius strip. start bold text, v, end bold text, with, vector, on top, left parenthesis, start color #0c7f99, t, end color #0c7f99, comma, start color #bc2612, s, end color #bc2612, right parenthesis, start color #0c7f99, t, end color #0c7f99, start color #bc2612, s, end color #bc2612, f, left parenthesis, x, comma, y, right parenthesis, f, left parenthesis, x, comma, y, comma, z, right parenthesis, start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis, start color #0c7f99, d, t, end color #0c7f99, start color #bc2612, d, s, end color #bc2612, d, \Sigma, equals, open vertical bar, start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, start color #0c7f99, t, end color #0c7f99, end fraction, times, start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, start color #bc2612, s, end color #bc2612, end fraction, close vertical bar, start color #0c7f99, d, t, end color #0c7f99, start color #bc2612, d, s, end color #bc2612, \iint, start subscript, S, end subscript, f, left parenthesis, x, comma, y, comma, z, right parenthesis, d, \Sigma, equals, \iint, start subscript, T, end subscript, f, left parenthesis, start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis, right parenthesis, open vertical bar, start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, start color #0c7f99, t, end color #0c7f99, end fraction, times, start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, start color #bc2612, s, end color #bc2612, end fraction, close vertical bar, start color #0c7f99, d, t, end color #0c7f99, start color #bc2612, d, s, end color #bc2612. Choose point \(P_{ij}\) in each piece \(S_{ij}\). We arrived at the equation of the hypotenuse by setting \(x\) equal to zero in the equation of the plane and solving for \(z\). To parameterize a sphere, it is easiest to use spherical coordinates. . 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 . Let the lower limit in the case of revolution around the x-axis be a. , the upper limit of the given function is entered. 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. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. Figure 16.7.6: A complicated surface in a vector field. Lets start off with a sketch of the surface \(S\) since the notation can get a little confusing once we get into it. 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. Here is the parameterization of this cylinder. we can always use this form for these kinds of surfaces as well. \end{align*}\]. What about surface integrals over a vector field? ; 6.6.2 Describe the surface integral of a scalar-valued function over a parametric surface. A cast-iron solid cylinder is given by inequalities \(x^2 + y^2 \leq 1, \, 1 \leq z \leq 4\). 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, \[\begin{align*} \int_C \vecs F \cdot \vecs N\, dS &= \iint_S \vecs F \cdot \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \,dS \\[4pt] Use surface integrals to solve applied problems. Example 1. However, unlike the previous example we are putting a top and bottom on the surface this time. Clicking an example enters it into the Integral Calculator. \end{align*}\], \[ \begin{align*}||\vecs t_{\phi} \times \vecs t_{\theta} || &= \sqrt{r^4\sin^4\phi \, \cos^2 \theta + r^4 \sin^4 \phi \, \sin^2 \theta + r^4 \sin^2 \phi \, \cos^2 \phi} \\[4pt] &= \sqrt{r^4 \sin^4 \phi + r^4 \sin^2 \phi \, \cos^2 \phi} \\[4pt] &= r^2 \sqrt{\sin^2 \phi} \\[4pt] &= r \, \sin \phi.\end{align*}\], Notice that \(\sin \phi \geq 0\) on the parameter domain because \(0 \leq \phi < \pi\), and this justifies equation \(\sqrt{\sin^2 \phi} = \sin \phi\). &= - 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] By Equation \ref{scalar surface integrals}, \[\begin{align*} \iint_S 5 \, dS &= 5 \iint_D \sqrt{1 + 4u^2} \, dA \\ Let \(\vecs v(x,y,z) = \langle x^2 + y^2, \, z, \, 4y \rangle\) m/sec represent a velocity field of a fluid with constant density 100 kg/m3. In a similar way, to calculate a surface integral over surface \(S\), we need to parameterize \(S\). Similarly, the average value of a function of two variables over the rectangular Also, dont forget to plug in for \(z\). In this case the surface integral is. The "Checkanswer" feature has to solve the difficult task of determining whether two mathematical expressions are equivalent. Main site navigation. What if you are considering the surface of a curved airplane wing with variable density, and you want to find its total mass? 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. Thank you! The fact that the derivative is the zero vector indicates we are not actually looking at a curve. Since the surface is oriented outward and \(S_1\) is the bottom of the object, it makes sense that this vector points downward. The Surface Area Calculator uses a formula using the upper and lower limits of the function for the axis along which the arc revolves. Therefore, the mass flow rate is \(7200\pi \, \text{kg/sec/m}^2\). Enter the function you want to integrate into the Integral Calculator. Therefore we use the orientation, \(\vecs N = \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle \), \[\begin{align*} \iint_S \rho v \cdot \,dS &= 80 \int_0^{2\pi} \int_0^{\pi/2} v (r(\phi, \theta)) \cdot (t_{\phi} \times t_{\theta}) \, d\phi \, d\theta \\ Not strictly required, but useful for intuition and analogy: (This is analogous to how computing line integrals is basically the same as computing arc length integrals, except that you throw a function inside the integral itself. In addition to parameterizing surfaces given by equations or standard geometric shapes such as cones and spheres, we can also parameterize surfaces of revolution. Surface integral of vector field calculator 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_ydzdx+f_zdxdy Solve Now. &= 4 \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2\phi}. 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. \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. Flux = = S F n d . \[\begin{align*} \vecs t_x \times \vecs t_{\theta} &= \langle 2x^3 \cos^2 \theta + 2x^3 \sin^2 \theta, \, -x^2 \cos \theta, \, -x^2 \sin \theta \rangle \\[4pt] &= \langle 2x^3, \, -x^2 \cos \theta, \, -x^2 \sin \theta \rangle \end{align*}\], \[\begin{align*} \vecs t_x \times \vecs t_{\theta} &= \sqrt{4x^6 + x^4\cos^2 \theta + x^4 \sin^2 \theta} \\[4pt] &= \sqrt{4x^6 + x^4} \\[4pt] &= x^2 \sqrt{4x^2 + 1} \end{align*}\], \[\begin{align*} \int_0^b \int_0^{2\pi} x^2 \sqrt{4x^2 + 1} \, d\theta \,dx &= 2\pi \int_0^b x^2 \sqrt{4x^2 + 1} \,dx \\[4pt] \end{align*}\], Therefore, to compute a surface integral over a vector field we can use the equation, \[\iint_S \vecs F \cdot \vecs N\, dS = \iint_D (\vecs F (\vecs r (u,v)) \cdot (\vecs t_u \times \vecs t_v)) \,dA. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. 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. However, if we wish to integrate over a surface (a two-dimensional object) rather than a path (a one-dimensional object) in space, then we need a new kind of integral that can handle integration over objects in higher dimensions. Figure-1 Surface Area of Different Shapes. Dont forget that we need to plug in for \(x\), \(y\) and/or \(z\) in these as well, although in this case we just needed to plug in \(z\). We can drop the absolute value bars in the sine because sine is positive in the range of \(\varphi \) that we are working with. &= \dfrac{2560 \sqrt{6}}{9} \approx 696.74. To place this definition in a real-world setting, let \(S\) be an oriented surface with unit normal vector \(\vecs{N}\). To embed this widget in a post, install the Wolfram|Alpha Widget Shortcode Plugin and copy and paste the shortcode above into the HTML source. \nonumber \]. The tangent vectors are \(\vecs t_u = \langle 1,-1,1\rangle\) and \(\vecs t_v = \langle 0,2v,1\rangle\). Surfaces can sometimes be oriented, just as curves can be oriented. What if you have the temperature for every point on the curved surface of the earth, and you want to figure out the average temperature? The formula for calculating the length of a curve is given as: L = a b 1 + ( d y d x) 2 d x. 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 \(\PageIndex{2}\)). where \(D\) is the range of the parameters that trace out the surface \(S\). Surface integrals of vector fields. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces. \nonumber \]. Surface integrals are important for the same reasons that line integrals are important. You can accept it (then it's input into the calculator) or generate a new one. 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 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 \(2\pi\). Specifically, here's how to write a surface integral with respect to the parameter space: The main thing to focus on here, and what makes computations particularly labor intensive, is the way to express. This results in the desired circle (Figure \(\PageIndex{5}\)). \nonumber \]. &= 7200\pi.\end{align*} \nonumber \]. Let's take a closer look at each form . GLAPS Model: Sea Surface and Ground Temperature, http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx. Imagine what happens as \(u\) increases or decreases. Follow the steps of Example \(\PageIndex{15}\). 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". This page titled 16.6: Surface Integrals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin Jed Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. Introduction. 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. 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. Computing surface integrals can often be tedious, especially when the formula for the outward unit normal vector at each point of \(\) changes. https://mathworld.wolfram.com/SurfaceIntegral.html. When we've been given a surface that is not in parametric form there are in fact 6 possible integrals here. A surface integral of a vector field. It is the axis around which the curve revolves. The Integral Calculator will show you a graphical version of your input while you type. &= \int_0^{\sqrt{3}} \int_0^{2\pi} u \, dv \, du \\ In doing this, the Integral Calculator has to respect the order of operations. We need to be careful here. Since the surface is oriented outward and \(S_1\) is the top of the object, we instead take vector \(\vecs t_v \times \vecs t_u = \langle 0,0,v\rangle\). 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) &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54 (1 - \cos^2\phi) \, \sin \phi + 27 \cos^2\phi \, \sin \phi \, d\phi \, d\theta \\ Notice that the corresponding surface has no sharp corners. &= 2\pi \sqrt{3}. Stokes' theorem is the 3D version of Green's theorem. It relates the surface integral of the curl of a vector field with the line integral of that same vector field around the boundary of the surface: The image of this parameterization is simply point \((1,2)\), which is not a curve. 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 \(\Delta v \vecs t_v(P_{ij})\). They have many applications to physics and engineering, and they allow us to develop higher dimensional versions of the Fundamental Theorem of Calculus. Direct link to Andras Elrandsson's post I almost went crazy over , Posted 3 years ago. This surface has parameterization \(\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 1 \leq v \leq 4\). The result is displayed after putting all the values in the related formula. Let the lower limit in the case of revolution around the x-axis be a.