surface integral calculator

If piece \(S_{ij}\) is small enough, then the tangent plane at point \(P_{ij}\) is a good approximation of piece \(S_{ij}\). The "Checkanswer" feature has to solve the difficult task of determining whether two mathematical expressions are equivalent. Double integral calculator with steps help you evaluate integrals online. You can use this calculator by first entering the given function and then the variables you want to differentiate against. 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. Let's take a closer look at each form . Surface integral of a vector field over a surface. Surface Area Calculator Surface integral of a vector field over a surface - GeoGebra Recall that to calculate a scalar or vector line integral over curve \(C\), we first need to parameterize \(C\). ), If you understand double integrals, and you understand how to compute the surface area of a parametric surface, you basically already understand surface integrals. 2. &= -55 \int_0^{2\pi} du \\[4pt] Surface Area and Surface Integrals - Valparaiso University Physical Applications of Surface Integrals - math24.net &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \, d\phi \\ 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. We see that \(S_2\) is a circle of radius 1 centered at point \((0,0,4)\), sitting in plane \(z = 4\). Notice that the corresponding surface has no sharp corners. All common integration techniques and even special functions are supported. We assume here and throughout that the surface parameterization \(\vecs r(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle\) is continuously differentiablemeaning, each component function has continuous partial derivatives. Then the heat flow is a vector field proportional to the negative temperature gradient in the object. 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. MathJax takes care of displaying it in the browser. The little S S under the double integral sign represents the surface itself, and the term d\Sigma d represents a tiny bit of area piece of this surface. \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. This division of \(D\) into subrectangles gives a corresponding division of surface \(S\) into pieces \(S_{ij}\). &= \int_0^3 \pi \, dv = 3 \pi. 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. Now, because the surface is not in the form \(z = g\left( {x,y} \right)\) we cant use the formula above. \nonumber \]. We now show how to calculate the ux integral, beginning with two surfaces where n and dS are easy to calculate the cylinder and the sphere. The same was true for scalar surface integrals: we did not need to worry about an orientation of the surface of integration. Integral Calculator | The best Integration Calculator 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. So, lets do the integral. Use the standard parameterization of a cylinder and follow the previous example. x-axis. 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 \(\PageIndex{11}\)). If you don't specify the bounds, only the antiderivative will be computed. This is not an issue though, because Equation \ref{scalar surface integrals} does not place any restrictions on the shape of the parameter domain. \end{align*}\]. Taking a normal double integral is just taking a surface integral where your surface is some 2D area on the s-t plane. With the idea of orientable surfaces in place, we are now ready to define a surface integral of a vector field. Suppose that i ranges from 1 to m and j ranges from 1 to n so that \(D\) is subdivided into mn rectangles. The mass of a sheet is given by Equation \ref{mass}. Calculus III - Surface Integrals - Lamar University The surface in Figure \(\PageIndex{8a}\) can be parameterized by, \[\vecs r(u,v) = \langle (2 + \cos v) \cos u, \, (2 + \cos v) \sin u, \, \sin v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v < 2\pi \nonumber \], (we can use technology to verify). &= \langle 4 \, \cos \theta \, \sin^2 \phi, \, 4 \, \sin \theta \, \sin^2 \phi, \, 4 \, \cos \phi \, \sin \phi \rangle. 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. Note that \(\vecs t_u = \langle 1, 2u, 0 \rangle\) and \(\vecs t_v = \langle 0,0,1 \rangle\). Note that we can form a grid with lines that are parallel to the \(u\)-axis and the \(v\)-axis in the \(uv\)-plane. The arc length formula is derived from the methodology of approximating the length of a curve. \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 \\ 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 sense, surface integrals expand on our study of line integrals. &= 80 \int_0^{2\pi} \Big[-54 \, \cos \phi + 9 \, \cos^3 \phi \Big]_{\phi=0}^{\phi=2\pi} \, d\theta \\ Notice that if \(u\) is held constant, then the resulting curve is a circle of radius \(u\) in plane \(z = u\). 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. The rate of heat flow across surface S in the object is given by the flux integral, \[\iint_S \vecs F \cdot dS = \iint_S -k \vecs \nabla T \cdot dS. &= - 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] Since we are only taking the piece of the sphere on or above plane \(z = 1\), we have to restrict the domain of \(\phi\). Then I would highly appreciate your support. We'll first need the mass of this plate. A flat sheet of metal has the shape of surface \(z = 1 + x + 2y\) that lies above rectangle \(0 \leq x \leq 4\) and \(0 \leq y \leq 2\). Recall that scalar line integrals can be used to compute the mass of a wire given its density function. 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. Free online 3D grapher from GeoGebra: graph 3D functions, plot surfaces, construct solids and much more! At the center point of the long dimension, it appears that the area below the line is about twice that above. The fact that the derivative is the zero vector indicates we are not actually looking at a curve. Remember, I don't really care about calculating the area that's just an example. We have seen that a line integral is an integral over a path in a plane or in space. Use a surface integral to calculate the area of a given surface. Again, notice the similarities between this definition and the definition of a scalar line integral. Notice that this cylinder does not include the top and bottom circles. we can always use this form for these kinds of surfaces as well. 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. This makes a=23.7/2=11.85 and b=11.8/2=5.9, if it were symmetrical. 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. Similarly, points \(\vecs r(\pi, 2) = (-1,0,2)\) and \(\vecs r \left(\dfrac{\pi}{2}, 4\right) = (0,1,4)\) are on \(S\). The definition of a surface integral of a vector field proceeds in the same fashion, except now we chop surface \(S\) into small pieces, choose a point in the small (two-dimensional) piece, and calculate \(\vecs{F} \cdot \vecs{N}\) at the point. Not what you mean? Integrals can be a little daunting for students, but they are essential to calculus and understanding more advanced mathematics. Finally, the bottom of the cylinder (not shown here) is the disk of radius \(\sqrt 3 \) in the \(xy\)-plane and is denoted by \({S_3}\). 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. With the standard parameterization of a cylinder, Equation \ref{equation1} shows that the surface area is \(2 \pi rh\). Lets start off with a sketch of the surface \(S\) since the notation can get a little confusing once we get into it. The surface integral of the vector field over the oriented surface (or the flux of the vector field across the surface ) can be written in one of the following forms: Here is called the vector element of the surface. Let \(S\) be a smooth orientable surface with parameterization \(\vecs r(u,v)\). We used a rectangle here, but it doesnt have to be of course. In this video we come up formulas for surface integrals, which are when we accumulate the values of a scalar function over a surface. Let the upper limit in the case of revolution around the x-axis be b. button to get the required surface area value. Similarly, if \(S\) is a surface given by equation \(x = g(y,z)\) or equation \(y = h(x,z)\), then a parameterization of \(S\) is \(\vecs r(y,z) = \langle g(y,z), \, y,z\rangle\) or \(\vecs r(x,z) = \langle x,h(x,z), z\rangle\), respectively. The Divergence Theorem A specialty in mathematical expressions is that the multiplication sign can be left out sometimes, for example we write "5x" instead of "5*x". Therefore, \(\vecs r_u \times \vecs r_v\) is not zero for any choice of \(u\) and \(v\) in the parameter domain, and the parameterization is smooth. In "Options", you can set the variable of integration and the integration bounds. Solutions Graphing Practice; New Geometry; Calculators; Notebook . How do you add up infinitely many infinitely small quantities associated with points on a surface? An extremely well-written book for students taking Calculus for the first time as well as those who need a refresher. Surface integrals are a generalization of line integrals. This idea of adding up values over a continuous two-dimensional region can be useful for curved surfaces as well. Describe surface \(S\) parameterized by \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u^2 \rangle, \, 0 \leq u < \infty, \, 0 \leq v < 2\pi\). The general surface integrals allow you to map a rectangle on the s-t plane to some other crazy 2D shape (like a torus or sphere) and take the integral across that thing too! Area of an ellipse Calculator - High accuracy calculation \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] The difference between this problem and the previous one is the limits on the parameters. Let \(\vecs r(u,v)\) be a parameterization of \(S\) with parameter domain \(D\). 4.4: Surface Integrals and the Divergence Theorem We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. is given explicitly by, If the surface is surface parameterized using Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface. \nonumber \]. 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. &= -110\pi. The vendor states an area of 200 sq cm. Here is the evaluation for the double integral. In case the revolution is along the x-axis, the formula will be: \[ S = \int_{a}^{b} 2 \pi y \sqrt{1 + (\dfrac{dy}{dx})^2} \, dx \]. Arc Length Calculator - Symbolab This surface has parameterization \(\vecs r(u,v) = \langle v \, \cos u, \, v \, \sin u, \, 4 \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq 1.\). Informally, a curve parameterization is smooth if the resulting curve has no sharp corners. Describe the surface integral of a scalar-valued function over a parametric surface. Well, the steps are really quite easy. It's just a matter of smooshing the two intuitions together. Solution First we calculate the outward normal field on S. This can be calulated by finding the gradient of g ( x, y, z) = y 2 + z 2 and dividing by its magnitude. \[\begin{align*} \vecs t_x \times \vecs t_{\theta} &= \langle 2x^3 \cos^2 \theta + 2x^3 \sin^2 \theta, \, -x^2 \cos \theta, \, -x^2 \sin \theta \rangle \\[4pt] &= \langle 2x^3, \, -x^2 \cos \theta, \, -x^2 \sin \theta \rangle \end{align*}\], \[\begin{align*} \vecs t_x \times \vecs t_{\theta} &= \sqrt{4x^6 + x^4\cos^2 \theta + x^4 \sin^2 \theta} \\[4pt] &= \sqrt{4x^6 + x^4} \\[4pt] &= x^2 \sqrt{4x^2 + 1} \end{align*}\], \[\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] Calculate the mass flux of the fluid across \(S\). Vector representation of a surface integral - Khan Academy Therefore, a parameterization of this cone is, \[\vecs s(u,v) = \langle kv \, \cos u, \, kv \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq h. \nonumber \]. 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. The flux of a vector field F F across a surface S S is the surface integral Flux = =SF nd. A surface integral of a vector field. The second step is to define the surface area of a parametric surface. &= \iint_D (\vecs F(\vecs r(u,v)) \cdot (\vecs t_u \times \vecs t_v))\,dA. . Stokes' theorem is the 3D version of Green's theorem. 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. The parameterization of the cylinder and \(\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|\) is. Set integration variable and bounds in "Options". How to calculate the surface integral of a vector field Here are the ranges for \(y\) and \(z\). You can use this calculator by first entering the given function and then the variables you want to differentiate against. The surface area of a right circular cone with radius \(r\) and height \(h\) is usually given as \(\pi r^2 + \pi r \sqrt{h^2 + r^2}\). The \(\mathbf{\hat{k}}\) component of this vector is zero only if \(v = 0\) or \(v = \pi\). The Divergence Theorem can be also written in coordinate form as. \label{surfaceI} \]. Thus, a surface integral is similar to a line integral but in one higher dimension. 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. surface integral - Wolfram|Alpha Solve Now. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Step 2: Compute the area of each piece. Double Integral calculator with Steps & Solver It can be also used to calculate the volume under the surface. ; 6.6.3 Use a surface integral to calculate the area of a given surface. 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 \(2\pi\). The surface area of \(S\) is, \[\iint_D ||\vecs t_u \times \vecs t_v || \,dA, \label{equation1} \], where \(\vecs t_u = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle\), \[\vecs t_v = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle. \label{mass} \]. Calculate the area of a surface of revolution step by step The calculations and the answer for the integral can be seen here. This is not the case with surfaces, however. What Is a Surface Area Calculator in Calculus? This is called the positive orientation of the closed surface (Figure \(\PageIndex{18}\)). d S, where F = z, x, y F = z, x, y and S is the surface as shown in the following figure. How does one calculate the surface integral of a vector field on a surface? Notice that we do not need to vary over the entire domain of \(y\) because \(x\) and \(z\) are squared. Now that we are able to parameterize surfaces and calculate their surface areas, we are ready to define surface integrals. The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. \end{align*}\], \[\iint_S z^2 \,dS = \iint_{S_1}z^2 \,dS + \iint_{S_2}z^2 \,dS, \nonumber \], \[\iint_S z^2 \,dS = (2\pi - 4) \sqrt{3} + \dfrac{32\pi}{3}. then This division of \(D\) into subrectangles gives a corresponding division of \(S\) into pieces \(S_{ij}\). Free Arc Length calculator - Find the arc length of functions between intervals step-by-step. 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. &= \iint_D \left(\vecs F (\vecs r (u,v)) \cdot \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \right) || \vecs t_u \times \vecs t_v || \,dA \\[4pt] For a vector function over a surface, 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, \[\begin{align*} \vecs t_{\phi} \times \vecs t_{\theta} &= \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \nonumber \\ 2 \cos \theta \cos \phi & 2 \sin \theta \cos \phi & -2\sin \phi \\ -2\sin \theta\sin\phi & 2\cos \theta \sin\phi & 0 \end{vmatrix} \\[4 pt] Step #5: Click on "CALCULATE" button. \end{align*}\], Therefore, the rate of heat flow across \(S\) is, \[\dfrac{55\pi}{2} - \dfrac{55\pi}{2} - 110\pi = -110\pi. In the first grid line, the horizontal component is held constant, yielding a vertical line through \((u_i, v_j)\). How can we calculate the amount of a vector field that flows through common surfaces, such as the . Without loss of generality, we assume that \(P_{ij}\) is located at the corner of two grid curves, as in Figure \(\PageIndex{9}\). If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. The way to tell them apart is by looking at the differentials. Gauss's Law Calculator - Calculate the Electric Flux 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 \(S\). \nonumber \], Therefore, the radius of the disk is \(\sqrt{3}\) and a parameterization of \(S_1\) is \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, 1 \rangle, \, 0 \leq u \leq \sqrt{3}, \, 0 \leq v \leq 2\pi\). 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 \(\vecs v(x,y,z) = \langle 2x, \, 2y, \, z\rangle\) represent a velocity field (with units of meters per second) of a fluid with constant density 80 kg/m3. The corresponding grid curves are \(\vecs r(u_i, v)\) and \((u, v_j)\) and these curves intersect at point \(P_{ij}\). We can start with the surface integral of a scalar-valued function. It follows from Example \(\PageIndex{1}\) that we can parameterize all cylinders of the form \(x^2 + y^2 = R^2\). In particular, surface integrals allow us to generalize Greens theorem to higher dimensions, and they appear in some important theorems we discuss in later sections. Assume for the sake of simplicity that \(D\) is a rectangle (although the following material can be extended to handle nonrectangular parameter domains). Also note that, for this surface, \(D\) is the disk of radius \(\sqrt 3 \) centered at the origin. Use parentheses! Math Assignments. 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. Double Integral Calculator An online double integral calculator with steps free helps you to solve the problems of two-dimensional integration with two-variable functions. Dont forget that we need to plug in for \(z\)! Verify result using Divergence Theorem and calculating associated volume integral. tothebook. If we only care about a piece of the graph of \(f\) - say, the piece of the graph over rectangle \([ 1,3] \times [2,5]\) - then we can restrict the parameter domain to give this piece of the surface: \[\vecs r(x,y) = \langle x,y,x^2y \rangle, \, 1 \leq x \leq 3, \, 2 \leq y \leq 5. Divide rectangle \(D\) into subrectangles \(D_{ij}\) with horizontal width \(\Delta u\) and vertical length \(\Delta v\). First, we calculate \(\displaystyle \iint_{S_1} z^2 \,dS.\) To calculate this integral we need a parameterization of \(S_1\). Take the dot product of the force and the tangent vector. &= 5 \int_0^2 \int_0^u \sqrt{1 + 4u^2} \, dv \, du = 5 \int_0^2 u \sqrt{1 + 4u^2}\, du \\