Surface Integral – General Form, Techniques, and Examples
The surface integral allows us to generalize line integrals to account for surfaces in three dimensions. Surface integrals are important when dealing with quantities in either of the three media: solid, liquid, and gas. These integrals are also significant when working with vector fields such as gravitational fields or flux.The surface integral allows us to extend our understanding of double integrals – by extending the concept to account for surfaces in three or even higher dimensions. Through surface integrals, we can extend Green’s theorem in higher dimensions as well.In this article, we’ll show you what curls represent in the physical world and how we can apply the formulas to calculate the curl of a vector field. We’ll also provide examples and problems for you to work on and better understand this important measure.
What Is a Surface Integral?
The surface integral represents the generalization of integrals evaluated over surfaces. A great way to understand surface integrals is to know that the process of evaluating is similar to evaluating double integrals. This time, however, we’re adding up the points of in $\mathbb{R}^3$. Similar to double integrals, surface integrals have two kinds: 1) those of the scalar-valued functions and 2) surface integrals for vector fields.\begin{aligned}\int \int_{S} f(x, y, z) \phantom{x}dS &= \int \int_{T} f(\textbf{v}(t, s)) \left|\dfrac{\partial \textbf{v}}{\partial t} \times \dfrac{\partial \textbf{v}}{\partial s}\right| \phantom{x}dtds\end{aligned}Let’s slowly understand how these expressions were derived- beginning with seeing the surface, $S$, in space as a finite surface area. Suppose that we have a scalar-valued function, $f$, is defined on the surface, $S$, and we can divide the surface into smaller portions.The image above shows a divided surface, $S$, into smaller portions: $\{S_1, S_2, S_3, …, S_n\}$ with each portion having an area of $\{\Delta S_1, \Delta S_2, \Delta S_3, …, \Delta S_n\}$, respectively. Keep in mind that for this example, $(x_k, y_k, z_k) \in S_k$, are selected randomly. Similar to how we’ve established definite and line integrals in the past, adding these infinitesimal portions’ area will lead us to the actual value of the surface integral. By evaluating the limit of $\Delta S_k$’s maximum value as it approaches zero, we’ll have the expression for the surface integral: $\int \int_{S} f(x,y, z) \phantom{x}dS$.
Surface Integral Formula
There are different ways to interpret surface integrals and their formulas, so let’s begin by breaking down how the formula for the surface integral for parametric surfaces. In the past, we’ve learned that we can write the smooth curve, $C$, can be parametrized by the vector expression, $\textbf{r}(t)$.\begin{aligned}\textbf{r}(t) &= x(t) \textbf{i} + y(t) \textbf{j} + z(t) \textbf{k}, a\leq t \leq b\end{aligned}This means that the arc length of $C$ can be expressed as $\int_{a}^{b} |\textbf{r}^{\prime}(t)| \phantom{x}dt$. We’ve established before that the integral of $C$ is given by the equation shown below.\begin{aligned}\int_{C} f \phantom{x}ds &= \int_{a}^{b} f(x(t), y(t), z(t)) |\textbf{r}^{\prime} (t)| \phantom{x}dt\\&= \int_{a}^{b} f(\textbf{r}(t)) |\textbf{r}^{\prime} (t)| \phantom{x}dt\end{aligned}Now, what happens when $f(x, y, z) = 1$? We can rewrite the surface integral as $\int \int_{D} |\textbf{r}_u \times \textbf{r}_v|\phantom{x} dA= A(S)$. From this, we can establish the formula for the surface integral of parametrized surfaces:
SURFACE INTEGRAL FOR A PAREMETRIZED SURFACESuppose that our surface is defined the vector function, $\textbf{r}(u, v) = x(u, v)\textbf{i} + y(u, v)\textbf{j} + z(u, v)\textbf{k}$. We can calculate the surface integral using the equation below.\begin{aligned}\int \int_{S} f(x,y ,z) \phantom{x}dS &= \int \int_{D} f(\textbf{r}(u, v)) |\textbf{r}_u \times \textbf{r}_v| \phantom{x} dA\end{aligned}Keep in mind that $\textbf{r}_u = \dfrac{\partial x}{\partial u}\textbf{i} + \dfrac{\partial y}{\partial u}\textbf{j} +\dfrac{\partial z}{\partial u}\textbf{k}$ and $\textbf{r}_v = \dfrac{\partial x}{\partial v}\textbf{i} + \dfrac{\partial y}{\partial v}\textbf{j} +\dfrac{\partial z}{\partial v}\textbf{k}$.
Before we move on to the next section, let’s see how we can use the formula shown above to find the surface integral of surfaces defined by $z = g(x, y)$.\begin{aligned}\textbf{r}(x, y) &= x\textbf{i} + y\textbf{j} + g(x,y) \textbf{k}\end{aligned}This means that $\textbf{r}_x \times \textbf{r}_y$ is equal to $-\dfrac{\partial g}{\partial x}\textbf{i} – \dfrac{\partial g}{\partial y}\textbf{j} + 1 \textbf{k}$.\begin{aligned}|\textbf{r}_x \times \textbf{r}_y| &= \sqrt{\left(\dfrac{\partial g}{\partial x} \right)^2 + \left(\dfrac{\partial g}{\partial y} \right)^2 + 1} \phantom{x}dA\end{aligned}This means that the surface integral of surfaces such as $z = g(x, y)$ can be expressed as shown below.\begin{aligned}\int \int_{S} f(x,y,z) \phantom{x}dS &= \int \int_{D} f(x, y, g(x, y)) \sqrt{\left(\dfrac{\partial g}{\partial x} \right)^2 + \left(\dfrac{\partial g}{\partial y} \right)^2 + 1} \phantom{x}dA \end{aligned}We’ll apply similar approaches when evaluating surfaces of the forms, $z = g(y, z)$ and $z = g(x, z)$.
We can evaluate surface integrals by identifying the equation representing the plane then applying the appropriate equations.
Whenever needed, divide the surface integrals into smaller chunks of surfaces.
Find the cross product of the two parametrized functions (or vectors) then use the result to set the equation of the plane.
Write down the expression for $dS$ by taking the magnitude of the cross products’ partial derivatives.
Set the appropriate limits of integration then evaluate the resulting expression.
These are general guidelines to remember when evaluating surface integral, but different cases may require a more creative approach. This is why it’s helpful to try out different examples when learning about surface integrals.Let’s try to evaluate the surface integral, $\int \int_{S} xy \phantom{x}dS$, where $S$ is a triangular region that has the following vertices: $(2, 0, 0)$, $(0, 4, 0)$, and $(0, 0, 4)$.We still don’t have the equation of the plane, so let’s use the fact that $\textbf{u} = 4\textbf{i} – 2 \textbf{j}$ and $\textbf{v} =4\textbf{i} – 4 \textbf{k}$, so the plane we’re integrating over has the following equation.\begin{aligned}8(x – 4) + 16y + 8z &= 0\\z &= 4 – x – 2y \end{aligned}Graph the region of $D$ to guide us in finding the limits of integration.From the sketch of our triangular region, we can see that we’ll be integrating over the following limits of integration: $0 \leq x \leq 4$ and $0 \leq y \leq -\dfrac{1}{2}x + 2$. Before we apply the formula for the surface integral, let’s write down the rest of the components that we’ll need:\begin{aligned}f(x, y, g(x, y)) &= xy\\ \dfrac{\partial g}{\partial x} &= -1\\\dfrac{\partial g}{\partial y} &= -2\end{aligned}Rewrite $\int \int_{S} f(x, y, z) \phantom{x}dS$ as a double integral: $\int \int_{D} f(x, y, g(x, y)) \sqrt{\left(\dfrac{\partial g}{\partial x}\right)^2 + \left(\dfrac{\partial g}{\partial y}\right)^2 + 1} \phantom{x}dA$. Evaluate the resulting double integral.\begin{aligned} \int \int_{S} f(x, y, z) \phantom{x}dS &= \int \int_{D} f(x, y, g(x, y)) \sqrt{\left(\dfrac{\partial g}{\partial x}\right)^2 + \left(\dfrac{\partial g}{\partial y}\right)^2 + 1} \phantom{x}dA\\&= \int_{0}^{4} \int_{0}^{-x/2 + 2} xy \sqrt{(-1)^2 + (-2)^2 + 1}dydx\\&= \sqrt{6} \int_{0}^{4} \int_{0}^{-\frac{x}{2} + 2} xy \phantom{x}dydx\\&=\dfrac{\sqrt{6}}{2}\int_{0}^{4} x\left[y^2 \right ]_{0}^{-\frac{x}{2} + 2} dx\\&=\dfrac{\sqrt{6}}{2} \int_{0}^{4} x\left(-\dfrac{x}{2} + 2 \right )^2 \phantom{x}dx\\&=\dfrac{\sqrt{6}}{2} \int_{0}^{4} \dfrac{x^3}{4} -2x^2 + 4x\phantom{x}dx\\&=\dfrac{\sqrt{6}}{2} \left[\dfrac{x^4}{16} – \dfrac{2x^3}{3} + 2x^2\right ]_{0}^{4}\\&= \dfrac{\sqrt{6}}{2}\left(\dfrac{16}{3} \right )\\&= \dfrac{8\sqrt{6}}{3} \end{aligned}Hence, we have $\int \int_{S} xy \phantom{x}dS = \dfrac{8\sqrt{6}}{3}$.
How To Do Surface Integrals by Converting The Coordinates?
There are instances when it’s easier to evaluate surface integrals when we rewrite the expression in Cartesian coordinate to a different form. Here are some common coordinates we might use to simplify the expression of the surface integral:
Sphere with a radius of $\boldsymbol{r}$:\begin{aligned}x^2 + y^2 &+ z^2 = r^2\\0 \leq &\theta \leq 2\pi\\0 \leq &\phi \leq \pi \end{aligned}
\begin{aligned}x &= r \sin \phi\cos \theta\\y&= r \sin \phi\sin \theta\\z&= r \cos \phi\\\\\textbf{r}(\theta, \phi) &= < r \sin \phi\cos \theta,r \sin \phi\sin \theta,r \cos \phi>\end{aligned}
Cylinder with a radius of $\boldsymbol{r}$:\begin{aligned}x^2 &+ y^2 = r^2\\0 \leq &\theta \leq 2\pi\\0 \leq &z \leq b \end{aligned}
\begin{aligned}x &= r \cos \theta\\y &= r \sin \theta\\z &= z\\\\\textbf{r}(z , \theta) &= <r \sin \theta,r \cos \theta,z>\end{aligned}
Let’s say we want to evaluate the surface integral, $\int \int_{S} y \phantom{x}dS$, where $S$ is half a sphere with a radius of $4$ units. We can parametrize the equation of the hemisphere using our parametrization for spheres.\begin{aligned}\textbf{r}(\theta, \phi) = < 4\sin \phi\cos \theta,4 \sin \phi\sin \theta,4 \cos \phi>\end{aligned}Since we’re working on the upper hemisphere, our limits of integration will be limited to:\begin{aligned} 0 \leq &\theta \leq 2\pi\\0 \leq &\phi \leq \dfrac{\pi}{2} \end{aligned}Take the partial derivatives of $\textbf{r}$ with respect to $\theta$ and $\phi$ – we’ve summarized the calculation for you here:\begin{aligned}\textbf{r}_{\theta}&= \dfrac{\partial}{\partial \theta} < 4 \sin \phi\cos \theta,4 \sin \phi\sin \theta,4 \cos \phi>\\&=<-4 \sin \phi \sin \theta,4 \sin \phi \cos \theta, 0>\\\\\textbf{r}_{\phi}&= \dfrac{\partial}{\partial \phi} <4 \sin \phi\cos \theta,4 \sin \phi\sin \theta,4 \cos \phi>\\&=<4 \cos \phi\cos\theta,4 \cos \phi\sin \theta, -4 \sin \phi> \end{aligned}Now that we have the expressions for $\textbf{r}_{\theta}$ and $\textbf{r}_{\phi}$, we can now find their cross product.\begin{aligned}\textbf{r}_{\theta} \times \textbf{r}_{\phi} &= \begin{vmatrix}\textbf{i} & \textbf{j}& \textbf{k}\\ -4 \sin \phi \sin \theta& 4 \sin \phi \cos \theta & 0\\ 4 \cos \phi \sin \theta& 4 \cos \phi\sin \theta& -4 \sin \phi\end{vmatrix}\\&= (-16 \sin^2 \phi \cos\theta -0)\textbf{i} – (0 + 16\sin^2 \phi\sin \theta)\textbf{j}+ (-16\sin \phi\cos \phi\sin^2\theta -16\sin \phi\cos \phi\cos^2\theta)\textbf{k}\\&= -16 \sin^2 \phi \cos\theta\textbf{i}- 16\sin^2 \phi\sin \theta\textbf{j}-16\sin \phi\cos \phi(\sin^2 \theta +\cos^2 \theta)\textbf{k}\\&= -16 \sin^2 \phi \cos\theta\textbf{i}- 16\sin^2 \phi\sin \theta\textbf{j}-16\sin \phi\cos \phi\textbf{k} \end{aligned}This means that the magnitude of the two vectors’ cross product, $|\textbf{r}_{\theta} \times \textbf{r}_{\phi}|$, is as shown below.\begin{aligned}|\textbf{r}_{\theta} \times \textbf{r}_{\phi}| &=\sqrt{(-16 \sin^2 \phi \cos\theta)^2+(- 16\sin^2 \phi\sin \theta)^2+ (-16\sin \phi\cos \phi)^2}\\&= \sqrt{16^2\sin^4 \phi \cos^2 \theta + 16^2\sin^4 \phi\sin^2\theta+ 16^2\sin^2 \phi\cos^2\phi} \\&= \sqrt{256\sin^4 \phi(\sin^2 \theta + \cos^2 \theta) +256\sin^2 \phi\cos^2\phi } \\&= \sqrt{256\sin^4 \phi(1) +256\sin^2 \phi\cos^2\phi }\\&= \sqrt{256\sin^2 \phi(\sin^2 \phi + \cos^2 \phi) }\\&= 16|\sin \phi|\\&= 16 \sin \phi \end{aligned}We can now rewrite the surface integral, $\int \int_{S} \phantom{x}dS$, as a double integral over the region, $D$, where the limits of integrations are: $0\leq \theta 2\pi$ and $0 \leq \dfrac{\phi}{2}$ and $dS = |\textbf{r}_{\theta} \times \textbf{r}_{\phi}| dA$.\begin{aligned}\int \int_{S} y\phantom{x}dS&= \int \int_D (4 \sin \phi\sin \theta)(16 \sin \phi) \phantom{x}dA \\&=\int_{0}^{2\pi} \int_{0}^{\pi/2}(4 \sin \phi\sin \theta)(16 \sin \phi) \phantom{x} d\phi d\theta\\&= 64\int_{0}^{2\pi} \int_{0}^{\pi/2}\sin^2 \phi\sin \theta \phantom{x} d\phi d\theta\\&=64\int_{0}^{2\pi} \left[\int_{0}^{\pi/2}\sin^2 \phi\phantom{x} d\phi \right ]\sin \theta d\theta \\&= 64\int_{0}^{2\pi} \left[\dfrac{1}{2}\phi- \dfrac{1}{4}\sin 2\phi\right ]_{0}^{\pi/2}\sin \theta d\theta\\&= 64 \cdot \dfrac{\pi}{4}[-\cos \theta]_{0}^{2\pi}\\&= 16\pi(0)\\&= 0 \end{aligned}This means that the surface integral is equal to zero. We’ve prepared more problems for you to work on – apply a similar process when working with the surface integrals shown below!Example 1Evaluate the surface integral, $\int \int_{S} x \phantom{x}dS$, where $S$ represents the surface, $y = x + z^2$, and bounded by the following on the $x$ and $z$ axes: $0 \leq x \leq 1$ and $0 \leq z \leq 2$.SolutionWe’re integrating over the surface, $y = x+ z^2$, with the graph of the surface as shown below.The equation shown below will guide us in evaluating the surface integral:\begin{aligned}\int \int_{S} f(x,y,z) \phantom{x}dS &= \int \int_{D} f(x, g(x, z), z) \sqrt{\left(\dfrac{\partial g}{\partial x} \right)^2 + \left(\dfrac{\partial g}{\partial z} \right)^2 + 1} \phantom{x}dA \end{aligned}Lucky for us, we now have the equation of the plane, so we can go ahead and find their partial derivatives of $y$ with respect to $x$ and $z$.\begin{aligned}f(x, g(x, z), z) &= x + z^2\\ \dfrac{\partial g}{\partial x} &= 1\\\dfrac{\partial f}{\partial z} &= 2z\end{aligned}We can now rewrite our surface integral to a double integral where $dS = \sqrt{\left(\dfrac{\partial g}{\partial z}\right)^2 + \left(\dfrac{\partial g}{\partial x}\right)^2 + 1} dA$. Evaluate the resulting double integral by applying the fundamental integral properties you’ve learned in the past.\begin{aligned}\int \int_{S} z \phantom{x}dS &= \int \int_{D} z \sqrt{\left(\dfrac{\partial g}{\partial z}\right)^2 + \left( \dfrac{\partial g}{\partial x}\right)^2 + 1} \phantom{x}dA\\&= \int_{0}^{2} \int_{0}^{1} z \sqrt{\left(2z\right)^2 + \left(1\right)^2 + 1} \phantom{x}dxdz\\&= \int_{0}^{2} \int_{0}^{1} z \sqrt{4z^2 + 2} \phantom{x}dxdz\\&= \int_{0}^{2} \int_{0}^{1} \sqrt{2}z \sqrt{1 + 2z^2} \phantom{x}dxdz\\&= \sqrt{2}\int_{0}^{2} z \sqrt{1 + 2z^2}[x]_{0}^{1} \phantom{x}dz\\&= \sqrt{2}\int_{0}^{2} z \sqrt{1 + 2z^2} \phantom{x}dz\\&= \sqrt{2}\left[\dfrac{1}{6}\left(1+2z^2\right)^{\frac{3}{2}} \right ]_{0}^{2}\\&= \sqrt{2} \left(\dfrac{13}{3} \right )\\&= \dfrac{13\sqrt{2}}{3} \end{aligned}This means that the surface integral, $\int \int_{S} x \phantom{x}dS$, is equal to $\dfrac{13\sqrt{2}}{3}$.Example 2Evaluate the surface integral, $\int \int_{S} y \phantom{x}dS$, where $S$ a part of the cylinder, $x^2 + y^2 =4$, and bounded by $z= 0$ and $z =4$.SolutionWe’ve learned that we can parameterize cylinders in $\mathbb{R}^3$ using the following parametrization:\begin{aligned}x &= r \cos \theta\\y &= r \sin \theta\\z &= z\\\\\textbf{r}(z , \theta) &= <r \sin \theta,r \cos \theta,z>\end{aligned}This means that the parametrization of our cylinder, $x^2 +y^2 = 4$, is equal to $<2 \cos \theta, 2\sin \theta, z>$.Keep in mind that the surface is bounded by $z$ from $0$ to $4$ and by $\theta$ from $0$ to $2\pi$. Now that we have $\textbf{r} = <2 \cos \theta, 2\sin \theta, z>$, let’s find the expression for $\textbf{r}_{z}$ and $\textbf{r}_{\theta}$.\begin{aligned}\textbf{r}_{z}&= \dfrac{\partial}{\partial z} <2 \cos \theta, 2\sin \theta, z>\\&=<0, 0, 2>\\\\\textbf{r}_{\theta}&= \dfrac{\partial}{\partial \theta} <2 \cos \theta, 2\sin \theta, z>\\&=<-2\sin \theta, 2\cos \theta, 0> \end{aligned}Take their cross products and calculate the resulting vector’s magnitude to find $|\textbf{r}_{z} \times \textbf{r}_{\theta}|$.\begin{aligned} \textbf{r}_{z} \times \textbf{r}_{\theta} &= \begin{vmatrix}\textbf{i} &\textbf{j}&\phantom{xxx}\textbf{k} \\ 0 &0 &\phantom{xxx}1 \\-2 \sin \theta &\phantom{xx}2 \cos \theta &\phantom{xxx}0 \end{vmatrix}\\&= (0 – 2\cos \theta)\textbf{i}-(0 – 2\sin \theta)\textbf{j} + 0\textbf{k}\\&= -2\cos \theta \textbf{i}+ 2\sin \theta \textbf{j}\\\\|\textbf{r}_{z} \times \textbf{r}_{\theta}| &= \sqrt{(-2 \cos\theta)^2 + (2 \sin \theta)^2}\\&= \sqrt{4 \cos^2\theta + 4\sin^2 \theta}\\&= \sqrt{4(\sin^2 \theta + \cos^2 \theta)}\\&= 2 \end{aligned}Now that we have the expression for $|\textbf{r}_{z} \times \textbf{r}_{\theta}|$, we can rewrite the surface integral as a double integral with $dS= $|\textbf{r}_{z} \times \textbf{r}_{\theta}| \phantom{x}dA$.\begin{aligned}\int \int_{S} y\phantom{x}dS &= \int \int_{D}(2\sin \theta)(2) \phantom{x}dA\\&= \int_{0}^{2\pi}\int_{0}^{4} 4\sin \theta \phantom{x}dzd\theta\\&= 4\int_{0}^{2\pi}\sin \theta\left[\int_{0}^{4}1 \phantom{x}dz \right ]\phantom{x}d\theta\\&= 4\int_{0}^{2\pi} 4\sin \theta \phantom{x}d\theta\\&= 16[-\cos \theta]_{0}^{2\pi}\\&= 16(0)\\&= 0 \end{aligned}This means that the surface integral, $\int \int_{S} y \phantom{x}dS$, is equal to zero.
Practice Questions
1. Evaluate the surface integral, $\int \int_{S} y \phantom{x}dS$, where $S$ represents the surface, $x = z + y^2$, and bounded by the following on the $z$ and $y$ axes: $0 \leq z \leq 1$ and $0 \leq y \leq 4$.
2. Evaluate the surface integral, $\int \int_{S} x \phantom{x}dS$, where $S$ a part of the cylinder, $x^2 + y^2 =4$, and bounded by $z= 0$ and $z = 6$.
Answer Key
1. $\int \int_{S} y \phantom{x} dS = \int_{0}^{1} \int_{0}^{4} y\sqrt{2 + 4y^2} \phantom{x}dydz = \dfrac{33\sqrt{66} – \sqrt{2}}{6}$
2. $\int \int_{S} x \phantom{x} dS = \int_{0}^{2\pi} \int_{0}^{8} 16\cos \theta \phantom{x}dz d\theta = 0$Images/mathematical drawings are created with GeoGebra.
The image above shows a divided surface, $S$, into smaller portions: $\{S_1, S_2, S_3, …, S_n\}$ with each portion having an area of $\{\Delta S_1, \Delta S_2, \Delta S_3, …, \Delta S_n\}$, respectively. Keep in mind that for this example, $(x_k, y_k, z_k) \in S_k$, are selected randomly. Similar to how we’ve established definite and line integrals in the past, adding these infinitesimal portions’ area will lead us to the actual value of the surface integral. By evaluating the limit of $\Delta S_k$’s maximum value as it approaches zero, we’ll have the expression for the surface integral: $\int \int_{S} f(x,y, z) \phantom{x}dS$.
We still don’t have the equation of the plane, so let’s use the fact that $\textbf{u} = 4\textbf{i} – 2 \textbf{j}$ and $\textbf{v} =4\textbf{i} – 4 \textbf{k}$, so the plane we’re integrating over has the following equation.\begin{aligned}8(x – 4) + 16y + 8z &= 0\\z &= 4 – x – 2y \end{aligned}Graph the region of $D$ to guide us in finding the limits of integration.
From the sketch of our triangular region, we can see that we’ll be integrating over the following limits of integration: $0 \leq x \leq 4$ and $0 \leq y \leq -\dfrac{1}{2}x + 2$. Before we apply the formula for the surface integral, let’s write down the rest of the components that we’ll need:\begin{aligned}f(x, y, g(x, y)) &= xy\\ \dfrac{\partial g}{\partial x} &= -1\\\dfrac{\partial g}{\partial y} &= -2\end{aligned}Rewrite $\int \int_{S} f(x, y, z) \phantom{x}dS$ as a double integral: $\int \int_{D} f(x, y, g(x, y)) \sqrt{\left(\dfrac{\partial g}{\partial x}\right)^2 + \left(\dfrac{\partial g}{\partial y}\right)^2 + 1} \phantom{x}dA$. Evaluate the resulting double integral.\begin{aligned} \int \int_{S} f(x, y, z) \phantom{x}dS &= \int \int_{D} f(x, y, g(x, y)) \sqrt{\left(\dfrac{\partial g}{\partial x}\right)^2 + \left(\dfrac{\partial g}{\partial y}\right)^2 + 1} \phantom{x}dA\\&= \int_{0}^{4} \int_{0}^{-x/2 + 2} xy \sqrt{(-1)^2 + (-2)^2 + 1}dydx\\&= \sqrt{6} \int_{0}^{4} \int_{0}^{-\frac{x}{2} + 2} xy \phantom{x}dydx\\&=\dfrac{\sqrt{6}}{2}\int_{0}^{4} x\left[y^2 \right ]_{0}^{-\frac{x}{2} + 2} dx\\&=\dfrac{\sqrt{6}}{2} \int_{0}^{4} x\left(-\dfrac{x}{2} + 2 \right )^2 \phantom{x}dx\\&=\dfrac{\sqrt{6}}{2} \int_{0}^{4} \dfrac{x^3}{4} -2x^2 + 4x\phantom{x}dx\\&=\dfrac{\sqrt{6}}{2} \left[\dfrac{x^4}{16} – \dfrac{2x^3}{3} + 2x^2\right ]_{0}^{4}\\&= \dfrac{\sqrt{6}}{2}\left(\dfrac{16}{3} \right )\\&= \dfrac{8\sqrt{6}}{3} \end{aligned}Hence, we have $\int \int_{S} xy \phantom{x}dS = \dfrac{8\sqrt{6}}{3}$.
The equation shown below will guide us in evaluating the surface integral:\begin{aligned}\int \int_{S} f(x,y,z) \phantom{x}dS &= \int \int_{D} f(x, g(x, z), z) \sqrt{\left(\dfrac{\partial g}{\partial x} \right)^2 + \left(\dfrac{\partial g}{\partial z} \right)^2 + 1} \phantom{x}dA \end{aligned}Lucky for us, we now have the equation of the plane, so we can go ahead and find their partial derivatives of $y$ with respect to $x$ and $z$.\begin{aligned}f(x, g(x, z), z) &= x + z^2\\ \dfrac{\partial g}{\partial x} &= 1\\\dfrac{\partial f}{\partial z} &= 2z\end{aligned}We can now rewrite our surface integral to a double integral where $dS = \sqrt{\left(\dfrac{\partial g}{\partial z}\right)^2 + \left(\dfrac{\partial g}{\partial x}\right)^2 + 1} dA$. Evaluate the resulting double integral by applying the fundamental integral properties you’ve learned in the past.\begin{aligned}\int \int_{S} z \phantom{x}dS &= \int \int_{D} z \sqrt{\left(\dfrac{\partial g}{\partial z}\right)^2 + \left( \dfrac{\partial g}{\partial x}\right)^2 + 1} \phantom{x}dA\\&= \int_{0}^{2} \int_{0}^{1} z \sqrt{\left(2z\right)^2 + \left(1\right)^2 + 1} \phantom{x}dxdz\\&= \int_{0}^{2} \int_{0}^{1} z \sqrt{4z^2 + 2} \phantom{x}dxdz\\&= \int_{0}^{2} \int_{0}^{1} \sqrt{2}z \sqrt{1 + 2z^2} \phantom{x}dxdz\\&= \sqrt{2}\int_{0}^{2} z \sqrt{1 + 2z^2}[x]_{0}^{1} \phantom{x}dz\\&= \sqrt{2}\int_{0}^{2} z \sqrt{1 + 2z^2} \phantom{x}dz\\&= \sqrt{2}\left[\dfrac{1}{6}\left(1+2z^2\right)^{\frac{3}{2}} \right ]_{0}^{2}\\&= \sqrt{2} \left(\dfrac{13}{3} \right )\\&= \dfrac{13\sqrt{2}}{3} \end{aligned}This means that the surface integral, $\int \int_{S} x \phantom{x}dS$, is equal to $\dfrac{13\sqrt{2}}{3}$.Example 2Evaluate the surface integral, $\int \int_{S} y \phantom{x}dS$, where $S$ a part of the cylinder, $x^2 + y^2 =4$, and bounded by $z= 0$ and $z =4$.SolutionWe’ve learned that we can parameterize cylinders in $\mathbb{R}^3$ using the following parametrization:\begin{aligned}x &= r \cos \theta\\y &= r \sin \theta\\z &= z\\\\\textbf{r}(z , \theta) &= <r \sin \theta,r \cos \theta,z>\end{aligned}This means that the parametrization of our cylinder, $x^2 +y^2 = 4$, is equal to $<2 \cos \theta, 2\sin \theta, z>$.
Keep in mind that the surface is bounded by $z$ from $0$ to $4$ and by $\theta$ from $0$ to $2\pi$. Now that we have $\textbf{r} = <2 \cos \theta, 2\sin \theta, z>$, let’s find the expression for $\textbf{r}_{z}$ and $\textbf{r}_{\theta}$.\begin{aligned}\textbf{r}_{z}&= \dfrac{\partial}{\partial z} <2 \cos \theta, 2\sin \theta, z>\\&=<0, 0, 2>\\\\\textbf{r}_{\theta}&= \dfrac{\partial}{\partial \theta} <2 \cos \theta, 2\sin \theta, z>\\&=<-2\sin \theta, 2\cos \theta, 0> \end{aligned}Take their cross products and calculate the resulting vector’s magnitude to find $|\textbf{r}_{z} \times \textbf{r}_{\theta}|$.\begin{aligned} \textbf{r}_{z} \times \textbf{r}_{\theta} &= \begin{vmatrix}\textbf{i} &\textbf{j}&\phantom{xxx}\textbf{k} \\ 0 &0 &\phantom{xxx}1 \\-2 \sin \theta &\phantom{xx}2 \cos \theta &\phantom{xxx}0 \end{vmatrix}\\&= (0 – 2\cos \theta)\textbf{i}-(0 – 2\sin \theta)\textbf{j} + 0\textbf{k}\\&= -2\cos \theta \textbf{i}+ 2\sin \theta \textbf{j}\\\\|\textbf{r}_{z} \times \textbf{r}_{\theta}| &= \sqrt{(-2 \cos\theta)^2 + (2 \sin \theta)^2}\\&= \sqrt{4 \cos^2\theta + 4\sin^2 \theta}\\&= \sqrt{4(\sin^2 \theta + \cos^2 \theta)}\\&= 2 \end{aligned}Now that we have the expression for $|\textbf{r}_{z} \times \textbf{r}_{\theta}|$, we can rewrite the surface integral as a double integral with $dS= $|\textbf{r}_{z} \times \textbf{r}_{\theta}| \phantom{x}dA$.\begin{aligned}\int \int_{S} y\phantom{x}dS &= \int \int_{D}(2\sin \theta)(2) \phantom{x}dA\\&= \int_{0}^{2\pi}\int_{0}^{4} 4\sin \theta \phantom{x}dzd\theta\\&= 4\int_{0}^{2\pi}\sin \theta\left[\int_{0}^{4}1 \phantom{x}dz \right ]\phantom{x}d\theta\\&= 4\int_{0}^{2\pi} 4\sin \theta \phantom{x}d\theta\\&= 16[-\cos \theta]_{0}^{2\pi}\\&= 16(0)\\&= 0 \end{aligned}This means that the surface integral, $\int \int_{S} y \phantom{x}dS$, is equal to zero.