Surface Area Using Double Integrals

Direct Knowlege Contributor David A. Smith
  • By David A. Smith, Founder & CEO, Direct Knowledge
  • David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

Introduction to Surface Area

We apply double integrals to the problem of computing the surface area over a region. We demonstrate a formula that is analogous to the formula for finding the arc length of a one variable function and detail how to evaluate a double integral to compute the surface area of the graph of a differentiable function of two variables.

Surface Area of a Differentiable Function

Theorem. (Surface Area) Assume that the function $f(x,y)$ has continuous partial derivatives $f_x$ and $f_y$ in a region $R$ of the $x y$-plane. Then the portion of the surface $z=f(x,y)$ that lies over $R$ has surface area \begin{equation} S=\iint _R \sqrt{ \left[ f_x(x,y)\right]^2 + \left[ f_y(x,y) \right]^2 +1 }\, dA. \end{equation}

Proof. Consider a surface defined by $z=f(x,y)$ over a region $R$ of the $x y-\text{plane}.$ Enclose the region $R$ in a rectangle partitioned by a grid with lines parallel to the coordinate axes. This creates a number of cells, and we let $R_1,$ $R_2,$ …., $R_n$ denote those that lie entirely within $R.$ For $m=1,2,3,\ldots,n,$ let $P_m\left(x_m^*, y_m^*\right)$ be any corner of the rectangle $R_m,$ and let $T_m$ be the tangent plane above $P_m$ on the surface of $z=f(x,y).$ Let $\triangle S_m$ denote the area of the patch of the surface that lies directly above $R_m.$ The rectangle $R_m$ projects onto a parallelogram $A B D C$ in the tangent plane $T_m,$ and if $R_m$ is small we would expect the area of this parallelogram to approximate closely the element of the area $\triangle S_m.$ If $\triangle x_m$ and $\triangle y_m$ are the lengths of the sides of the rectangle $R_m,$ the approximating parallelogram will have sides determined by the vectors \begin{equation} {A B}=\triangle x_m {i} + \left[ f_x\left(x_m^*, y_m^* \right)\triangle x_m\right]{k} \end{equation} and \begin{equation} {A C}=\triangle y_m {j} + \left[f_y\left(x_m^*,y_m^*\right)\triangle y_m\right] {k}. \end{equation} If $K_m$ is the area of the approximating parallelogram, we have \begin{equation} K_m=|{A B}\times {A C}|. \end{equation} To determine $K_m,$ we first find the cross product: \begin{align}{A B} \times {A C} & =\left| \begin{array}{ccc}{i} & {j} & {k} \\ \triangle x_m & 0 & f_x \left(x_m^*, y_m^*\right) \triangle x_m \\ 0 & \triangle y_m & f_y \left(x_m^*, y_m^*\right)\triangle y_m \end{array} \right| \\ & =\triangle x_m\triangle y_m\left(-f_x\left(x_m^*,y_m^*\right)\, {i}-f_y \left(x_m^*,y_m^*\right)\, {j}+{k}\right). \end{align} Then we calculate the norm, \begin{equation} K_m = |{A B} \times {A C}| = \triangle x_m\triangle y_m \sqrt{ \left[f_x \left(x_m^*, y_m^*\right)\right|^2 + \left[f_y\left( x_m^*, y_m^*\right)\right]^2+1 }. \end{equation} Finally, summing over the entire partition, we see that the surface area over $R$ may be approximated by the sum \begin{equation} \triangle S_n=\sum_{m=1}^n \sqrt{\left[ f_x\left(x_m^*, y_m^*\right)\right]{}^2 + \left[f_y \left(x_m^*, y_m^*\right)\right]{}^2+1}\left(\triangle A_m\right) \end{equation} where $\triangle A_m=\triangle x_m\triangle y_m.$ This is a Riemann sum, and by taking an appropriate limit (as the partition becomes more refined), we find that the surface area $S,$ satisfies \begin{align} S & =\lim_{n\to \infty }\sum_{m=1}^n \triangle A_m \sqrt{\left[f_x \left(x_m^*,y_m^*\right) \right]^2 + \left[f_y \left(x_m^*, y_m^*\right) \right]^2+1} \\ & = \iint_R\sqrt{ \left[f_x(x,y) \right]^2 + \left[f_y(x,y)\right]^2+1} \, d A. \end{align}

Example. Find the surface area of the part of the surface $z=x^2+2y$ that lies above the triangular region $T$ in the $x y-\text{plane}$ with vertices $(0,0),$ $(1,0),$ and $(1,1).$

Solution. The region $T$ is described by $T=\{(x,y) \mid 0\leq x\leq 1,0\leq y\leq x\}.$ We have \begin{align} S& =\iint_R\sqrt{(2x)^2+(2)^2+1} \, d A =\int_0^1\int_0^x\sqrt{4x^2+5}\, dy\, dx \\ & =\int_0^1x\sqrt{4x^2+5}\, dx =\frac{1}{12}\left(27-5\sqrt{5}\right). \end{align}

Example. Find the surface area of a sphere of radius $a.$

Solution. Let $z=f(x,y)=\sqrt{a^2-x^2-y^2}.$ Then \begin{equation}
f_x=\frac{-x}{\sqrt{a^2-x^2-y^2}}, \qquad f_y=\frac{-y}{\sqrt{a^2-x^2-y^2}},\end{equation} and \begin{equation} \sqrt{f_x^2+f_y^2+1} = \sqrt{\frac{x^2+y^2+\left(a^2-x^2-y^2\right)}{a^2-x^2-y^2}} = \frac{a}{\sqrt{a^2-x^2-y^2}}. \end{equation} In polar form we have
\begin{align} & 8\int_0^{\pi /2}\int_0^a\frac{a}{\sqrt{a^2-r^2}}r \, drd\theta \\ & =-4a\int_0^{\pi /2}\int_0^a\frac{(-2r)}{\sqrt{a^2-r^2}} \, drd\theta =-8a\int_0^{\pi /2} (-a) \, d\theta =4\pi a^2. \end{align}

Example. Find the surface area of a cylinder of radius $a$ and height $h.$

Solution. Let $z=f(x,y)=\sqrt{a^2-x^2}.$ Then $ f_x=\frac{-x}{\sqrt{a^2-x^2}}, f_y=0, $ and \begin{equation} \sqrt{f_x^2+f_y^2+1}=\sqrt{\frac{x^2+a^2-x^2}{a^2-x^2}}=\frac{a}{\sqrt{a^2-x^2}}.\end{equation} We find the surface area as
\begin{align} 4\int_0^a\int_0^h\frac{a}{\sqrt{a^2-x^2}} \, dy\, dx & =4a\int_0^a\frac{h a}{\sqrt{a^2-x^2}} dx \\ & =\left. 4 a h \sin ^{-1}\left(\frac{x}{a}\right) \right|_0^a =2\pi a h. \end{align}

Example. Find the surface area the part of the paraboloid $z=x^2+y^2$ that lies under the plane $z=9.$

Solution. The plane intersects the paraboloid in the circle $x^2+y^2=9$ when $z=9.$ Therefore, the given surface lies above the disk $D$ with center at the origin and radius 3. Converting to polar coordinates we have \begin{align} S& = \iint_R \sqrt{(2x)^2+(2y)^2+1} \, d A =\int_0^{2\pi }\int_0^3\sqrt{4r^2+1}r \, dr \, d\theta \\ & =\int_0^{2\pi }d \theta \int_0^3\frac{1}{8}\sqrt{4r^2+1}(8r) \, dr =\frac{\pi }{6}\left(37\sqrt{37}-1\right). \end{align}

Example. Find the surface area the portion of the sphere $x^2+y^2+z^2=4$ that lies inside the cylinder $x^2+y^2=2y.$

Solution. Let $z=f(x,y)=\sqrt{4-x^2-y^2}.$ Then \begin{equation}f_x=\frac{-x}{\sqrt{4-x^2-y^2}}, \qquad f_y=\frac{-y}{\sqrt{4-x^2-y^2}}, \end{equation} and \begin{equation} \sqrt{f_x^2+f_y^2+1}=\sqrt{\frac{x^2+y^2+\left(4-x^2-y^2\right)}{4-x^2-y^2}}=\frac{2}{\sqrt{4-x^2-y^2}}. \end{equation} The projected region in polar form is $r=2 \sin \theta .$ Since half the surface is above the $x y-\text{plane}$ and half the surface is below we have
\begin{align} S& =2\int_0^{\pi }\int_0^{2 \sin \theta }\frac{2r}{\sqrt{4-r^2}}drd\theta =-8\int_0^{\pi } \left(\sqrt{4-4\sin ^2\theta }-2\right) \, d\theta \\ & =-8\int_0^{\pi /2} (2 \cos \theta -2) \, d\theta -8\int{\pi /2}^{\pi } (-2 \cos \theta -2) \, d\theta =8(\pi -2). \end{align}

Surface Area Defined Parametrically

We show a way to find the surface area of the graph of a surface given a parameterization of the surface. We also illustrate how to find the surface area with two examples and then show how to find the surface area of a torus with given inner and outer radii.

Theorem. (Surface Area Parametrically) Let $S$ be a surface defined parametrically by \begin{equation} {R}(u,v) = x(u,v){i} + y(u,v){j} + z(u,v){k} \end{equation} on the region $D$ in the $u v$-plane, and assume that $S$ is smooth in the sense that ${R}_u$ and ${R}_v$ are continuous with ${R}_u\times {R}_v\neq {0}$ on $D.$ Then the surface area $S$ is given by \begin{equation} S=\underset{D}{\iint }\left|{R}_u\times {R}_v\right| d u d v. \end{equation} The quantity $\left|{R}_u\times {R}_v\right|$ is called the fundamental cross product.

Proof. Suppose a surface $S$ is defined parametrically by the vector function \begin{equation} {R}(u,v)=x(u,v){i}+y(u,v){j}+z(u,v){k} \end{equation} for parameters $u$ and $v.$ Let $D$ be a region in the $x y$-plane on which $x,$ $y,$ and $z,$ as well as their partial derivatives with respect to $u$ and $v$ are continuous. The partial derivatives of $R(u,v)$ are given by \begin{equation} {R}_u=\frac{\partial x}{\partial u}{i}+\frac{\partial y}{\partial u}{j}+\frac{\partial z}{\partial u}{k} \qquad \text{ and } \qquad {R}_v=\frac{\partial x}{\partial v}{i}+\frac{\partial y}{\partial v}{j}+\frac{\partial z}{\partial v}{k}. \end{equation} Suppose the region $D$ is subdivided into cells. Consider a typical rectangle in this partition, dimension $\triangle x$ and $\triangle y$, where $\triangle x$ and $\triangle y$ are small. If we project this rectangle onto the surface ${R}(u,v),$ we obtain a curvilinear parallelogram with adjacent sides ${R}_u(u,v)\triangle u$ and ${R}_v(u,v)\triangle v.$ The area of this rectangle is approximated by \begin{equation} \triangle S=\left|{R}_u(u,v)\triangle u\times {R}_v(u,v)\triangle v\right| =\left|{R}_u(u,v)\times {R}_v(u,v)\right|\triangle u \triangle v. \end{equation} By taking an appropriate limit, we find the surface area to be a double integral.

Example. Find the area of the surface given parametrically by the equation \begin{equation} {R}(u,v)= u v{i}+(u-v) {j}+(u+v) {k} \end{equation} for $u^2+v^2\leq 1.$

Solution. We find \begin{equation} {R}_u\times {R}_v=\left| \begin{array}{ccc} {i} & {j} & {k} \\ v & 1 & 1 \\ u & -1 & 1 \end{array} \right|=2{i}+(u-v){j}-(u+v){k}. \end{equation} Therefore the fundamental cross product is \begin{equation} \left|{R}_u\times {R}_v\right|=\sqrt{4+(u-v)^2+(-u-v)^2} = \sqrt{4+2u^2+2v^2}. \end{equation} Using polar coordinates the surface area is \begin{align} S & =\int_0^{2\pi }\int_0^1\sqrt{4+2r^2}rdrd\theta =\frac{1}{4}\int_0^{2\pi }\int_0^1\sqrt{4+2r^2}4rdrd\theta \\ & =\frac{1}{6}\int_0^{2\pi } \left(6\sqrt{6}-8\right) \, d\theta =\frac{2\pi }{3}\left(3\sqrt{6}-4\right). \end{align}

Example. Find the area of the surface given parametrically by the equation \begin{equation} {R}(u,v)=(u \sin v){i}+(u \cos v){j}+ v {k} \end{equation} for $0\leq u\leq a$ and $0\leq v\leq b.$

Solution. We find \begin{equation} {R}_u=(\sin v){i}+(\cos v){j}, \quad {R}_v=(u \cos v){i}+(-u \sin v){j}+{k},\end{equation}
and \begin{equation} {R}_u\times {R}_v=\left| \begin{array}{ccc} {i} & {j} & {k} \\ \sin v & \cos v & 0 \\ u \cos v & -u \sin v & 1 \end{array} \right|=(\cos v){i}-(\sin v){j}-u {k}.\end{equation} Therefore the fundamental cross product is $\left|{R}_u\times {R}_v\right| = \sqrt{1+u^2}.$ So the surface area is \begin{align} S& =\int_0^b\int_0^a\sqrt{1+u^2} \, du dv =\int_0^b \left[\frac{\ln \left| u + \sqrt{1+u^2}\right|}{2}+\frac{u\sqrt{1+u^2}}{2} |_0^a\right] \, dv \\ & =\frac{b}{2} \left[\ln \left(a+\sqrt{1+a^2}\right)+a\sqrt{1+a^2}\right]. \end{align}

Example. Find the area of the torus which is given parametrically by \begin{equation} {R}(u,v)=(a+b \cos v)\cos u{i}+(a+b \cos v)\sin u {j}+ (b \sin v) {k} \end{equation} for $0<b<a,$ $0\leq u\leq 2\pi ,$ and $0\leq v\leq 2\pi.$

Solution. We find \begin{align} & {R}_u=-(a+b \cos v)\sin u {i}+(a+b \cos v)\cos u{j}, \\ & {R}_v=-b \sin v \cos u {i}-b \sin v \sin u{j} +b \cos v {k}, \text{ and} \\ & {R}_u\times {R}_v =\left| \begin{array}{ccc} {i} & {j} & {k} \\ -(a+b \cos v) \sin u & (a+b \cos v)\cos u & 0 \\ -b \sin v \cos u & -b \sin v \sin u & b \cos v \end{array} \right| \\ & \qquad \qquad =\left(b^2\cos ^2v+a b \cos v\right)(\cos u){i} +\left(b^2\cos ^2v+a b \cos v\right)(\sin u){j} \\ & \qquad \qquad \qquad \qquad +\left(b^2\sin v \cos v+a b \sin v\right){k}.\end{align} Therefore
$ \left|{R}_u\times {R}_v\right|=\left|a b+b^2\cos v\right|.$ So the surface area is \begin{align} S& =\int_0^{2\pi }\int_0^{2\pi }\left|a b+b^2\cos v\right| du dv =\int_0^{2\pi }2\pi \left(a b+b^2\cos v\right)dv =4\pi ^2 a b. \end{align}

Exercises on Surface Area

Exercise. Find the surface area of the part of the paraboloid $z=x^2+y^2$ that lies under the plane $z=9.$

Exercise. Find the surface area of the surface of the portion of the plane $4x+y+z=9$ that lies in he first octant.

Exercise. Find the surface area of the surface of the portion of the paraboloid $z=3x^2+3y^2$ that lies inside the cylinder $x^2+y^2=1.$

Exercise. Find the surface area of the surface of the portion of the plane $2x+2y-z=0$ that is above the square in the plane with vertices $(0,1,0),$
$(0,0,0),$ $(1,0,0),$ and $(1,1,0).$

Exercise. Find the surface area of the surface of the portion of the surface $z=x^2$ that lies over the triangular region in the plane with vertices
$(0,0,0),$ $(0,1,0),$ and $(1,0,0).$

Exercise. Find the surface area of the surface of the portion of the sphere $x^2+y^2+z^2=25$ inside the cylinder $x^2+y^2=9.$

Exercise. Find the surface area of the surface of the portion of the cone $z=2\sqrt{x^2+y^2}$ inside the cylinder $x^2+y^2=4.$

Exercise. Find the surface area of the surface of the portion of the sphere $x^2+y^2+z^2=a^2$ inside the cylinder $x^2+y^2=a x$ and above the $x y$-plane.

Exercise. Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by $z=e^{-x}\sin y$ over the triangle with vertices $(0,0,0),$ $(0,1,1),$ $(0,1,0).$

Exercise. Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by $x=z^3-y z+y^3$ over the square $(0,0,0),$ $(0,0,2),$ $(0,2,0),$ and $(0,2,2).$

Exercise. Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by $z=\cos \left(x^2+y^2\right)$ over the disk $x^2+y^2\leq \frac{\pi }{2}.$

Exercise. Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by $z=e^{-x}\cos y$ over the disk $x^2+y^2\leq 2.$

Exercise. Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by $z=x^2+5 x y+y^2$ over the region in the $x y$-plane bounded by the curve $x y=5$ and the line $x+y=6.$

Exercise. Setup, but do not evaluate, the iterated integral for the surface area of the portion of the surface given by $z=x^2+3 x y+y^2$ over the region
in the $x y$-plane bounded by $0\leq x\leq 4,$ $0\leq y\leq x.$

Exercise. Compute the magnitude of the fundamental cross product for the surface defined parametrically by ${R}(u,v)=(2 u \sin v)\, {i}+(2 u \cos v)\, {j}+u^2\, {k}.$

Exercise. Compute the magnitude of the fundamental cross product for the surface defined parametrically by ${R}(u,v)=(4 \sin u \cos v ){i}+(4 \sin u \sin v) \, {j}+(5 \cos u) {k}.$

Exercise. Compute the magnitude of the fundamental cross product for the surface defined parametrically by ${R}(u,v)=u \, {i}+ v^2 \, {j}+u^3 \, {k}.$

Exercise. Compute the magnitude of the fundamental cross product for the surface defined parametrically by ${R}(u,v)=(2 u \sin v)\, {i}+(2 u \cos v)\, {j}+\left(u^2\sin 2 v\right){k}.$

Exercise. Find the area of the surface given parametrically by the equation
${R}(u,v)=u v \, {i}+(u-v)\,{j}+(u+v){k}$ for $u^2+v^2\leq 1.$

Exercise. A spiral ramp has the vector parametric equation ${R}(u,v)=(u \cos v) \, {i}+(u \sin v) \,{j}+v \,{k}$ for $0\leq u\leq 1,$ $0\leq v\leq \pi .$ Find the surface area of this ramp.