OBJECTIVE

When you finish this section, you should be able to:

1. Find the surface area that lies above a region $R$ (p. 938)

NEED TO REVIEW?

Vectors are discussed in Section 10.2, pp. 699-702, and properties of the cross product are discussed in Section 10.5, pp. 724-730.

Surface Area Above a region $R$

Let $z=f(x, y)$ be a function of two variables defined on a closed, bounded region $R$. If $f_{x}$ and $f_{y}$ are continuous on $R$, then the surface area $S$ of the part of the surface that lies above $R$ is given by $$\bbox[5px, border:1px solid black, #F9F7ED]{ S=\displaystyle\iint\limits_{\kern-3ptR}\displaystyle\sqrt{ [f_{x}(x,y)]^{2}+[f_{y}(x,y)]^{2}+1}\, {\it dA} }$$

Find the surface area of the part of the surface $z=f(x, y)=\dfrac{2}{3} (x^{3/2}+y^{3/2})$ that lies above the region $R,$ a rectangle enclosed by the lines $x=0$, $x=1$, $y=0$, and $y=2$.

Figure 38 $z=\dfrac{2}{3}( x^{3/2}+y^{3/2})$

Solution Figure 38 shows the surface $z=f(x, y)=\dfrac{2}{3} (x^{3/2}+y^{3/2})$ and the surface area $S$ above the rectangle $R$. We begin by finding the partial derivatives of $z=f(x, y)$: $$f_{x}(x,y)=x^{1/2} \qquad f_{y}(x,y)=y^{1/2}$$

Since $f_x$ and $f_y$ are continuous on $R$, the surface area $S$ above $R$ is $$\begin{eqnarray*} S &=&\displaystyle\iint\limits_{\kern-3ptR}\sqrt{[f_{x}(x, y)]^{2}+[f_{y}(x,y)]^{2}+1} \,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}\sqrt{( x^{1/2}) ^{2}+( y^{1/2}) ^{2}+1}\,{\it dA} \notag \\ &=& \int_{0}^{1}\int_{0}^{2}\sqrt{x+y+1}\,{\it dy}\,{\it dx}=\int_{0}^{1}\left[ \dfrac{2}{3 }(x+y+1)^{3/2}\right] _{0}^{2}\,{\it dx} \notag \\ &=&\dfrac{2}{3}\int_{0}^{1}\big[ (x+3)^{3/2}-(x+1)^{3/2}\big] \,{\it dx}=\left( \dfrac{2}{3}\right) \left( \dfrac{2}{5}\right) \big[ (x+3)^{5/2}-(x+1)^{5/2}\big] _{0}^{1} \notag \\ &=&\dfrac{4}{15}( 32-9\sqrt{3}-4\sqrt{2}+1) = \dfrac{4}{15} (33-9\sqrt{3}-4\sqrt{2}) \end{eqnarray*}$$

NOW WORK

Problem 5.

Find the surface area of the part of the paraboloid $z=f(x,y)=1-x^{2}-y^{2}$ that lies above the $xy$-plane.

Figure 39 $z=1-x^{2}-y^{2}, z\geq 0$

Solution Figure 39 shows the part of the surface $z=1-x^{2}-y^{2}$ that lies above the $xy$-plane and its projection onto the $xy$-plane, the region $R$ enclosed by the circle $x^{2}+y^{2}=1$. We begin by finding the partial derivatives of $z=f(x,y)$. $$f_{x}(x,y)=-2x\qquad f_{y}(x,y)=-2y$$

Since $f_x$ and $f_y$ are continuous on $R$, the surface area $S$ of the part of $z=f(x,y) =1-x^{2}-y^{2}$ that lies above $R$ is $$\begin{eqnarray*} S&=&\displaystyle\iint\limits_{\kern-3ptR}\sqrt{[f_{x}(x,y)]^{2}+[f_{y} (x,y)]^{2} +1} \,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}\sqrt{ (-2x)^{2}+(-2y)^{2}+1}\,{\it dA}\\ &=&\displaystyle\iint\limits_{\kern-3ptR}\sqrt{4(x^{2}+y^{2})+1} \,{\it dA} \end{eqnarray*}$$

Since both the region $R$ (a circle) and the integrand involve $x^{2}+y^{2}$ , we use polar coordinates $(r,\theta)$. For the region $R,$ we have $0\leq r\leq 1$ and $0\leq \theta \leq 2\pi$. Then $$\begin{equation*} \begin{array}{rcl} S& =&\displaystyle\iint\limits_{\kern-3ptR}\sqrt{4(x^{2}+y^{2})+1}\,d\!A\underset{\underset{\underset{{\color{#0066A7}{\hbox {\({dx}\, {dy}=r\,{dr} d\theta\)}}}} {\color{#0066A7}{\hbox{\(x^{2}+y^{2}=r^{2}\)}}}} {\color{#0066A7}{\uparrow}}}{=} \int_{0}^{2\pi }\int_{0}^{1}\sqrt{4r^{2}+1}r\,dr\,d\theta \notag \\ &=&\int_{0}^{2\pi }\dfrac{1}{12}\big[ ( 4r^{2}+1) ^{3/2}\big] _{0}^{1}\,d\theta =\dfrac{1}{12}(5\sqrt{5}-1)\int_{0}^{2\pi }\,d\theta = \dfrac{1}{12}(5\sqrt{5}-1)2\pi\\[9pt] &=&\dfrac{\pi }{6}(5\sqrt{5}-1) \end{array} \end{equation*}$$

NOW WORK

Problem 9.

Find the surface area of the part of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$ that lies above the $xy$-plane and is contained within the cylinder $x^{2}+y^{2}=ax,$ $a\gt 0$.

Figure 40 $x^{2}+y^{2}+z^{2}=a^{2},$ $z\geq 0$

Solution Figure 40 shows the sphere and the cylinder. The equation of the surface in explicit form is $$z=f(x,y)=\sqrt{a^{2}-x^{2}-y^{2}}$$

The partial derivatives of $z$ $=f(x,y)$ are $$f_{x}(x,y)=-\dfrac{x}{\sqrt{a^{2}-x^{2}-y^{2}}}\qquad f_{y}(x,y)=- \dfrac{y}{\sqrt{a^{2}-x^{2}-y^{2}}}$$

The projection of the cylinder onto the $xy$-plane is the region $R$ enclosed by the circle $x^{2}+y^{2}=ax$, $a\gt 0.$ Then the surface area $S$ that we seek is $$S=\displaystyle\iint\limits_{\kern-3ptR}\sqrt{\dfrac{x^{2}}{a^{2}-x^{2}-y^{2}}+\dfrac{y^{2}}{ a^{2}-x^{2}-y^{2}}+1}\,\,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}\sqrt{\dfrac{a^{2}}{ a^{2}-x^{2}-y^{2}}}\,{\it dA}$$

Since the integrand involves the expression $x^{2}+y^{2}$ and the boundary of $R$ is the circle $x^{2}+y^{2}=ax$, we use polar coordinates $(r,\theta)$. We begin by finding the limits of integration. Since $$\begin{eqnarray*} x^{2}+y^{2} &=&ax \\ r^{2} &=&ar\cos \theta \\ r &=&a\cos \theta \end{eqnarray*}$$

we find that $r$ varies from $0$ to $a\cos \theta$.

Since half the area is to the right of the $xz$-plane and half is to the left of the plane, we can let $\theta$ vary from $0$ to $\dfrac{\pi }{2}$ and double the area. Then the surface area $S$ is $$\begin{array}{rcl} S& =&\int\limits\int_{\kern-11ptR}\sqrt{\dfrac{a^{2}}{a^{2}-x^{2}-y^{2}}}\,d\!A \underset{\underset{{\color{#0066A7}{\hbox {a>0}}}}{\color{#0066A7}{\uparrow }}}{=} \int\limits\int_{\kern-11ptR}\dfrac{a}{\sqrt{a^{2}-r^{2}}} \,r\,dr\,d\theta \underset{\underset{{\color{#0066A7}{\hbox{Double the area}}}}{\color{#0066A7}{\uparrow}}}{=} 2\int_{0}^{\pi /2}\!\!\int_{0}^{a\cos \theta }\dfrac{a}{\sqrt{a^{2}-r^{2}}} \,r\,dr\,d\theta \notag \\ \quad \\ &=&-2a\int_{0}^{\pi /2}\Big[ \sqrt{a^{2}-r^{2}}\Big] _{0}^{a\cos \theta }d\theta =-2a\int_{0}^{\pi /2}(a\sin \theta -a)\,d\theta\\ &=&-2a^{2}\big[-\cos \theta -\theta \big] _{0}^{\pi /2}=a^{2}(\pi -2) \end{array}$$

NOW WORK

Problem 13.