OBJECTIVES

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

1. Find the area of a region enclosed by the graph of a polar equation (p. 678)
2. Find the area of a region enclosed by the graphs of two polar equations (p. 681)
3. Find the surface area of a solid of revolution obtained from the graph of a polar equation (p. 682)

RECALL

The area \(A\) of the sector of a circle of radius \(r\) formed by a central angle of \(\theta \) radians is \(A=\dfrac{1}{2} r^{2} \theta \).

THEOREM Area in Polar Coordinates

If \(r=f(\theta )\) is nonnegative and continuous on the closed interval \([ \alpha ,\beta ] ,\) where \(\alpha \lt\beta \) and \(\beta -\alpha \leq 2\pi \), then the area \(A\) of the region enclosed by the graph of \(r=f(\theta )\) and the rays \(\theta =\alpha \) and \(\theta =\beta \) is given by \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{{A=\int_{\alpha }^{\beta }\dfrac{1}{2}r^{2}~d\theta}}}\]

NEED TO REVIEW?

The definite integral is discussed in Section 5.2, pp. 353-359.

Finding the Area Enclosed by a Part of a Cardioid

Find the area of the region enclosed by the cardioid \(r=1-\sin \theta ,\; 0\leq \theta \leq \dfrac{\pi }{2}.\)

Solution The cardioid represented by \(r=1-\sin \theta \) is shown in Figure 51(a) and the area enclosed by the cardioid from \(\theta = 0\) to \(\theta = \dfrac{\pi}{2}\) is shaded.

Figure 51 \(r=1-\sin \theta ,\;0\leq \theta \leq \dfrac{\pi }{2}\)

The region is swept out beginning with the ray \( \theta =0\) and ending with the ray \(\theta =\dfrac{\pi }{2}\), as shown in Figure 51(b). The limits of integration are \(0\) and \(\dfrac{\pi }{2}\) and the area \(A\) is \[ \begin{eqnarray*} A&=&\int_{\alpha }^{\beta }\frac{1}{2}r^{2}\,d\theta =\int_{0}^{\pi /2}\frac{1 }{2}(1-\sin \theta )^{2}\,d\theta =\frac{1}{2}\int_{0}^{\pi /2}(1-2\sin \theta +\sin ^{2}\theta )~d\theta \\[6pt] &=&\dfrac{1}{2}\int_{0}^{\pi /2}\left\{ 1-2\sin \theta +\frac{1}{2}[ 1-\cos ( 2\theta ) ] \right\} d\theta \qquad {\color{#0066A7}{\sin ^{2}\theta =\dfrac{1-\cos ( 2\theta )}{2}}} \\\\[5pt] &=&\frac{1}{2}\int_{0}^{\pi /2}\left[ \frac{3}{2}-2\sin \theta -\frac{1}{2}\cos ( 2\theta ) \right] d\theta\\[5pt] &=&\dfrac{1}{2}\left[ \dfrac{3}{2}\,\theta +2\cos \theta -\dfrac{1}{4}\sin ( 2\theta ) \right] _{0}^{\pi /2}=\dfrac{ 3\pi -8}{8} \end{eqnarray*} \]

NOW WORK

Problem 9.

Finding the Area Enclosed by a Cardioid

Find the area enclosed by the cardioid \(r=1-\sin \theta \).

Solution Look again at the cardioid in Figure 51(a). The region enclosed by the cardioid is swept out beginning with the ray \(\theta =0\) and ending with the ray \(\theta =2\pi \). So, the limits of integration are \(0\) and \(2\pi ,\) and the area \(A\) is \[ A=\int_{0}^{2\pi }\frac{1}{2}(1-\sin \theta )^{2}~d\theta =\frac{1}{2} \left[ \frac{3}{2}\theta +2 \cos \theta -\frac{1}{4}~\sin ( 2\theta ) \right] _{0}^{2\pi }=\frac{3\pi }{2} \]

NOW WORK

Problem 13.

Finding the Area Enclosed by a Rose

Find the area enclosed by the graph of \(r=2\cos ( 3\theta ) \), a rose with three petals.

Figure 52 \(r=2\cos ( 3\theta )\)

Solution Figure 52 shows the rose. The area of the blue shaded region in quadrant I equals one-sixth of the area \(A\) enclosed by the graph.* The shaded region in quadrant I is swept out beginning with the ray \(\theta =0.\) It ends at the point \(( 0,\theta ) ,\) \(0\lt\theta \lt\dfrac{\pi }{2},\) where \(\theta \) is the solution of the equation. \[ \begin{eqnarray*} 2\cos ( 3\theta ) &=&0\qquad 0\lt\theta \lt\dfrac{\pi }{2} \\[4pt] \cos ( 3\theta ) &=&0 \\[4pt] 3\theta &=&\dfrac{\pi }{2}+2k\pi \\[4pt] \theta &=&\dfrac{\pi }{6}+\dfrac{2k\pi }{3} \end{eqnarray*} \]

Since \(0\lt\theta \lt\dfrac{\pi }{2},\) we have \(\theta =\dfrac{\pi }{6}.\) The area of the shaded region in quadrant I swept out by the rays \(\theta =0\) and \(\theta =\dfrac{\pi }{6}\) is given by \(\int_{0}^{\pi /6}\dfrac{1}{2} r^{2}d\theta ,\) and the area \(A\) of the region we seek is \(6\) times this area. \[ \begin{eqnarray*} A &=& 6\int_{0}^{\pi /6}\dfrac{1}{2}r^{2}\,d\theta =3\int_{0}^{\pi /6} 4\cos ^{2}( 3\theta ) ~d\theta =12\int_{0}^{\pi /6}\cos ^{2}( 3\,\theta )~d\theta\\[4pt] &=& 12\int_{0}^{\pi /6}\frac{1+\cos ( 6\theta ) }{2}~d\theta\qquad {\color{#0066A7}{\hbox{\(\cos ^{2}( 3\theta ) =\dfrac{1+\cos ( 6\theta ) }{2}\)}}}\\\\[4pt] &=&6 \left[ \theta +\dfrac{1}{6}\sin ( 6\theta ) \right] _{0}^{\pi /6}=6\left( \frac{\pi }{6}\right) =\pi \end{eqnarray*} \]

NOW WORK

Problem 17.

Finding the Area Enclosed by a Limaçon

Figure 53 \(r=2+\cos \theta \)

Find the area of the region enclosed by the limaçon \(r=2+\cos \theta \).

Solution Figure 53 shows the graph of \(r=2+\cos \theta\), a limaçon without an inner loop.

We see that the region above the polar axis equals the region below it, so the area \(A\) of the region enclosed by the limaçon equals twice the area of the region enclosed by \(r=2+\cos \theta \) and swept out by the rays \(\theta =0\) and \(\theta =\pi \). \[ \begin{eqnarray*} A& =&2\int_{0}^{\pi }\dfrac{1}{2}r^{2}d\theta =\int_{0}^{\pi }(2+\cos \theta )^{2}\,d\theta =\int_{0}^{\pi }(4+4\cos \theta +\cos ^{2}\theta )\,d\theta \notag \\[4pt] &=& \int_{0}^{\pi }\left[ 4+4\cos \theta +\dfrac{1+\cos ( 2\theta ) }{2}\right] d\theta \qquad{\color{#0066A7}{\cos ^{2}\theta =\dfrac{1+\cos ( 2\theta ) }{2}}} \\[4pt]\\[4pt] & =&\left[ 4\theta +4\sin \theta +\frac{\theta }{2}+\frac{1}{4}\sin ( 2\theta ) \right] _{0}^{\pi }=\frac{9\pi }{2} \end{eqnarray*} \]

NOW WORK

Problem 21.

Finding the Area of the Region Enclosed by the Graphs of Two Polar Equations

Find the area of the region that lies outside the cardioid \(r=1+\cos \theta \) and inside the circle \(r=3\cos \theta \).

Solution We begin by graphing each equation. See Figure 54(a). Then we find the points of intersection of the two graphs by solving the equation, \[ \begin{eqnarray*} 3\cos \theta & =&1+\cos \theta \\[3pt] 2\cos \theta & =&1 \\[3pt] \cos \theta & =&\frac{1}{2} \end{eqnarray*} \] \[ \begin{equation*} \theta =-\frac{\pi }{3}\;\hbox{or}\;\theta =\frac{\pi }{3} \end{equation*} \]

The graphs intersect at the points \(\left( \dfrac{3}{2},-\dfrac{\pi }{3} \right) \) and \(\left( \dfrac{3}{2},\dfrac{\pi }{3}\right) \).

Figure 54

The area \(A\) of the region that lies outside the cardioid and inside the circle is shown as the shaded portion in Figure 54(b). Notice that the area \( A\) is the difference between the area of the region enclosed by the circle \( r=3\cos \theta \) swept out by the rays \(\theta =-\dfrac{\pi }{3}\) and \( \theta =\dfrac{\pi }{3}\), and the area of the region enclosed by the cardioid \(r=1+\cos \theta \) swept out by the same rays. So, \[ \begin{array}{@rcl} A& =&\int_{-\pi /3}^{\pi /3}\frac{1}{2}(3\cos \theta )^{2}d\theta -\int_{-\pi /3}^{\pi /3}\frac{1}{2}(1+\cos \theta )^{2}d\theta =\frac{1}{2} \int_{-\pi /3}^{\pi /3}[ 9\cos ^{2}\theta -( 1+2\cos \theta +\cos^{2}\theta ) ] \,d\theta \notag \\[12pt] & =&\frac{1}{2}\int_{-\pi /3}^{\pi /3}(8\cos ^{2}\theta -1-2\cos \theta )~d\theta = \frac{1}{2}\int_{-\pi /3}^{\pi /3} \left[ 8\left( \frac{1+\cos ( 2\theta ) }{2}\right) -1-2\cos \theta \right] d\theta \notag \\[12pt] & =&\frac{1}{2}\int_{-\pi /3}^{\pi /3}[ 3+4\cos ( 2\theta ) -2\cos \theta ] ~d\theta =\frac{1}{2}\Big[ 3\theta +2\sin ( 2\theta ) -2\sin \theta \Big] _{-\pi /3}^{\pi /3}=\pi \end{array} \]

NOW WORK

Problem 23.

Finding the Surface Area of a Solid of Revolution

Find the surface area of the solid of revolution generated by revolving the arc of the circle \(r=a,\) \(a>0,\) \(0\leq \theta \leq \dfrac{\pi }{4}\), about the polar axis.

Solution See Figure 55. We find the surface area \(S\) using formula (1). Since \(r=f(\theta )=a,\;f' ( \theta) =0.\) Then \[ \begin{eqnarray*} S& =&2\pi \int_{\alpha }^{\beta }f(\theta )\sin \theta \sqrt{[ f(\theta )] ^{2}+[ f' ( \theta) ] ^{2}}~d\theta\\[4pt] &=&2\pi \int_{0}^{\pi /4}a\sin \theta \sqrt{a^{2}}\,d\theta =2\pi a^{2}\int_{0}^{\pi /4}\sin \theta \,d\theta \notag \\[4pt] & =&2\pi a^{2}\big[ -\cos \theta \big] _{0}^{\pi /4}=2\pi a^{2}\left( -\frac{\sqrt{2}}{2}+1\right) =\pi a^{2}( 2-\sqrt{2}) \end{eqnarray*} \]

Figure 55

NOW WORK

Problem 27.