7.9: Area in Polar Coordinates
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- 155881
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section we derive a formula for the area of a region in polar coordinates. Section 6.3 on the length of a curve in rectangular coordinates should be studied before this and the following section.
Our starting point for areas in rectangular coordinates was the formula for the area of a rectangle. In polar coordinates our starting point is the formula for the area of a sector of a circle.
A sector of a circle with radius \(r\) and central angle \(\theta\) has area
\[A=\frac{1}{2} r^{2} \theta \nonumber \]
An arc of a circle with radius \(r\) and central angle \(\theta\) has length
\[s=r \theta \text {. } \nonumber \]
PROOF Consider a sector POQ shown in Figure 7.9.1. To simplify notation let \(O\) be the origin, and put the sector \(P O Q\) in the first quadrant with \(P\) on the
Figure 7.9.1
\(x\)-axis. Then
\[P=(r, 0), \quad Q=(r \cos \theta, r \sin \theta) \nonumber \]
The arc \(Q P\) has the equation
\[y=\sqrt{r^{2}-x^{2}}, \quad r \cos \theta \leq x \leq r \nonumber \]
We see from the figure that
\[A=\frac{1}{2} r^{2} \sin \theta \cos \theta+\int_{r \cos \theta}^{r} \sqrt{r^{2}}-x^{2} d x \nonumber \]
Integrating by the trigonometric substitution \(x=r \sin \phi\), we get
Therefore \(A=\frac{1}{2} r^{2} \theta\). By definition, \(A=\frac{1}{2} r s\), so
\[s=\frac{2 A}{r}=r \theta \nonumber \]
The next theorem gives the formula for area in polar coordinates.
Let \(r=f(\theta)\) be continuous and \(r \geq 0\) for \(a \leq \theta \leq b\), where \(b \leq a+2 \pi\). Then the region \(R\) bounded by the curve \(r=f(\theta)\) and the lines \(\theta=a\) and \(\theta=b\) has area
\[A=\frac{1}{2} \int_{a}^{b} f(\theta)^{2} d \theta \nonumber \]
Discussion Imagine a point \(P\) moving along the curve \(r=f(\theta)\) from \(\theta=a\) to \(\theta=b\). The line \(O P\) will sweep out the region \(R\) in Figure 7.9.2. Since \(b \leq a+2 \pi\), the line will complete at most one revolution, so no point of \(R\) will be counted more than once.
The formula for area can be seen intuitively by considering an infinitely small wedge \(\Delta A\) of \(R\) between \(\theta\) and \(\theta+\Delta \theta\). (Figure 7.9.3). The wedge is almost a sector
of a circle of radius \(f(\theta)\) with central angle \(\Delta 0\), so
\[\Delta A \approx \frac{1}{2} f(\theta)^{2} \Delta \theta \quad \text { (compared to } \Delta \theta \text { ). } \nonumber \]
By the Infinite Sum Theorem,
\[A=\frac{1}{2} \int_{a}^{b} f(\theta)^{2} d \theta \nonumber \]
The actual proof follows this intuitive idea but the area of \(\Delta A\) must be computed more carefully.
PROOF Let \(\triangle \theta\) be positive infinitesimal and let \(\theta\) be a hyperreal number between \(a\) and \(b-\Delta \theta\). Consider the wedge of \(R\) with area \(\Delta A\) between \(\theta\) and \(\theta+\Delta 0\). Since \(f(\theta)\) is continuous, it has a minimum value \(m\) and maximum value \(M\) between \(\theta\) and \(\theta+\Delta \theta\), and furthermore,
\[m \approx f(\theta), \quad M \approx f(\theta) \nonumber \]
The sector between \(\theta\) and \(\Delta \theta\) of radius \(m\) is inscribed in \(\Delta A\) while the sector of radius \(M\) is circumscribed about \(\Delta A\).
Figure 7.9.4
(Figure 7.9.4 shows the inscribed and circumscribed sectors for real \(\Delta \theta\) and infinitesimal \(\Delta 0\).) By Theorem 1, the two sectors have areas \(\frac{1}{2} m^{2} \Delta 0\) and \(\frac{1}{2} M^{2} \Delta \theta\). Moreover, \(\triangle A\) is between those two areas,
\[\begin{aligned} \frac{1}{2} m^{2} \Delta \theta & \leq \Delta A \leq \frac{1}{2} M^{2} \Delta \theta \\ \frac{1}{2} m^{2} & \leq \Delta A / \Delta \theta \leq \frac{1}{2} M^{2} \end{aligned} \nonumber \]
Taking standard parts:
\[\frac{1}{2} f(\theta)^{2} \leq s t(\Delta A / \Delta \theta) \leq \frac{1}{2} f(\theta)^{2} \nonumber \]
Therefore
\[\Delta A / \Delta \theta \approx \frac{1}{2} f^{\prime}(\theta)^{2} \nonumber \]
and by the Infinite Sum Theorem,
\[A=\frac{1}{2} \int_{a}^{b} f(\theta)^{2} d \theta \nonumber \]
Theorem 1 is also true in the case that \(r=f(\theta)\) is continuous and \(r \leq 0\).
\[A=\frac{1}{2} \int_{a}^{b} f(\theta)^{2} d \theta=\frac{1}{2} \int_{a}^{b}(-f(\theta)\}^{2} d \theta \nonumber \]
the region \(R\) bounded by the curve \(r=f(\theta)\) has the same area as the region \(S\) bounded by the curve \(r=-f(\theta)\). Both areas are positive. As we can see from Figure 7.9.5, \(S\) looks exactly like \(R\) but is on the opposite side of the origin.
Find the area of one loop of the "four-leaf clover" \(r=\sin 2 \theta\). From Figure 7.9.6, we see that one loop is traced out when \(\theta\) goes from 0 to \(\pi / 2\). Therefore the area is
Solution
\[\begin{aligned} A & =\frac{1}{2} \int_{0}^{\pi / 2} \sin ^{2}(2 \theta) d \theta=\frac{1}{2} \int_{0}^{\pi} \frac{1}{2} \sin ^{2} \phi d \phi \\ & \left.=\frac{1}{4} \int_{0}^{\pi} \sin ^{2} \phi d \phi=\frac{1}{4}\left(-\frac{1}{2} \sin \phi \cos \phi+\frac{1}{2} \phi\right)\right]_{0}^{\pi}=\frac{1}{8} \pi \end{aligned} \nonumber \]
As one would expect, all four loops have the same area.
On the loop from \(\theta=\pi / 2\) to \(\theta=\pi\), the value of \(r=\sin 2 \theta\) is negative. However, the area is again
\[A=\frac{1}{2} \int_{\pi / 2}^{\pi} \sin ^{2}(2 \theta) d \theta=\frac{1}{8} \pi \nonumber \]
Our next example shows why the hypothesis that \(r\) has the same sign for \(a \leq \theta \leq b\) is needed in Theorem 2 .
Find the area of the region inside the circle \(r=\sin \theta\) (Figure 7.9.7).
Solution
The point \((r, \theta)\) goes around the circle once when \(0 \leq \theta \leq \pi\) with \(r\) positive, and again when \(\pi \leq \theta \leq 2 \pi\) with \(r\) negative. The theorem says that we will get the correct area if we take either 0 and \(\pi\), or \(\pi\) and \(2 \pi\), as the limits of
integration. Thus
\[\left.A=\int_{0}^{\pi} \frac{1}{2} \sin ^{2} \theta d \theta=\frac{1}{2}\left(-\frac{1}{2} \sin \theta \cos \theta+\frac{1}{2} \theta\right)\right]_{0}^{\pi}=\frac{1}{4}(\pi-0)=\pi / 4 \nonumber \]
Alternatively,
\[\left.A=\int_{\pi}^{2 \pi} \frac{1}{2} \sin ^{2} \theta d \theta=\frac{1}{2}\left(-\frac{1}{2} \sin \theta \cos \theta+\frac{1}{2} \theta\right)\right]_{\pi}^{2 \pi}=\frac{1}{4}(2 \pi-\pi)=\pi / 4 \nonumber \]
Since the curve is a circle of radius \(\frac{1}{2}\), our answer \(\pi / 4\) agrees with the usual formula \(A=\pi r^{2}\).
Integrating from 0 to \(2 \pi\) would count the area twice and give the wrong answer.
Find the area of the region inside both the circles \(r=\sin \theta\) and \(r=\cos \theta\).
Solution
The first thing to do is draw the graphs of both curves. The graphs are shown in Figure 7.9.8.
Figure 7.9 .8
We see that the two circles intersect at the origin and at \(\theta=\pi / 4\). The region is divided into two parts, one bounded by \(r=\sin \theta\) for \(0 \leq \theta \leq \pi / 4\) and the other bounded by \(r=\cos \theta\) for \(\pi / 4 \leq \theta \leq \pi / 2\). Thus
\[\begin{aligned} A & =\int_{0}^{\pi / 4} \frac{1}{2} \sin ^{2} \theta d \theta+\int_{\pi / 4}^{\pi / 2} \frac{1}{2} \cos ^{2} \theta d \theta \\ & \left.\left.=\frac{1}{2}\left(-\frac{1}{2} \sin \theta \cos \theta+\frac{1}{2} \theta\right)\right]_{0}^{\pi / 4}+\frac{1}{2}\left(\frac{1}{2} \sin \theta \cos \theta+\frac{1}{2} \theta\right)\right]_{\pi / 4}^{\pi / 2} \\ & =\frac{1}{2}\left[\left(-\frac{1}{4}-0\right)+\left(\frac{\pi}{8}-0\right)+\left(0-\frac{1}{4}\right)+\left(\frac{\pi}{4}-\frac{\pi}{8}\right)\right]=\frac{\pi}{8}-\frac{1}{4} \end{aligned} \nonumber \]
PROBLEMS FOR SECTION 7.9
In Problems 1-13, find the area of the regions bounded by the following curves in polar coordinates.
| 1 | \(r=2 a \cos \theta\) | \(\mathbf{2}\) | \(r=1+\cos \theta\) |
| \(\mathbf{3}\) | \(r=\sqrt{\sin \theta}\) | \(\mathbf{4}\) | \(r=2+\cos \theta\) |
| \(\mathbf{5}\) | The loop in \(r=\tan \left(\frac{1}{2} \theta\right)\) | 6 | One loop of \(r=\cos (3 A)\) |
7 One loop of \(r=\sin ^{2} \theta\)
8 The large loop of \(r=\frac{1}{2}+\cos \theta\)
9 The small loop of \(r=\frac{1}{2}+\cos \theta\)
10 One loop of \(r^{2}=\cos (2 \theta)\)
\(11 \quad \theta=0, \quad \theta=\pi / 3, \quad r=\cos \theta\)
\(12 \theta=\pi / 6, \quad \theta=\pi / 3, \quad r=\sec \theta\)
\(13 \quad r=\tan \theta, \quad r=\frac{1}{\sqrt{2}} \csc \theta\)
14 Find the area of the region inside the curve \(r=2 \cos \theta\) and outside the curve \(r=1\).
15 Find the area of the region inside the curve \(r=2 \sin \theta\) and above the line \(r=\frac{3}{2} \csc \theta\)
16 Find the area of the region inside the spiral \(r=\theta, 0 \leq \theta \leq 2 \pi\).
17 Find the area of the region inside the spiral \(r=\sqrt{\theta}, 0 \leq \theta \leq 2 \pi\).
18 Find the area of the region inside both of the curves \(r=\sqrt{3} \cos \theta, r=\sin \theta\).
19 Find the area of the region inside both of the curves \(r=1-\cos \theta, r=\cos \theta\).
\(\mathbf{2 0}\) The center of a circle of radius one is on the circumference of a circle of radius two. Find the area of the region inside both circles.
21 Find a formula for the area of the region between the curves \(r=f(\theta)\) and \(r=g(\theta)\), \(a \leq \theta \leq b\), when \(0 \leq f(\theta) \leq g(\theta)\).