Calculus III 14.03 Double Integrals in Polar Coordinates

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14.03 Double Integrals in Polar Coordinates

• Write and evaluate double integrals in polar coordinates.

Double Integrals in Polar Coordinates

Some double integrals are much easier to evaluate in polar form than in rectangular form. This is especially true for regions such as circles, cardioids, and rose curves, and for integrands that involve \(x^{2}+y^{2}\). Section 10.4 described how polar coordinates \((r, \theta)\) are related to cartesian coordinates \((x,y)\) as follows.

\(x= r \cos \theta \)	and	\( \:\:\:\: y= r \sin \theta\)
\( r^{2} = x^{2}+y^{2} \:\:\:\:\)	and	\( \:\:\:\: \tan \theta= y/x\)

Example 14.3.1 Using Polar Coordinates to Describe a Region

Use polar coordinates to describe each region shown in Figure 14..3.1.

Figure 14.3.1

Solution
a. The region \(R\) is a quarter circle with radius 2. The description in polar coordinates is

\( R=\{(r,\theta):0 \leqslant r \leqslant 2, \: 0 \leqslant \theta \leqslant \pi/2\}.\)

b. The region \(R\) contains all points between concentric circles with radii 1 and 3. The description in polar coordinates is

\( R=\{(r,\theta):1 \leqslant r \leqslant 3, \: 0 \leqslant \theta \leqslant 2\pi\}.\)

c. The region \(R\) is a cardioid with \(a=b=3.\) The description in polar coordinates is

\( R=\{(r,\theta):0 \leqslant r \leqslant 3 + 3 \sin \theta, \: 0 \leqslant \theta \leqslant 2\pi\}.\)

Polar sector Figure 14.3.2 Polar grid superimposed over region \(R\). Figure 14.3.3 The polar sector \(R_{i}\) is the set with all points \((r,\theta)\) such that \(r_{1} \leqslant r \leqslant r_{2}\) and \(\theta_{1} \leqslant \theta \leqslant \theta_{2}\). Figure 14.3.4 Horizontally simple region \(S\). Figure 14.3.5

The regions in Example 14.3.1 are special cases for polar sectors \( R=\{(r,\theta): r_{1} \leqslant r \leqslant r_{2}, \: \theta_{1} \leqslant \theta \leqslant \theta_{2}\}\:\:\:\: \color{red}{ \text{Polar sector }}\) as shown in Figure 14.3.2. To define a double integral for a continuous function \(z=f(x,y)\) in polar coordinates, consider a region \(R\) bounded by the the graphs for \(r = g_{1}(\theta) \text{ and } r = g_{2}(\theta) \) and the lines \(\theta = \alpha \) and \(\theta = \beta \). Partition the region \(R\) into small polar sectors and not into small rectangles. On \(R\), superimpose a polar grid made from rays and circular arcs, as shown in Figure 14.3.3. The polar sectors \(R_{i}\) lying entirely within \(R\) form an inner polar partition \(\Delta \), whose norm \( \| \Delta \| \) is the length for the longest diagonal among the \(n\) polar sectors. Consider a specific polar sector \(R_{i}\), as shown in Figure 14.3.4. The area for \(R_{i}\) is \( \Delta A_{i} = r_{i} \Delta r_{i} \Delta \theta_{i} \:\:\:\: \color{red}{ \text{Area for } R_{i}}\) where \(\Delta r_{i} =r_{2} - r_{1}\) and \( \Delta \theta_{i} = \theta_{2} - \theta_{1} \). This implies that the volume for a solid with a height \(f(r_{i} \cos \theta_{i}, \: r_{i} \sin \theta_{i})\) above \(R_{i}\) is approximately \(f(r_{i} \cos \theta_{i}, \: r_{i} \sin \theta_{i})r_{i} \Delta r_{i} \Delta \theta_{i}\) which produces $$\int_{R} \int f(x,y) \: dA \approx \sum_{i=1}^{n} f(r_{i} \cos \theta_{i}, \: r_{i} \sin \theta_{i})r_{i} \Delta r_{i} \Delta \theta_{i}. $$ The sum on the right can be interpreted as a Riemann sum for \(f(r \cos \theta, \: r \sin \theta)r. \) The region \(R\) corresponds to a horizontally simple region \(S\) in the \(r \theta\)-plane, as shown in Figure 14.3.5. The polar sectors \(R_{i}\) correspond to rectangles \(S_{i}\), and the area \(\Delta A_{i}\) for \(S_{i}\) is \(\Delta r_{i} \Delta \theta_{i}\). The equation's right side corresponds to the double integral $$\int_{S} \int f(r \cos \theta, \: r \sin \theta)r \: dA. $$ Applying Theorem 14.2.2 produces $$\int_{R} \int f(x,y) \: dA $$ $$= \int_{S} \int f(r \cos \theta, \: r \sin \theta)r \: dA$$ $$=\int_{\alpha}^{\beta} \int_{g_{1}(\theta)}^{g_{2}(\theta)} f(r \cos \theta, \: r \sin \theta)r \: dr \: dA. $$ This leads to Theorem 14.3.1 below, which is described in Section 14.8.

Theorem 14.3.1 Changing Variables from Cartesian to Polar Form

Let \(R\) be a plane region with all points \((x,y)=( r \cos \theta, r \sin \theta) \) satisfying the conditions \(0 \leqslant g_{1}(\theta) \leqslant r \leqslant g_{2}(\theta), \: \alpha \leqslant \theta \leqslant \beta\), where \(0 \leqslant ( \beta - \alpha ) \leqslant 2 \pi\). If \(g_{1}\) and \(g_{2}\) are continuous on \([\alpha, \beta]\) and \(f\) is continuous on \(R\), then

$$\int_{R} \int f(x,y) \: dA = \int_{\alpha}^{\beta} \int_{g_{1}(\theta)}^{g_{2}(\theta)} f(r \cos \theta, \: r \sin \theta)r \: dr \: d \theta.$$

If \(z=f(x,y)\) is nonnegative on \(R\), then the integral can be interpreted as the volume for the solid region between the graph for \(f\) and the region \(R\). Always include the extra factor \(r\) in the integrand.

The region \(R\) is restricted to two basic types, \(r\)-simple and \(\theta\)-simple regions, as shown in Figures 14.3.6, and 14.3.7.

\(r\)-Simple region. Figure 14.3.6

\(\theta\)-Simple region. Figure 14.3.7

Example 14.3.2 Evaluating a Double Polar Integral

\(r\)-Simple region Figure 14.3.8

The area for a polar sector has the formula \(dA=r \: dr \: d\theta\) which increases as \(r\) moves from the origin. Note the extra \(r\). Let \(R\) be the annular region lying between the two circles \(x^{2}+y^{2}=1\) and \(x^{2}+y^{2}=5.\) Evaluate the integral $$ \int_{R} \int (x^{2}+y) \:dA.$$ Solution The polar boundaries are \( 1 \leqslant r \leqslant \sqrt{5}\) and \( 0 \leqslant \theta \leqslant 2\pi\), as shown in Figure 14.3.8. Let \(x^{2} = (r \cos \theta)^{2}\) and \(y= r \sin \theta \). This produces $$ \int_{R} \int (x^{2}+y) \:dA $$ $$= \int_{0}^{2\pi} \int_{1}^{\sqrt{5}} (r^{2} \cos^{2} \theta + r \sin \theta)r \:dr \:d\theta $$ $$=\int_{0}^{2\pi} \int_{1}^{\sqrt{5}} (r^{3} \cos^{2} \theta + r^{2} \sin \theta) \:dr \:d\theta $$ $$= \left. \int_{0}^{2\pi} \left( \frac{r^{4}}{4} \cos^{2} \theta + \frac{r^{3}}{3} \sin \theta \right) \right]_{1}^{\sqrt{5}} \: d\theta $$ $$= \int_{0}^{2\pi} \left( 6 \cos^{2} \theta + \frac{5 \sqrt{5}-1}{3} \sin \theta \right) \: d\theta $$ $$= \int_{0}^{2\pi} \left( 3 + 3 \cos 2\theta + \frac{5 \sqrt{5}-1}{3} \sin \theta \right) \: d\theta $$ $$= \left. \left( 3 \theta + \frac{3 \sin 2 \theta}{2} - \frac{5 \sqrt{5}-1}{3} \cos \theta \right) \right]_{0}^{2\pi} = 6 \pi $$

Example 14.3.3 Changing From Cartesian To Polar Coordinates

Figure 14.3.9

Use polar coordinates to find the volume for the solid region bounded by the hemisphere \(z=\sqrt{16-x^{2}-y^{2}}\:\:\:\: \color{red}{ \text{Hemisphere forms upper surface.}} \) and below by the circular region \(R\) given by \( x^{2}+y^{2} \leqslant 4 \:\:\:\: \color{red}{ \text{Circular region forms lower surface.}} \) as shown in Figure 14.3.9. Solution Figure 14.3.9 shows \(R\)'s bounds \( - \sqrt{4-y^{2}} \leqslant x \leqslant \sqrt{4-y^{2}}, \: -2 \leqslant y \leqslant 2\) and that \(0 \leqslant z \leqslant \sqrt{16-x^{2}-y^{2}}\). In polar coordinates the bounds are \( 0 \leqslant r \leqslant 2 \text{ and } 0 \leqslant \theta \leqslant 2 \pi \) with height \(z=\sqrt{16-x^{2}-y^{2}} = \sqrt{16 - r^{2}}\). The volume \(V\) is $$V $$ $$= \int_{R} \int f(x,y) \:dA $$ $$= \int_{0}^{2\pi} \int_{0}^{2} \sqrt{16 - r^{2}} \:r \:dr \: d\theta $$ $$= \left. -\frac{1}{3} \int_{0}^{2\pi} (16-r^{2})^{3/2} \right]_{0}^{2} \: d\theta$$ $$= -\frac{1}{3} \int_{0}^{2\pi} (24\sqrt{3}-64) \: d\theta $$ $$= \left. -\frac{8}{3} (3\sqrt{3}-8) \theta \right]_{0}^{2\pi} $$ $$= \frac{16\pi}{3} (8-3\sqrt{3}) \approx 46.979 $$

Example 14.3.4 Finding Areas for Polar Regions

Figure 14.3.10

Find the area enclosed by the graph for \(r=3 \cos 3\theta\). Let \(R\) be one petal for the curve shown in Figure 14.3.10. This region is \(r\)-simple with boundaries \(-\pi \leqslant \theta \leqslant \pi/6\) and \(0 \leqslant r \leqslant 3 \cos 3 \theta\). The total area is the area for one petal times 3. $$ \frac{1}{3}A $$ $$= \int_{R} \int dA $$ $$=\int_{-\pi/6}^{\pi/6} \int_{0}^{3 \cos 3 \theta} r \: dr \: d\theta$$ $$= \left. \int_{-\pi/6}^{\pi/6} \frac{1}{2} r^{2} \right]_{0}^{3 \cos 3 \theta} \: d\theta $$ $$= \frac{9}{2} \int_{-\pi/6}^{\pi/6} \cos^{2} 3 \theta \: d \theta $$ $$= \frac{9}{4} \int_{-\pi/6}^{\pi/6} (1+\cos 6 \theta) \: d \theta $$ $$= \frac{9}{4} \left[ \theta+ \frac{1}{6}\sin 6 \theta \right]_{-\pi/6}^{\pi/6} = \frac{3 \pi}{4} $$ $$=(3)\frac{3 \pi}{4} = \frac{9\pi}{4}\:\:\:\: \color{red}{ \text{Multiply by 3 }} $$

Example 14.3.5 Changing the Integration Order

\(\theta)-Simple region Figure 14.3.11

Find the area for the region bounded by the spiral $$ r= \frac{\pi}{3\theta}$$ and below by the polar axis between \(r=1\) and \(r=2\), as shown in Figure 14.3.11. Solution Let the polar boundaries be \(1 \leqslant r \leqslant 2\) and $$ 0 \leqslant \theta \leqslant \frac{\pi}{3r}. $$ Evaluating the double integral produces, $$A $$ $$=\int_{1}^{2} \int_{0}^{\pi/3r}r \: d\theta \: dr $$ $$= \left. \int_{1}^{2} r\theta \right]_{0}^{\pi/(3r)} \: dr $$ $$= \int_{1}^{2} \frac{\pi}{3} \:dr$$ $$= \left. \frac{\pi r}{3} \right]_{1}^{2} = \frac{\pi}{3}$$

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Parent Article: Calculus III 14 Multiple Integration