Double Integrals in Polar Coordinates
- Page ID
- 21041
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As with rectangular coordinates, we can also use polar coordinates to find areas of certain regions using a double integral. As before, we need to understand the region whose area we want to compute. Sketching a graph and identifying the region can be helpful to realize the limits of integration. Generally, the area formula in double integration will look like
\[\text{Area of} \, A = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} 1 \,r \, dr \, d\theta. \nonumber \]
Example \(\PageIndex{6A}\): Finding an Area Using a Double Integral in Polar Coordinates
Evaluate the area bounded by the curve \(r = \cos \, 4\theta\).
Solution
Sketching the graph of the function \(r = \cos \, 4\theta\) reveals that it is a polar rose with eight petals (see the following figure).
Using symmetry, we can see that we need to find the area of one petal and then multiply it by 8. Notice that the values of \(\theta\) for which the graph passes through the origin are the zeros of the function \(\cos \, 4\theta\), and these are odd multiples of \(\pi/8\). Thus, one of the petals corresponds to the values of \(\theta\) in the interval \([-\pi/8, \pi/8]\). Therefore, the area bounded by the curve \(r = \cos \, 4\theta\) is
\[\begin{align*} A &= 8 \int_{\theta=-\pi/8}^{\theta=\pi/8} \int_{r=0}^{r=\cos \, 4\theta} 1\,r \, dr \, d\theta \\ &= 8 \int_{\theta=-\pi/8}^{\theta=\pi/8}\left.\left[\frac{1}{2}r^2\right|_0^{\cos \, 4\theta}\right] d\theta \\ &= 8 \int_{-\pi/8}^{\pi/8} \frac{1}{2} \cos^24\theta \, d\theta \\&= 8\left. \left[\frac{1}{4} \theta + \frac{1}{16} \sin \, 4\theta \, \cos \, 4\theta \right|_{-\pi/8}^{\pi/8}\right] \\&= 8 \left[\frac{\pi}{16}\right] = \frac{\pi}{2}\; \text{units}^2. \end{align*}\]