Examples

Example 4.2.3   Find the area enclosed by one leaf of the four-leaved rose $r=\cos(2\theta)$.

Solution: We need the boundaries of integration. Start at $\theta=-\pi/4$ and go to $\theta=\pi/4$. As a check, note that $\cos((-\pi/4) \cdot 2) = 0 = \cos((\pi/4) \cdot 2).$ We evaluate

$$\frac{1}{2} \cdot \int_{-\pi/4}^{\pi/4} \cos(2\theta)^2 d\theta = \int_{0}^{\pi/4} \cos(2\theta)^2 d\theta$$

$$= \frac{1}{2} \int_{0}^{\pi/4} (1+\cos(4\theta)) d\theta$$

$$= \frac{1}{2} \left[ \theta + \frac{1}{4}\cdot \sin(4\theta)\right]_{0}^{\pi/4}$$

$$= \frac{\pi}{8}.$$

We used that

$$\cos^2(x) = (1+\cos(2x))/2 \qquad \sin^2(x) = (1-\sin(2x))/2,$$ (4.6)

which follow from

$$\cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x) - 1 = 1-2\sin^2(x).$$

Example 4.2.4   Find area of region inside the curve $r=3\cos(\theta)$ and outside the cardiod curve $r=1+\cos(\theta)$.

Figure: Graph of $r=3\cos(\theta)$ and $r=1+\cos(\theta)$

Solution: This is the same as before. It's the difference of two areas. Figure out the limits, which are where the curves intersect, i.e., the $\theta$ such that

$$3\cos(\theta) = 1+\cos(\theta).$$

Solving, $2\cos(\theta) = 1$, so $\cos(\theta) = 1/2$, hence $\theta = \pi/3$ and $\theta = -\pi/3$. Thus the area is

$$A = \frac{1}{2} \int_{-\pi/3}^{\pi/3} (3\cos(\theta))^2 - (1+\cos(\theta))^2 d\theta$$

$$= \int_{0}^{\pi/3} (3\cos(\theta))^2 - (1+\cos(\theta))^2 d\theta$$

$$= \int_{0}^{\pi/3} (8 \cos^2(\theta) -2 \cos(\theta) -1) d\theta$$

$$= \int_{0}^{\pi/3} \left(8 \cdot \frac{1}{2}\left(1+\cos(2\theta)\right) -2 \cos(\theta) -1\right) d\theta$$

$$= \int_{0}^{\pi/3} 3 + 4 \cos(2\theta) - 2\cos(\theta) d\theta$$

$$= \Bigl[3\theta + 2 \sin(2\theta) - 2\sin(\theta)\Bigr]_{0}^{\pi/3}$$

$$= \pi + 2\cdot \sqrt{\frac{3}{2}} - 2\sqrt{\frac{3}{2}} - 0 - 2\cdot 0 - 2\cdot 0$$

$$= \pi$$