What is the area enclosed by #r=2cos((5theta)/3-(13pi)/8)-3sin((5theta)/8+(pi)/4) # between #theta in [0,pi]#?

Question

Ezekiel Alexander · Accepted Answer

See the answer below:

Adam Adkins · Answer

undefined To find the area enclosed by the curve $r = 2\cos\left(\frac{5\theta}{3} - \frac{13\pi}{8}\right) - 3\sin\left(\frac{5\theta}{8} + \frac{\pi}{4}\right)$ between $\theta$ in $[0, \pi]$, you need to integrate the expression $r$ with respect to $\theta$ and then take half of the result due to symmetry.

The formula for the area enclosed by a polar curve is given by:

$$A = \frac{1}{2}\int_{\alpha}^{\beta} r^2 d\theta$$

where $r$ is the polar equation of the curve, and $\alpha$ and $\beta$ represent the limits of integration.

In this case, $r = 2\cos\left(\frac{5\theta}{3} - \frac{13\pi}{8}\right) - 3\sin\left(\frac{5\theta}{8} + \frac{\pi}{4}\right)$ and the limits of integration are $\alpha = 0$ and $\beta = \pi$.

You need to compute $r^2$ and then integrate it with respect to $\theta$ from 0 to $\pi$. After integrating, multiply the result by $1/2$ to find the area enclosed by the curve within the given limits.