Find the area enclosed by curve with polar coordinates

Question

I am having a little difficulty finding the area enclosed by the curve, $r(\theta) = 4 + sin\theta + cos\theta$ with $0 \le \theta \le 2\pi$. I tried integrating over $0 \le \theta \le 2\pi$ and $0 \le r \le 4 + \sqrt2$, but I keep getting 16$\pi$, while the correct answer should be 17$\pi$. Anyone has an idea what might have gone wrong and what the correct method is? Would appreciate it!

Answer 1

Answer:

Area $ = \int_{0}^{2\pi} \int_{0}^{4+sin\theta + cos\theta} rdrd\theta$

This reduces to

$$A = \int_{0}^{2\pi} \frac{r^2}{2}d\theta$$ $$A = \int_{0}^{2\pi}\frac{1}{2} \left[16+sin^2\theta + cos^2\theta + sin2\theta + 8sin\theta+8cos\theta\right]d\theta$$ $$A = \int_{0}^{2\pi}\frac{1}{2} \left[17+ sin2\theta + 8sin\theta+8cos\theta\right]d\theta$$ $$A = \frac{1}{2} \left[17(2\pi+\frac{1}{2}cos2\theta|_{2\pi}^{0} + 8cos\theta|_{2\pi}^{0} +8sin\theta|_{0}^{2\pi}\right]$$

If you evaluate this you get A $ = 17\pi$