To find the area of the region outside $r = 3 + 4 \sin(θ)$ and inside $r = 5 + 2 \sin(θ)$ from $0 \leq θ \leq π/2$

Question

The answer I got is wrong. I started by finding the derivative and then substituting my values for theta, not sure if that is right.

Answer 1

Please see the graph between $0$ and $\frac{\pi}{2}$. You need to simply consider the lower limit and upper limit of the radius between two curves.

The integral is simply

$\displaystyle \int_0^{\pi/2}\int_{3+4\sin\theta}^{5+2\sin\theta} r \ dr \ d\theta$