Here's my given problem:
Compute the area of the region of the graph of r=4sinθ and the graph of r=4cosθ.
I know from
r=2acosθ and r=2asinθ
that the centers are (2,0) and (0,2). I also know graphically that these two equations are overlapping circles. I set the equations equal to one another to get the points of intersection, which are 0 or 2π and 3π/2 for 4sinθ and π/2 and π for 4cosθ (frankly I am still not entirely sure how to choose which function's angles for integration bounds).
Just broadly/conceptually, what should I try to look for or integrate to find this overlapping area? I'm having trouble figuring out how to get those properties of that shape. Several of the integrations I tried just equaled zero, and common sense tells me there's more to it than getting the area of a sector of one of the circles because the zone of interest doesn't go all the way to the centers of either of the circles.
You want to find the area of the gray part in the following figure :
$\qquad\qquad\qquad$
Since $A,B$ is a point for $\theta=\frac{\pi}{4},\frac{\pi}{2}$ in $r=4\cos\theta$ respectively, and $A,B$ is a point for $\theta=\frac{\pi}{4},0$ in $r=4\sin\theta$ respectively, the area is $$\int_{\pi/4}^{\pi/2}\frac 12(4\cos\theta)^2d\theta+\int_{0}^{\pi/4}\frac 12(4\sin\theta)^2d\theta=\color{red}{2\pi -4}.$$