calculusintegrationpolar coordinates
Find the area enclosed by circles $r=3\cos\theta$ and $r=3\sin\theta$
I know that the graph of $r=3\sin\theta$ intersects at $\theta=0$ and at $\theta=\pi$ why aren't these two bounds used and why only $\theta$ equaling $\pi$? Also the graph $r=3\cos\theta$ intersects the pole at $\theta=\frac{3\pi}{2},\frac{1\pi}{2}$. So why do they only use $\frac{\pi}{2}$?
I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.
There are two distinct regions where your curves overlap. They do overlap for the intervals of $\theta$ that you give, but that gives only the large overlap at the upper right of the origin. You left out the smaller overlap at the lower left of the origin.
That overlap is covered by $\frac{5\pi}4\le \theta\le \frac{3\pi}2$ for the $r=1+\cos\theta$ curve and by $\pi\le \theta\le \frac{5\pi}4$ for the $r=1+\sin\theta$ curve. To get the correct answer you will need to include that little piece.
Best Answer
The two circles are intersecting at $\theta=\dfrac{\pi}{4}$ and $r=0$
The area region can be divided into two parts.
Note that we are bounded by the region of the red circle $($the lower part of the half$)$ by $\theta=0\mbox{ to }\theta=\dfrac{\pi}{4}$
Similarly, the region of the purple circle $($the upper part of the half$)$ is bounded by $\theta=\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$
So, we get $$\dfrac12\int_{0}^{\dfrac{\pi}{4}}(3\sin\theta)^2d\theta+\dfrac12\int_{\frac{\pi}{4}}^{\dfrac{\pi}{2}}(3\cos\theta)^2d\theta$$