Find all points of intersection $(r,\theta )$ of the curves $r=5\cos(\theta),\ \ r= 4 \sin(\theta)$.
In this problem the curves intersect at the pole and one other point. Only enter the answer for nonzero $r$ in the form $(r,\theta)$ with $\theta$ measured in radians.
Find the area inclosed in the intersection of the two graphs.
I understand that if I make them equal I can find the intersection in radians which is $4/5*tan\theta=1$ but I dont understand the $r$ part or how to find the area.
To complement the answer by @Ross Millikan.
It is very important for such a problem to have a figure; I have displayed one below.
The unique non zero solution modulo $2\pi$ to equation $5 \cos(\theta)=4\sin(\theta)$ is
$$\tag{1}\theta_0:=\arctan(5/4).$$
Using the usual formula $\frac12\int r^2(\theta)d\theta$ for the area enclosed by a curve described by a polar equation (take care that we use the red curve first, then the blue curve):
$$A=\frac12\int_{0}^{\pi/2} min (4 \sin(\theta),5 \cos(\theta))^2d\theta$$
$$A=\frac12\int_0^{\theta_0} (4 \sin(\theta))^2d\theta+\frac12\int_{\theta_0}^{\pi/2} (5 \cos(\theta))^2d\theta.$$
Using formulas $cos^2(\theta)=\frac12(1+\cos(2\theta))$ and $sin^2(\theta)=\frac12(1-\cos(2\theta))$, one obtains:
$$A=\frac{16}{4}\int_0^{\theta_0} (1-\cos(2\theta))d\theta+\frac{25}{4}\int_{\theta_0}^{\pi/2} (1+\cos(2\theta))d\theta.$$
$$A=\frac{16}{4}(\theta_0-\frac12 \sin(2\theta_0))+\frac{25}{4}(\frac{\pi}{2}-\theta_0-\frac12 \sin(2\theta_0)).$$
$$A=\frac{25\pi}{8} - \frac{9}{4}\theta_0-\frac{41}{8} \sin(2\theta_0) .$$
Using (1): $A \approx 3.6888$
Once you have $\frac 45 \tan \theta=1$ you can determine $\theta$. You can plug that $\theta$ into either of your original equations to get $r$.
For the area, have you plotted the two graphs? You have two circles. You are to find the area of the intersection of the two circles. You can do that with geometry, knowing the radii and distance between the centers. Draw a line through the two intersection points and you get two circular segments.