Consider the unit circle $x^2 + y^2 = 1$ and the line $y = 3x$. I want to compute the area of the portion of the circle enclosed above the line $y = 0$ but to the left of the line $y = 3x$. In other words, I require $y > 3x$ and $y > 0$ to hold.
NOTE: Although the shaded blue area in the image above goes outside of the circle, I only want the area of the portion enclosed inside of the circle.
Is there any nice geometric way to solve this problem? I tried to solve it with a double integral, but I can't quite figure out the bounds. I'm happy with any solution. I know that the entire second quadrant has area $\pi/4$, but figuring out the last bit is difficult.
I thought about finding the angle from the line $x = 0$ to $y = 3x$ by drawing a right triangle with side-lengths $3$ and $1$. This means that the angle is $\theta = \arctan(1/3)$, but how can I figure out the area from here?
So one easy way is to find the angle the line hits the positive $x$-axis at, which is $t = \arctan 3$ and in polar, your area becomes $$ A = \int_{\theta = t}^{\theta = \pi} \int_{r=0}^{r=1} rdrd\theta $$ and volume where the height is given by $h(r,\theta)$ is $$ V = \int_{\theta = t}^{\theta = \pi} \int_{r=0}^{r=1} h(r,\theta) rdrd\theta. $$
If you are not interested in the volume and just need the area, it is proportional to area of the circle (which is $A_\circ = \pi$) like the ratio of angles, i.e. $$ A = \frac{\pi - t}{2\pi} A_\circ = \frac{\pi - t}{2}. $$