Area of a region bounded by a circle and vertical line using integration

Question

What is the area of the right region bounded by $x^2 + y^2 =25$ and $x = -3$?

My attempt solution: I solved first for the area of the left region and subtracted it from the area of the circle, which is

$$ 25 \pi - \int_{-4}^4 -3 + \sqrt {25-y^2} \,dy $$ $$ 78.5 - 11.18... = 67.36 $$ So, the area of the right region is 67.36.

My other solution is: $$ \int_{0}^5 \sqrt {25-y^2} \,dy + \int_{0}^4 -(-3)\,dy + \int_{4}^5 -(-\sqrt {25-y^2})= 67.36$$

Are the area (67.36) and the integration process all correct? Is there more precise integration for the area of the right bounded region, without using double or triple integration?

Answer 1

I think both of your approaches are correct.

I went about it the following way:

$25\pi - 2 \cdot \int_{-5}^{-3} \sqrt{(-5)^2 - x^2} = 67.36$. When integrating I got $arcsin(3/5)$, which is where I rounded my values. I do not think, there is a way to avoid this other than leaving arcsin(3/5) in your reslut, if you want it to be more precise.