What is the area of the portion of the circle defined by x^2 - 12x + y^2 = 28 that lies above the x-axis and to the right of the line y = 8 - x?

Guest Jan 28, 2021

#1**+1 **

We first find the intersection point of x^2 - 12x + y^2 = 28 and y = 8 - x.

\(\begin{cases} x^2 - 12x + y^2 = 28\\ y = 8-x \end{cases}\)

After some work, we get (x, y) = \((7-\sqrt{31}, 1+\sqrt{31})\). We don't need the other intersection point.

We also get that the region is represented by the compound inequality \(\max(0, 8- x)\leq y \leq \sqrt{28 + 12x - x^2}\).

We use the straight line x = 8 to divide the region into 2 parts.

Using integration, the area is

\(\displaystyle \int^{8}_{7-\sqrt{31}}\left(\sqrt{28 +12x - x^2}-(8-x)\right)\,dx+\int^{14}_8 \sqrt{28 + 12x - x^2}\,dx\)

which would be \(16\pi + 32\arcsin\dfrac{\sqrt{31} - 1}{8} - \sqrt{31} - 1 \approx 63.14\text{ sq.units}\)

The computation is too long so I didn't include it.

Here's a verification of my answer by Wolfram Alpha:

This is the diagram I used:

MaxWong Jan 28, 2021