I have been trying to solve this question for a while and I seem to get stuck although it does not seem hard Given $r\cos\theta=6$ and $r=10$ we need to find the area inside the circle and on the right of $r\cos\theta=6$. I am self-learning calculus with analytic geometry so I don't know if my ways are correct.
first I took $r=10$ and made it $x^2+y^2=100$ the center is $(0,0)$ and radius is $10$ and also $r\cos\theta=6$ to $x=6$, I found where they intersect and got $6^2+y^2=100$ and we get $y=8$ so the intersection point is $(6,8)$. since we know that we have $x=6$ and the radius of the circle is $10$ we can use Pythagorean theory to find the area of a triangle inside the circle, we get that the area of the triangle is $0.5\times8\times4=16$.
which leaves a small area between the triangle and the circle which I cannot seem to find. I am still not sure how to find the needed $\theta$ for the integral and I am not sure how to use it.
The final answer according to the book should be $100\left(\cos^{-1}\left(\frac3{5}\right)\right)-48$ thanks for any tips and help!
Let $A_\Delta=\frac{1}{2} \cdot 6 \cdot 8 = 24=$ area of triangle and $A_\text{circle}=100\pi$ be the area of the circle.
The area of the circle to the left of the line is thus
$$A_\text{L} = 100\left(\pi - \cos^{-1} \frac{3}{5} \right) + 2 A_\Delta$$
The area to the right is
$$A_\text{circle} - A_\text{L} = 100 \cos^{-1}\frac{3}{5} - 48.$$