Question

Let R be the region bounded by the graph of x^2 + y^2 = 49. Find the positive value of c (nearest 0.001) such that the vertical line x=c divides R into two regions where one region has exactly twice the area of the other.

Answer 1

\[y = f(x) =\sqrt{ 49-x^2}; 0\le x\le C\]
\[\int_{0}^{C}{\sqrt{ 49-x^2}dx} = \frac{49\pi}{12}\]