Which integral gives the area of the region in the first quadrant that lies below the line $y = 3$ and inside the circle $x^{2} + y^{2} = 25$?

Explanation

1. 01

   The region lies against the $y$-axis from $y = 0$ to $y = 3$; at height $y$ a strip runs from $x = 0$ to the circle, $x = \sqrt{25 - y^{2}}$.

2. 02

   So the area is $\int_0^{3} \sqrt{25 - y^{2}}\,dy = 6 + \frac{25}{2}\sin^{-1}\frac{3}{5}$.