Question:

The area of the region in the first quadrant inside the circle $x^2+y^2=8$ and outside the parabola $y^2=2x$ is equal to:

$\dfrac{\pi}{2}-\dfrac{1}{3}$

$\pi-\dfrac{1}{3}$

$\dfrac{\pi}{2}-\dfrac{2}{3}$

$\pi-\dfrac{2}{3}$

Step-by-Step Solution

Key Concept: $\int_0^2(\sqrt{8-x^2}-\sqrt{2x})dx=\pi+2-8/3=\pi-2/3$.

Area $=\pi-2/3$.

Correct Answer: 4