Question

The area bounded by the curves $y^2=4x$ and $x^2=4y$ is :

• A. $\frac{32}{3}$
• B. $\frac{16}{3}$
• C. $\frac{8}{3}$
• D. $0$

Answer 1

Answer: (B)

Area $=\underset{0}{\overset{4}{\int }}\left(2\sqrt{x}-\frac{x^2}{4}\right)dx$
$={\left(2\left(\frac{x^{3/2}}{3/2}\right)-\frac{x^3}{12}\right)}_0^4$
$=\frac{4}{3}\times 8-\frac{64}{12}$
$=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$