Question

The area (in square units) bounded by the curves $y^2=4x$ and $x^2=4y$ is

• A. $\frac{64}{3}$
• B. $\frac{16}{3}$
• C. $\frac{8}{3}$
• D. $\frac{2}{3}$

Answer 1

Answer: (B)

$\frac{16}{3}$

$y^2=4x$

$x^2=4y$

solving two equations

$\frac{y^2}{4}=x$

${\left(\frac{y^2}{4}\right)}^2=4y$

$\frac{y^4}{4\times 4}=4y$

$y^4=64y$

$y(y^2-64)=0$

$y=0,y^3=64$

$y=4$

${\int }_0^4(\frac{y^2}{4}-2\sqrt{y})dy$

$={[\frac{y^3}{4\times 3}-\frac{y^{\frac{B}{2}}}{3}]}_0^4$

$=\left|\frac{-16}{3}\right|=\frac{16}{3}$