Question

The area included between the parabolas $x^2=4y$ and $y^2=4x$ is (in square units)

• A. $\frac{4}{3}$
• B. $\frac{1}{3}$
• C. $\frac{16}{3}$
• D. $\frac{8}{3}$

Answer 1

Answer: (C)

$\frac{16}{3}$

One intersection point of the two parabolas is the origin, while the other can be found out by substituting either equation into the other.

Thus we have ${\left(\frac{x^2}{4}\right)}^2=4x$

which yields $x^4=64x$ i.e. $x=0,4$

Area included between the two curves can be found out by ${\int }_0^4(2\sqrt{x}-\frac{x^2}{4})dx$

$={[\frac{4x^{1.5}}{3}-\frac{x^3}{12}]}_0^4$

$=\frac{32}{3}-\frac{64}{12}$

$=\frac{32}{6}$