Question

The area bounded by $x^2=4ay$ and $y=2a$ is?

• A. $\frac{16\sqrt{2}{a}^2}{3}$
• B. $\frac{16a^2}{3}$
• C. $\frac{8a^2}{3}$
• D. $\frac{8\sqrt{2}{a}^2}{3}$

Answer 1

The correct option is A $\frac{16\sqrt{2}{a}^2}{3}$

REF.Image

$x^2=4ay,y=2a$

$A=A_1+A_2$

$=2A_2$

$=2{\int }_0^{2\sqrt{2}a}(2a-\frac{x^2}{4a})dx$

$=2[2ax-\frac{x^3}{4a\times 3}{]}_0^{2\sqrt{2}a}$

$=2[4\sqrt{2}{a}^2-\frac{4a^2\sqrt{2}}{3}]$

$=\frac{16}{3}\sqrt{2}{a}^2$