Question

If the area enclosed by the parabolas $y=a-x^2$ and $y=x^2$ is $18\sqrt{2}$ sq.units. Find the value of 'a'

• A. 1
• B. 2
• C. 5
• D. 9

Answer 1

The correct option is D 9

The intersection point is $(\sqrt{\frac{a}{2}},\frac{a}{2}),(-\sqrt{\frac{a}{2}},\frac{a}{2})$
$\int y_1-y_2.dx$
$\int a-2x^2.dx$
$=[ax-\frac{2x^3}{3}{]}_{-\sqrt{\frac{a}{2}}}^{\sqrt{\frac{a}{2}}}$
$=2[ax-\frac{2x^3}{3}{]}_0^{\sqrt{\frac{a}{2}}}$
$2[\frac{a\sqrt{a}}{\sqrt{2}}-\frac{2a\sqrt{a}}{6\sqrt{2}}]$
$=\sqrt{2}[a\sqrt{a}-\frac{a\sqrt{a}}{3}]$
$=\frac{2\sqrt{2}a\sqrt{a}}{3}=18\sqrt{2}$
$2a\sqrt{a}=54$
$a\sqrt{a}=27=3^3$
Squaring both sides, we get
$a^3=3^6$
$a=3^2$
Hence
$a=9$.