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. $a=-9$
• B. $a=6$
• C. $a=9$
• D. $a=-6$

Answer 1

The correct option is C $a=9$

For the point of intersection
$x^2=a-x^2$
$2x^2=a$
$x=\pm \sqrt{\frac{a}{2}}$
Hence the required area will be
$={\int }_{-\sqrt{\frac{a}{2}}}^{\sqrt{\frac{a}{2}}}{x}^2-(a-x^2).dx$
$={\int }_{-\sqrt{\frac{a}{2}}}^{\sqrt{\frac{a}{2}}}2x^2-a.dx$
$=2{\int }_0^{\sqrt{\frac{a}{2}}}2x^2-a.dx$
$=2[\frac{2x^3}{3}-ax{]}_0^{\sqrt{\frac{a}{2}}}$
$=2[\frac{2.a\sqrt{a}}{6.\sqrt{2}}-\frac{a\sqrt{a}}{\sqrt{2}}]$
$=2[\frac{a\sqrt{a}}{3\sqrt{2}}-\frac{a\sqrt{a}}{\sqrt{2}}]$
$=18\sqrt{2}$
$|\frac{a\sqrt{a}}{3\sqrt{2}}-\frac{a\sqrt{a}}{\sqrt{2}}|=9\sqrt{2}$
$\frac{2a\sqrt{a}}{3\sqrt{2}}=9\sqrt{2}$
$a\sqrt{a}=27$
$a^{\frac{3}{2}}=3^3$
$a=3^2$
Hence $a=9$.