91Ó°ÊÓ

We are given$y=x^2$and$y=0$and$[-3,3]$.

The region of integration is as below:

First ${\mathrm{Î\copyright }}_1$is bounded by $x=-\sqrt{y},x=-3$and $x$axis.

Hence, $-3≤x≤\sqrt{y},0≤y≤9$

Similarly ${\mathrm{Î\copyright }}_2$is bounded by $x=\sqrt{y},x=-3$and $x$axis.

Hence,$\sqrt{y}≤x≤3,0≤y≤9$

It is written as:

${âˆ\neg }_{\mathrm{Î\copyright }}dA={âˆ\neg }_{{\mathrm{Î\copyright }}_1}dA+{âˆ\neg }_{{\mathrm{Î\copyright }}_2}dA$

$⇒{âˆ\neg }_{\mathrm{Î\copyright }}dA={∫}_0^{9-\sqrt{y}}{∫}_{-3}^{9}dxdy+{∫}_0^3{∫}_{\sqrt{y}}dxdy$

The region where first integration is done is

From above graph

for ${\mathrm{Î\copyright }}_1$, role="math" localid="1654028615727" $0≤y≤x^2,-3≤x≤0$

for role="math" localid="1654028593706" ${\mathrm{Î\copyright }}_2$,$0≤y≤x^2,0≤x≤3$

Area is written as

${âˆ\neg }_{\mathrm{Î\copyright }}dA={âˆ\neg }_{{\mathrm{Î\copyright }}_4}dA+{âˆ\neg }_{{\mathrm{Î\copyright }}_2}dA$

Using values ${âˆ\neg }_{\mathrm{Î\copyright }}dA={∫}_{-3}^0{∫}_0^{x^2}dydx+{∫}_0^3{∫}_0^{x^2}dydx$

Area of region is given by

${∫}_{-3}^0{∫}_0^{x^2}dydx+{∫}_0^3{∫}_0^{x^2}dydx$

Solving integrals

${âˆ\neg }_{\mathrm{Î\copyright }}dA={∫}_{-3}^0[y{]}_0^{x^2}dx+{∫}_0^3[y{]}_0^{x^2}dx$

$={∫}_{-3}^0{x}^{2}dx+{∫}_0^3{x}^{2}dx$

$={\left[\frac{x^3}{3}\right]}_{-3}^0+{\left[\frac{x^3}{3}\right]}_0^3$

$=18$

Area of region is$18$units