Given that y = x and ${\mathit{x}}^2+{\mathit{y}}^2$ = 32
Put y = x in ${\mathit{x}}^2+{\mathit{y}}^2$ = 32
${\mathit{x}}^2+{\mathit{y}}^2$ = 32
⇒ $2{\mathit{x}}^2$ = 32
⇒ x = ± 4 = y
Hence, point s of intersection are A (4,4) and B (4,-4) .
Required shaded area
= $\underset{0}{\overset{4}{\int }}$ x dx + $\underset{4}{\overset{\sqrt{32}}{\int }}$ $\sqrt{32-{\mathit{x}}^2}$ dx (Since Region in the first quadrant)
= ${\left(\frac{{\mathit{x}}^2}{4}\right)}_0^4$ +
= 4π sq . units