Question

Find the area of the region bounded by the curves
$x^2+y^2=16$ and $x^2=6y$.

Answer 1

We have,

$x^2+y^2=16......\left(1\right)$

$x^2=6y......\left(2\right)$

From equation (1) and (2) to,

$y^2+6y=16$

$y^2+6y-16=0$

$y^2+(8-2)y-16=0$

$y^2+8y-2y-16=0$

$y(y+8)-2(y+8)=0$

$(y+8)(y-2)=0$

If

$y+8=0$

$y=-8$

$y-2=0$

$y=2$

$y=-8$ as parabola lies in first or fourth quadrant.

We get,

When, $y=2$ put in equation (2) and we get,

$x^2=6y$

$x^2=6\times 2$

$x^2=12$

$x=2\sqrt{3}$

By equation (1) and (2) intersect $p(2,2\sqrt{3})$

$=16\pi -2{\int }_0^{2}ydx-{\int }_2^{4}ydx$

$=16\pi -2{\int }_0^2\sqrt{6}{x}^{\frac{1}{2}}dx-2{\int }_2^4\sqrt{16-x^2}dx$

$=16\pi -2\sqrt{6}{{\left[\frac{2}{3}{x}^{\frac{3}{2}}\right]}_0}^2-2[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}]$

$=16\pi -\frac{4}{3}\sqrt{6}{\left(2\right)}^{\frac{3}{2}}-2[\frac{4}{2}\sqrt{16-16}+\frac{16}{2}{\sin }^{-1}1-\frac{2}{2}\sqrt{16-4}-\frac{16}{2}{\sin }^{-1}\frac{2}{4}]$

$=16\pi -\frac{16}{\sqrt{3}}-16\times \frac{\pi }{2}+4\sqrt{3}+\frac{8\pi }{3}$

$=\frac{-4}{\sqrt{3}}+\frac{32\pi }{3}$

$=\frac{4}{3}(8\pi -\sqrt{3})$

Hence, this is the answer.