Mathematics - Area Under Curves Question with Solution | TestHub

To find the points of intersection of the circle $x^2+y^2=9$ and the parabola $y^2=8x$, substitute $y^2=8x$ into the equation of the circle:

$x^2+8x-9=0$

$(x+9)(x-1)=0$

Since $x\ge 0$ for the parabola, we get $x=1$.

The required region is bounded by $x^2+y^2\le 9$ and $y^2\le 8x$. This region is symmetric with respect to the x-axis. The total area $A$ is given by:

$A=2\left({\int }_0^1\sqrt{8x}dx+{\int }_1^3\sqrt{9-x^2}dx\right)$

Evaluating the first integral:

${\int }_0^1\sqrt{8x}dx=2\sqrt{2}{\left[\frac{2}{3}{x}^{3/2}\right]}_0^1=\frac{4\sqrt{2}}{3}$

Evaluating the second integral:

${\int }_1^3\sqrt{9-x^2}dx={\left[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{x}{3}\right)\right]}_1^3$

Substituting the limits:

$=\left(0+\frac{9}{2}{\sin }^{-1}(1)\right)-\left(\frac{1}{2}\sqrt{8}+\frac{9}{2}{\sin }^{-1}\left(\frac{1}{3}\right)\right)$

$=\frac{9\pi }{4}-\sqrt{2}-\frac{9}{2}{\sin }^{-1}\left(\frac{1}{3}\right)$

Adding the two integrals and multiplying by 2:

$A=2\left(\frac{4\sqrt{2}}{3}+\frac{9\pi }{4}-\sqrt{2}-\frac{9}{2}{\sin }^{-1}\left(\frac{1}{3}\right)\right)$

$A=\frac{8\sqrt{2}}{3}+\frac{9\pi }{2}-2\sqrt{2}-9{\sin }^{-1}\left(\frac{1}{3}\right)$

Answer: $\frac{8\sqrt{2}}{3}+\frac{9\pi }{2}-2\sqrt{2}-9{\sin }^{-1}\frac{1}{3}$