# How do you find the volume bounded by y^2=x^3 and y=x^2 revolved about the y-axis?

Jun 10, 2018

$\frac{\pi}{20}$ units ^3

#### Explanation:

We need to find where these two curves intersect to find the bounds of integration.

${y}^{2} = {x}^{3} \mathmr{and} y = {x}^{2}$, squaring the second expression, ${y}^{2} = {x}^{4}$ Solving for ${y}^{2}$,.......... [${x}^{4} = {x}^{3}$] i.e, ${x}^{3} \left[x - 1\right] = 0$.

So $x = 1 , x = 0$ are the points of intersection.

From the graphs of these expressions in can be seen that $y = \sqrt{{x}^{3}}$ has a greater area than $y = {x}^{2}$ so we must find the area under $y = {x}^{2}$ and subtract it from the area under $y = \sqrt{{x}^{3}}$ and then revolve this area about the $x$ axis between the bounds $x = 1 , x = 0$

Volume of revolution is given by $\pi {\int}_{a}^{b} {y}^{2} \mathrm{dx}$ So the volume of revolution = [piint_0^1x^3dx -  piint_0^1x^4dx].= $\pi \left[{x}^{4} / 4 - {x}^{5} / 5\right]$ [ after integration by the general power rule] and evaluated for $x = 1 , x = 0$ will result in $\pi \left[\frac{1}{4} - \frac{1}{5}\right]$ = $\frac{\pi}{20}$.