# How do you find the volume of region bounded by graphs of y = x^2 and y = sqrt x about the x-axis?

##### 1 Answer
Mar 15, 2018

$\textcolor{b l u e}{\frac{\pi}{3} \text{cubic units.}}$

#### Explanation:

From the graph we can see that the volume we seek is between the two functions. In order to find this, we must find the volume of revolution of $f \left(x\right) = \sqrt{x}$ and subtract the volume of revolution of $f \left(x\right) = {x}^{2}$. This is shown as the shaded area.

First we need to find the upper and lower bounds. We know the lower bound is $0$ since $f \left(x\right) = \sqrt{x}$ is undefined for $x < 0$. The upper bound is where the functions intersect:

$\therefore$

${x}^{2} = \sqrt{x}$

${x}^{2} / {x}^{\frac{1}{2}} = 1$

${x}^{\frac{3}{2}} = 1$

Squaring:

${x}^{3} = 1$

$x = \sqrt[3]{1} = 1$

Volume of $\boldsymbol{f \left(x\right) = \sqrt{x}}$:

$\pi {\int}_{0}^{1} \left({x}^{\frac{1}{2}}\right) = \pi {\left[\frac{2}{3} {x}^{\frac{3}{2}}\right]}_{0}^{1}$

$= \pi \left\{{\left[\frac{2}{3} {x}^{\frac{3}{2}}\right]}^{1} - {\left[\frac{2}{3} {x}^{\frac{3}{2}}\right]}_{0}\right\}$

Plugging in upper and lower bounds:

$= \pi \left\{{\left[\frac{2}{3} {\left(1\right)}^{\frac{3}{2}}\right]}^{1} - {\left[\frac{2}{3} {\left(0\right)}^{\frac{3}{2}}\right]}_{0}\right\} = \frac{2 \pi}{3}$ cubic units

Volume of $\boldsymbol{f \left(x\right) = {x}^{2}}$

$\pi {\int}_{0}^{1} \left({x}^{2}\right) = \pi {\left[\frac{1}{3} {x}^{3}\right]}_{0}^{1}$

=pi{[[1/3x^3]^(1)-[1/3x^3]_(0)}

Plugging in upper and lower bounds:

=pi{[[1/3(1)^3]^(1)-[1/3(0)^3]_(0)}=pi/3 cubic units.

Required volume is:

$\frac{2 \pi}{3} - \frac{\pi}{3} =$$\textcolor{b l u e}{\frac{\pi}{3} \text{cubic units.}}$

Volume of revolution: