# How do you find the volume of the solid y=x^2, y=x^3 revolved about the x-axis?

Jan 18, 2017

Volume  (2pi)/35 " unit^3

#### Explanation:

graph{(y-x^2)(y-x^3)=0 [-0.1, 1.5, -0.1, 1.5]}

The Volume of Revolution about $O x$ is given by:

$V = {\int}_{x = a}^{x = b} \setminus \pi {y}^{2} \setminus \mathrm{dx}$

So, the volume of revolution bounded by two curves $f \left(x\right)$ and $g \left(x\right)$ with $f \left(x\right) > g \left(x\right)$ is given by the difference between the individual solids of revolution, thus:

$V = {\int}_{x = a}^{x = b} \setminus \pi {f}^{2} \left(x\right) \setminus \mathrm{dx} - {\int}_{x = a}^{x = b} \setminus \pi {g}^{2} \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus = {\int}_{x = a}^{x = b} \setminus \pi \left\{{f}^{2} \left(x\right) - {g}^{2} \left(x\right)\right\} \setminus \mathrm{dx}$

We can see by observation that the points of intersection of the curves $y = {x}^{2}$ and $y = {x}^{3}$ are $\left(0 , 0\right)$ and $\left(1 , 1\right)$.

So for for this problem:

$V = {\int}_{0}^{1} \setminus \pi \left\{{\left({x}^{2}\right)}^{2} - {\left({x}^{3}\right)}^{2}\right\} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \pi \setminus {\int}_{0}^{1} \setminus {x}^{4} - {x}^{6} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \pi \setminus {\left[{x}^{5} / 5 - {x}^{7} / 7\right]}_{0}^{1}$
$\setminus \setminus \setminus = \pi \setminus \left\{\left(\frac{1}{5} - \frac{1}{7}\right) - \left(0\right)\right\}$
$\setminus \setminus \setminus = \pi \cdot \frac{2}{35}$

$\setminus \setminus \setminus = \frac{2 \pi}{35}$