# What is the Volume of Revolution if the area bounded by the curve y=x^2-4x and the x-axis is is rotated about the x-axis?

Mar 5, 2017

$\frac{512 \pi}{15} \setminus u n i {t}^{3}$

#### Explanation:

I recommend that you always draw a sketch to clarify what needs calculating.

graph{x^2-4x [-10, 10, -5, 5]}

The curve intersects the $x$-axis, when:

${x}^{2} - 4 x = 0 \implies x \left(x - 4\right) = 0 \implies x = 0 , 4$

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

$V O R = {\int}_{\alpha}^{\beta} \setminus \pi {y}^{2} \setminus \mathrm{dx}$

So in this case:

$V O R = {\int}_{0}^{4} \setminus \pi \setminus {\left({x}^{2} - 4 x\right)}^{2} \setminus \mathrm{dx}$
$\text{ } = \pi \setminus {\int}_{0}^{4} \setminus \left({x}^{4} - 8 {x}^{3} + 16 {x}^{2}\right) \setminus \mathrm{dx}$
$\text{ } = \pi {\left[{x}^{5} / 5 - 2 {x}^{4} + 16 {x}^{3} / 3\right]}_{0}^{4}$
$\text{ } = \pi \left\{\left({4}^{5} / 5 - 2 \cdot {4}^{4} + 16 \cdot {4}^{3} / 3\right) - \left(0\right)\right\}$
 " " = pi (1024/5-512+1024/3)-(0) }
$\text{ } = \frac{512 \pi}{15}$