# Question #ae4f3

Mar 26, 2017

The volume is $53.617$.

#### Explanation:

There is only one curve being rotated, so we can use the disc method. The disc method says that for each value of $x$, the vertical cross section is a circle with an area of $\pi \cdot f {\left(x\right)}^{2}$, since $f \left(x\right)$ is the radius. Therefore:

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

First, we need to find our bounds. Since we are given no other bounds, the bounds must be the zeroes of $y = 4 - 4 {x}^{2}$. So, we set $y$ equal to zero and solve for our two $x$ values.

$0 = 4 - 4 {x}^{2}$
$4 {x}^{2} = 4$
${x}^{2} = 1$
$x = \pm 1$

So, our bounds are $- 1$ and $1$.

Now, all we have left to do is use the disc method formula to find the volume.

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

$= \pi \cdot {\int}_{-} {1}^{1} {\left(4 - 4 {x}^{2}\right)}^{2} \mathrm{dx}$

$= \pi \cdot {\int}_{-} {1}^{1} {4}^{2} \cdot {\left(1 - {x}^{2}\right)}^{2} \mathrm{dx}$

$= 16 \pi \cdot {\int}_{-} {1}^{1} \left({x}^{4} - 2 {x}^{2} + 1\right) \mathrm{dx}$

$= 16 \pi \cdot \left({x}^{5} / 5 - 2 {x}^{3} / 3 + x\right) {|}_{-} {1}^{1}$

$= 16 \pi \cdot \left(\left(\frac{1}{5} - \frac{2}{3} + 1\right) - \left(- \frac{1}{5} + \frac{2}{3} - 1\right)\right)$

$= 16 \pi \cdot \left(\frac{2}{5} - \frac{4}{3} + 2\right)$

$= \frac{256 \pi}{15}$

$= 53.617$