# How do you find the volume of the solid generated by revolving the region bounded by the curves y=x^(2)-x, y=3-x^(2) rotated about the y=4?

Jul 10, 2015

$V = \frac{685}{32} \pi$ cubic units

#### Explanation:

First, sketch the graphs.

${y}_{1} = {x}^{2} - x$
${y}_{2} = 3 - {x}^{2}$

$x$-intercept
${y}_{1} = 0 \implies {x}^{2} - x = 0$ And we have that $\left\{\begin{matrix}x = 0 \\ x = 1\end{matrix}\right.$
So intercepts are $\left(0 , 0\right)$ and $\left(1 , 0\right)$

Get the vertex:
${y}_{1} = {x}^{2} - x \implies {y}_{1} = {\left(x - \frac{1}{2}\right)}^{2} - \frac{1}{4} \implies {y}_{1} - \left(- \frac{1}{4}\right) = {\left(x - \frac{1}{2}\right)}^{2}$

So vertex is at $\left(\frac{1}{2} , - \frac{1}{4}\right)$

Repeat previous:
${y}_{2} = 0 \implies 3 - {x}^{2} = 0$ And we have that $\left\{\begin{matrix}x = \sqrt{3} \\ x = - \sqrt{3}\end{matrix}\right.$
So intercepts are $\left(\sqrt{3} , 0\right)$ and $\left(- \sqrt{3} , 0\right)$

${y}_{2} = 3 - {x}^{2} \implies {y}_{2} - 3 = - {x}^{2}$

So vertex is at $\left(0 , 3\right)$

Result:

How to get the volume? We shall use the disc method!

This method is simply that: $\text{Volume} = \pi {\int}_{a}^{b} {y}^{2} \mathrm{dx}$

The idea is simple, however you've got to use it smartly.

And that's what we're going to do.

Lets call our volume $V$
$\implies V = {V}_{1} - {V}_{2}$

${V}_{1} = \pi {\int}_{a}^{b} {\left(4 - {y}_{1}\right)}^{2} \mathrm{dx}$

${V}_{2} = \pi {\int}_{a}^{b} {\left(4 - {y}_{2}\right)}^{2} \mathrm{dx}$

NB: I'm taking $\left(4 - y\right)$ because $y$ is only the distance from the $x$-axis to the curve, whereas we want the distance from the line $y = 4$ to the curve!

Now to find $a$ and $b$, we equate ${y}_{1}$ and ${y}_{2}$ and then solve for $x$

${y}_{1} = {y}_{2} \implies 2 {x}^{2} - x + 3 = 0$
$\implies 2 {x}^{2} + 2 x - 3 x + 3 = 0$
$\implies \left(2 x - 3\right) \left(x + 1\right) = 0 \implies \left\{\begin{matrix}x = \frac{3}{2} = 1.5 \\ x = - 1\end{matrix}\right.$

Since $a$ comes before $b$, $\implies a = - 1$ and $b = 1.5$

$\implies {V}_{1} = \pi {\int}_{- 1}^{1.5} {\left(4 - {y}_{1}\right)}^{2} \mathrm{dx} = \pi {\int}_{-} {1}^{1.5} {\left(4 - {x}^{2} - x\right)}^{2} \mathrm{dx} = \pi {\int}_{- 1}^{1.5} {\left({x}^{2} + x - 4\right)}^{2} \mathrm{dx}$

$\implies \pi \int {\left(- 1\right)}^{1.5} \left({x}^{4} + 3 {x}^{3} - 7 {x}^{2} - 8 x + 16\right) \mathrm{dx} = \pi {\left[{x}^{5} / 5 + {x}^{4} / 2 - \frac{7 {x}^{3}}{3} - 4 {x}^{2} + 16 x\right]}_{-} {1}^{1.5}$

${V}_{1} = \frac{685 \pi}{24}$

Do same for ${V}_{2}$:

${V}_{2} = \pi {\int}_{-} {1}^{1.5} {\left(4 - {y}_{2}\right)}^{2} \mathrm{dx} = \pi {\int}_{-} {1}^{1.5} {\left(4 - 3 + {x}^{2}\right)}^{2} \mathrm{dx} = \pi {\int}_{- 1}^{1.5} {\left(1 + x - 4\right)}^{2} \mathrm{dx}$

$\implies \pi \int {\left(- 1\right)}^{1.5} \left(1 + 2 {x}^{2} + {x}^{4}\right) \mathrm{dx} = \pi {\left[x + \frac{2 {x}^{3}}{3} + {x}^{5} / 5\right]}_{-} {1}^{1.5}$

${V}_{1} = \frac{685 \pi}{96}$

$V = {V}_{1} - {V}_{2} = \frac{685}{24} - \frac{685}{96} = \textcolor{b l u e}{\frac{685 \pi}{32}}$