# Consider the solid obtained by rotating the region bounded by y=2x, y=2sqrtx about the line y = 2, how do you find the volume?

Jun 11, 2015

If you imagine this, it's like a cone that has been somewhat hollowed out, with a round lip.

Strange as it is, since the opening of the cone faces the left, I want to actually flip it horizontally. Also, I want to shift it down so that it is rotated around the x-axis ($y = 0$). Now, we are graphing according to the equation $y = - 2 \sqrt{1 - x} + 2$ instead of $y = 2 \sqrt{x}$.

It was flipped vertically ($2 \sqrt{x} \to - 2 \sqrt{x}$), horizontally ($x \to - x$), shifted right 1 ($x \to x - 1 \to 1 - x$), and up 2 ($- 2 \sqrt{1 - x} \to - 2 \sqrt{1 - x} + 2$).

$y = 2 x$ stays the same. The shape did not change, however---only in direction. It went from this to this.

Alright, now that it looks nicer, we can integrate from $0$ to $1$. You could have done it as it was, but I think better when the first circle starts out at a radius of $0$. Since we are layering circles, we would use a variant on volume:

$V = A \left(x\right) \cdot h = \pi {\left(r \left(x\right)\right)}^{2} \cdot x = {\int}_{0}^{1} \pi {\left[r \left(x\right)\right]}^{2} \mathrm{dx}$

where the horizontal height $x$ is represented by the accumulation of "circular" layers from $x = 0$ to $x = 1$. $r \left(x\right)$ is a special function that alters how the revolved shape is made.

We are using the idea that $\pi {r}^{2}$ implies a constant radius revolved around an axis going through the center of a soon-to-be-formed circle. Instead of a constant radius, it will be varying according to the two equations, $y = 2 x$ and $y = - 2 \sqrt{1 - x} + 2$. We will be acquiring the area enclosed below $y = 2 x$ and above $y = - 2 \sqrt{1 - x} + 2$, and then revolving that around the x-axis. Therefore, we need to subtract the two:

$y = 2 x - \left(- 2 \sqrt{1 - x} + 2\right) = 2 x + 2 \sqrt{1 - x} - 2$

Using the above equation in the volume formula:

${\int}_{0}^{1} \pi {\left[r \left(x\right)\right]}^{2} \mathrm{dx}$
$= \pi {\int}_{0}^{1} {\left(2 x + 2 \sqrt{1 - x} - 2\right)}^{2} \mathrm{dx}$

Expansion:

$= \pi {\int}_{0}^{1} 4 {x}^{2} - 12 x + 8 + 8 x \sqrt{1 - x} - 8 \sqrt{1 - x} \mathrm{dx}$

Splitting into separate integrals:

$= \pi \left[{\int}_{0}^{1} 4 {x}^{2} \mathrm{dx} - {\int}_{0}^{1} 12 x \mathrm{dx} + {\int}_{0}^{1} 8 \mathrm{dx} + {\int}_{0}^{1} 8 x \sqrt{1 - x} \mathrm{dx} - {\int}_{0}^{1} 8 \sqrt{1 - x} \mathrm{dx}\right]$

Evaluating some of them, and leaving some harder ones for next step:

$= \pi \left[\frac{4}{3} {x}^{3} - 6 {x}^{2} + 8 x\right] {|}_{0}^{1}$

$+ \pi \left[{\int}_{0}^{1} 8 x \sqrt{1 - x} \mathrm{dx} - {\int}_{0}^{1} 8 \sqrt{1 - x} \mathrm{dx}\right]$

These last two need some extra thinking.

Let's try integration by parts on the first one.

Let:
$u = 8 x$
$\mathrm{dv} = \sqrt{1 - x} \mathrm{dx}$
$\mathrm{du} = 8 \mathrm{dx}$
$v = - \frac{2}{3} {\left(1 - x\right)}^{\frac{3}{2}}$

$= u v - \int v \mathrm{du}$

$= - \frac{16 x}{3} {\left(1 - x\right)}^{\frac{3}{2}} + \frac{16}{3} \int {\left(1 - x\right)}^{\frac{3}{2}} \mathrm{dx}$

Evaluating the integral here:

$= \pi \left\{- \frac{16 x}{3} {\left(1 - x\right)}^{\frac{3}{2}} + \frac{16}{3} \left[- \frac{2}{5} {\left(1 - x\right)}^{\frac{5}{2}}\right]\right\} {|}_{0}^{1}$

Now the second one is not so bad. Just a variant of the integral right above:
$8 {\int}_{0}^{1} \sqrt{1 - x} \mathrm{dx} = \left[- \frac{16}{3} {\left(1 - x\right)}^{\frac{3}{2}}\right] {|}_{0}^{1}$

Overall:

$= \pi \left[\frac{4}{3} {x}^{3} - 6 {x}^{2} + 8 x + \left\{- \frac{16 x}{3} {\left(1 - x\right)}^{\frac{3}{2}} + \left[- \frac{32}{15} {\left(1 - x\right)}^{\frac{5}{2}}\right]\right\} - \left\{- \frac{16}{3} {\left(1 - x\right)}^{\frac{3}{2}}\right\}\right] {|}_{0}^{1}$

Distributing the negative signs (parentheses are important!):

$= \pi \left[\frac{4}{3} {x}^{3} - 6 {x}^{2} + 8 x - \frac{16 x}{3} {\left(1 - x\right)}^{\frac{3}{2}} - \frac{32}{15} {\left(1 - x\right)}^{\frac{5}{2}} + \frac{16}{3} {\left(1 - x\right)}^{\frac{3}{2}}\right] {|}_{0}^{1}$

Plugging in $1$ and $0$ and doing $F \left(1\right) - F \left(0\right)$, with $\pi$ still factored out:

$= \pi \left[\left(\frac{4}{3} - 6 + 8 \cancel{- \frac{16}{3} {\left(1 - 1\right)}^{\frac{3}{2}} - \frac{32}{15} {\left(1 - 1\right)}^{\frac{5}{2}} + \frac{16}{3} {\left(1 - 1\right)}^{\frac{3}{2}}}\right) - \left(\cancel{\frac{4}{3} \cdot 0 - 6 \cdot 0 + 8 \cdot 0 - \frac{16 \cdot 0}{3} {\left(1 - 0\right)}^{\frac{3}{2}}} - \frac{32}{15} {\left(1 - 0\right)}^{\frac{5}{2}} + \frac{16}{3} {\left(1 - 0\right)}^{\frac{3}{2}}\right)\right]$

Home free from here:

$= \pi \left[\frac{4}{3} - 6 + 8 + \frac{32}{15} - \frac{16}{3}\right]$

$= \pi \left[\frac{20}{15} - \frac{90}{15} + \frac{120}{15} + \frac{32}{15} - \frac{80}{15}\right]$

$= \frac{2 \pi}{15} = 0.1 \overline{33} \pi \approx 0.4189 {\text{u}}^{3}$