# How do you find the volume of the solid obtained by rotating the region bounded by: y=sqrt(x-1), y=0, x=5 rotated about y=7?

Dec 16, 2017

#### Explanation:

I would use shells. Here is a picture of the region with a slice taken parallel to the axis of rotation.
The slice is taken at a value of $y$, so we need to rewrite the curve $y = \sqrt{x - 1}$ as $x = {y}^{2} + 1$

The thickness of the slice and the shell is $\mathrm{dy}$
The radius is $r = 7 - y$
The height is $h = 5 - \left({y}^{2} + 1\right) = 4 - {y}^{2}$

The shell has volume $2 \pi r h \times \text{thickness} = 2 \pi \left(7 - y\right) \left(4 - {y}^{2}\right) \mathrm{dy}$

$y$ varies from $0$ to $2$ (That is the $y$ value on the curve at $x = 5$.)

The resulting solid has volume:

$V = 2 \pi {\int}_{0}^{2} \left(7 - y\right) \left(4 - {y}^{2}\right) \mathrm{dy}$

$= 2 \pi {\int}_{0}^{2} \left(28 - 4 y - 7 {y}^{2} + {y}^{3}\right) \mathrm{dy}$

$= 2 \pi {\left[28 y - 2 {y}^{2} - \frac{7}{3} {y}^{3} + {y}^{4} / 4\right]}_{0}^{2}$

$= \frac{200 \pi}{3}$

If you prefer washers

Then the thickness is $\mathrm{dx}$,
the greater radius is $7$, and
the lesser radius is $\sqrt{x - 1}$.

$x$ varies from $1$ to $5$, so the volume of the solid is

$V = \pi {\int}_{1}^{5} \left({7}^{2} - {\left(7 - \sqrt{x - 1}\right)}^{2}\right) \mathrm{dx}$

$= \pi {\int}_{1}^{5} \left(14 \sqrt{x - 1} - \left(x - 1\right)\right) \mathrm{dx}$

To integrate I would substitute $u = x - 1$ to get
$\pi {\int}_{0}^{4} \left(14 {u}^{\frac{1}{2}} - u\right) \mathrm{du}$

Dec 16, 2017

See below.

#### Explanation:

Looking at the graph, we can see that the area A rotated around $y = 7$ is the volume we seek.

We can find this volume by finding the volume of the area (A+B). This is easy and doesn’t require integration. The radius of the cylinder that will be formed has a radius of 7, this is just the height from the x axis. We square this and multiply by $\pi$ x the length of the interval $\left[1 , 5\right]$

Length of interval is $5 - 1 = 4$

$\therefore$

Volume (A + B ):

$V = {7}^{2} \pi \cdot 196 \pi$

From this we need to subtract the volume of B. From the graph we can see that the radius is $7 - \sqrt{x - 1}$. This will have to be found by integration in the normal way.

Volume of B:

${\left(7 - \sqrt{x - 1}\right)}^{2} = 48 - 14 \sqrt{x - 1} + x$

$V = \pi \cdot {\int}_{1}^{5} \left(48 - 14 \sqrt{x - 1} + x\right) \mathrm{dx} = 48 x + \frac{1}{2} {x}^{2} - \frac{28}{3} {\left(x - 1\right)}^{\frac{3}{2}}$

Area first:

${\left[48 x + \frac{1}{2} {x}^{2} - \frac{28}{3} {\left(x - 1\right)}^{\frac{3}{2}}\right]}^{5} - {\left[48 x + \frac{1}{2} {x}^{2} - \frac{28}{3} {\left(x - 1\right)}^{\frac{3}{2}}\right]}_{1}$
..................................................................................................................................

${\left[48 \left(5\right) + \frac{1}{2} {\left(5\right)}^{2} - \frac{28}{3} {\left(\left(5\right) - 1\right)}^{\frac{3}{2}}\right]}^{5} - {\left[48 \left(1\right) + \frac{1}{2} {\left(1\right)}^{2} - \frac{28}{3} {\left(\left(1\right) - 1\right)}^{\frac{3}{2}}\right]}_{1}$

$\pi \cdot \left[\frac{505}{2} - \frac{112}{3} \sqrt{4}\right] - \left[\frac{97}{2}\right] = \frac{388}{3} \pi$

Volume= $\frac{388}{3} \pi$

Volume of A:

$196 \pi - \frac{388}{3} \pi = \frac{200}{3} \pi = 66.667 \pi$

Volume of revolution: