How do you use the disk or shell method to find the volume of the solid generated by revolving the regions bounded by the graphs of y = x^(1/2), y = 2, and x = 0 about the line x = -1?

Oct 19, 2015

See the explanation section, below.

Explanation:

Here is a picture of the region with a thin slice taken vertically.

So thickness is $\mathrm{dx}$ and we will use cylindrical shells.

The representative shell will have volume
$2 \pi \text{radius" * "height" * "thickness} = 2 \pi \left(x + 1\right) \left(2 - \sqrt{x}\right) \mathrm{dx}$.

We see that the values of $x$ vary from $0$ to $4$

So the volume of the solid is found by evaluating:

${\int}_{0}^{4} 2 \pi \left(x + 1\right) \left(2 - \sqrt{x}\right) \mathrm{dx} = 2 \pi {\int}_{0}^{4} \left(x + 1\right) \left(2 - \sqrt{x}\right) \mathrm{dx}$

Expand the product and integrate term by term.

To use disks/washers , we need to take our slices perpendicular to the axis of rotation as shown below.

The curve is now expressed as $x = {y}^{2}$

The representative washer has volume: $\pi \left({R}^{2} - {r}^{2}\right) \mathrm{dy}$ Where $r$ is the greater radius ($x = - 1$ to $x = {y}^{2}$) and $r$ is the lesser radius ($x = - 1$ to $x = 0$) and the thickness is $\mathrm{dy}$.

$y$ goes from $0$ to $2$, so the Volume is

${\int}_{0}^{2} \pi \left({\left({y}^{2} + 1\right)}^{2} - {\left(1\right)}^{1}\right) \mathrm{dy} = \pi {\int}_{0}^{2} \left({\left({y}^{2} + 1\right)}^{2} - {\left(1\right)}^{1}\right) \mathrm{dy}$

Again, expand and integrate term by term.