# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y=x+2 and y=x^2 rotated about the x axis?

Oct 23, 2015

See the explanation section, below.

#### Explanation:

Graph the region, including the points of intersection, $\left(- 1 , 1\right)$ and $\left(2 , 4\right)$.

In order to use shells, we must take our representative slices parallel to the axis of rotation. So the thickness of each shell will be $\mathrm{dy}$.

The thickness of each shell will be $\mathrm{dy}$ and we will be integrating with respect to $y$. So, we need to express the boundaries as function of $y$:

$x = y - 2$ and $x = \pm \sqrt{y}$

And the limits of integration are $0$ to $4$

The volume of the representative shell is $2 \pi \cdot \text{radius"*"height} \cdot \mathrm{dy}$

The radius of each shell will be $y$

The "height" of each cylindrical shell will be the distance between the $x$ values. This will be the $x$ on the right (${x}_{\text{right}}$) minus the $x$ on the left (${x}_{\text{left}}$).

Notice in the picture, that as $y$ goes from $0$ to $4$, the values of ${x}_{\text{right}}$ and ${x}_{\text{left}}$ change.

From $y = 0$ to $y = 1$ , we have:
$\text{height" = x_"right" - x_"left} = \sqrt{y} - \left(- \sqrt{y}\right) = 2 \sqrt{y} = \frac{8 \pi}{5}$

So we need

${\int}_{0}^{1} 2 \pi y \left(2 \sqrt{y}\right) \mathrm{dy} = 4 \pi {\int}_{0}^{1} {y}^{\frac{3}{2}} \mathrm{dy}$

From $y = 1$ to $y = 4$ , we have:
$\text{height" = x_"right" - x_"left} = \sqrt{y} - \left(y - 2\right)$

So we need

${\int}_{1}^{4} 2 \pi y \left(\sqrt{y} - y + 2\right) \mathrm{dy} = 2 \pi {\int}_{1}^{4} \left({y}^{\frac{3}{2}} - {y}^{2} + 2 y\right) \mathrm{dy} = \frac{64 \pi}{5}$

The total volume is $\frac{8 \pi}{5} + \frac{84 \pi}{5} = \frac{72}{5} \pi$