# How do you find the volume of the solid generated by revolving the region bounded by the curves y = x^(1/2), y = 2, and x = 0 rotated about the x=-1?

Mar 24, 2016

$V = \frac{176 \pi}{15} \cong 36.86$[cubic units]

#### Explanation:

Refer to the figure below

The generator region is Region A limited by $y = {x}^{\frac{1}{2}}$, $y = 2$ and $x = 0$ (the y-axis).

Since the revolution is around an axis parallel to the y-axis, it's convenient to use the inverse function
$f \left(y\right) = {y}^{2}$

We can use a trick to solve this problem: translating the y-axis to the position of the axis of the revolution. Consequently
$y = {\left(X - 1\right)}^{\frac{1}{2}}$
$g \left(y\right) = {y}^{2} + 1$

We only have to remember to exclude Region B, since now the old $x = 0$ became $x = 1$

Applying the formula for volume of a solid of revolution (with a hole), with axis of revolution in y-axis:

$V = \pi {\int}_{a}^{b} \left[R {\left(y\right)}^{2} - r {\left(y\right)}^{2}\right] \cdot \mathrm{dy}$, where $R \left(y\right)$ is the external radius and $r \left(y\right)$ is the internal radius
$V = \pi {\int}_{0}^{2} \left[{\left({y}^{2} + 1\right)}^{2} - {1}^{2}\right] \cdot \mathrm{dy}$
$V = \pi {\int}_{0}^{2} \left({y}^{4} + 2 {y}^{2} + \cancel{1} - \cancel{1}\right) \cdot \mathrm{dy}$
$V = \pi {\int}_{0}^{2} \left({y}^{4} + 2 {y}^{2}\right) \cdot \mathrm{dy}$
$V = \pi \left({y}^{5} / 5 + \frac{2}{3} {y}^{2}\right) {|}_{0}^{2}$
$V = \pi \left(\frac{32}{5} + \frac{2}{3} \cdot 8\right) = \pi \frac{96 + 80}{15} = \frac{176 \pi}{15} \cong 36.86$ [cubic units]