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^3, y = 0, x = 2 rotated about the x axis?

May 23, 2015

The answer is $V = \frac{128 \pi}{7}$.

Firstly, you should graph your three functions to see clearly what is the delimited region :

We are rotating our plane region around the x-axis, which means we have to write our function $f \left(x\right) = {x}^{3}$ in the form $f \left(y\right) = \sqrt[3]{y}$.

The radius of any cylinder in our final volume will be given by $y$.

And the width will be $2 - \sqrt[3]{y}$.

(Look here for more details or diagrams about the cylinders)

Therefore, the cross-sectional area of any cylinder will be :

$A \left(y\right) = 2 \pi y \left(2 - \sqrt[3]{y}\right) = 4 \pi y - 2 \pi {y}^{\frac{4}{3}}$

The radius of the cylinders, according to the delimited region, goes from :

$f \left(x\right) = 0 = y$ (with $x = 0$) to $f \left(x\right) = 8 = y$ (with $x = 2$).

So the volume of our solid, which is the sum of all the cross-sectional area of the cylinders, is :

$V = {\int}_{0}^{8} A \left(y\right) \mathrm{dy}$.

The antiderivative of $a {y}^{n}$ is given by $a \cdot \frac{1}{n + 1} {y}^{n + 1}$.

Thus, the antiderivative of $A \left(y\right)$ is :

$\frac{1}{2} 4 \pi {y}^{2} - \frac{3}{7} 2 \pi {y}^{\frac{7}{3}} = 2 \pi {y}^{2} \left(1 - \frac{3}{7} \sqrt[3]{y}\right)$

Now we can calculate the integral :

$V = {\int}_{0}^{8} A \left(y\right) \mathrm{dy} = {\left[2 \pi {y}^{2} \left(1 - \frac{3}{7} \sqrt[3]{y}\right)\right]}_{0}^{8}$

$= 2 \pi {\left[{y}^{2} \left(1 - \frac{3}{7} \sqrt[3]{y}\right)\right]}_{0}^{8} = 2 \pi \left({8}^{2} \left(1 - \frac{3 \cdot 2}{7}\right) - 0\right)$

$= 2 \pi \cdot \frac{64}{7} = \frac{128 \pi}{7}$.