Solid of Revolution About y=2

Question

Find the volume of the solid of revolution bounded by $y=x^4$ and $y=1$ rotated about $y=2$.

Here's my attempt:

$\pi\displaystyle\int_{-1}^1(2-x^4)^2dx$

$\pi\displaystyle\int_{-1}^1(4-4x^4+x^8)dx$

$\pi(\displaystyle\int_{-1}^1(4)dx-4\displaystyle\int_{-1}^1(x^4)dx+\displaystyle\int_{-1}^1(x^8)dx)$

$\pi(8-4(\frac{2}{5})+\frac{2}{9})$

$\frac{298\pi}{45}$

But the book says $\frac{208\pi}{45}$

What did I do wrong? Thanks.

Answer 1

Hint:

What you are looking for

Your answer

\begin{align*} \text{Volume }&=\text{Volume of the solid without the "hole"}-\text{Volume of the cylinder }\\ &=\pi\displaystyle\int_{-1}^1(2-x^4)^2dx-\pi\displaystyle\int_{-1}^11dx \end{align*}