How do i solve this integration question using the washer and shell method? What is the volume of a solid enclosed by $y = (x-1)^2$ and $y = 4$ revolved around $x = - 3$?
I tried the washer method and the shell method and got different answers each time and I'm really confused please help!
My set up for the washer method was:
$$ \pi \int_0^4 {(4 + \sqrt{y} + 1)^2 - 4^2 \ \ dy} $$
My set up for the shell method was:
$$ 2 \pi \int_{-1}^3 {(x + 3)(4 - \sqrt{x} + 1)^2} \ \ dx $$
 A: Washer Method
The larger radius comes from the right side of the parabola $y = (x - 1)^2$, while the smaller radius comes from the left side of that parabola. Rewriting that parabola in terms of $x$, we have:
$$ y = (x - 1)^2 \Rightarrow x = 1 \pm \sqrt{y} . $$
Then, $ R(y) = (1 + \sqrt{y}) - (-3) $ and $ r(y) = (1 - \sqrt{y}) - (-3) $, where in both functions, the $ - (-3) $ comes from rotating about $x = -3$. These radii are also a bit easier to see graphically.
Then we have
$$ R(y) = 4 + \sqrt{y} , $$
$$ r(y) = 4 - \sqrt{y} . $$
Then, our formula for the volume is
$$ V_w = \pi \int_0^4 {R(y)^2 - r(y)^2 \ dy} $$
$$ = \pi \int_0^4 {(4 + \sqrt{y})^2 - (4 - \sqrt{y})^2 \ dy} $$
$$ = \frac{256\pi}{3} . $$
Shell Method
You've set up the radius of your cylindrical shells correctly:
$$ r(x) = x + 3 . $$
But, the height is just the difference in the $y$-coordinates of the two curves bounding your region:
$$ h(x) = 4 - (x - 1)^2 . $$
So, our volume is then
$$ V_s = 2 \pi \int_{-1}^3 {r(x)h(x) \ dx} $$
$$ = 2 \pi \int_{-1}^3 {(x + 3)(4 - (x - 1)^2) \ dx} $$
$$ = \frac{256\pi}{3} . $$
A: Substitute $x\mapsto x-3$, then $x\mapsto r$. Then the region in question is
$$
y=(r-4)^2\le4\tag1
$$
The shell method:
$$
\int_2^6\overbrace{\left(4-(r-4)^2\right)}^{\text{width of the shell}}\overbrace{\ \ 2\pi r\,\mathrm{d}r\ \ \vphantom{4^2}}^{\substack{\text{length$\,\times$}\\\text{thickness}\\\text{of the shell}}}\tag2
$$

The washer method:
For a given $y$, the outer radius is $4+\sqrt{y}$ and the inner radius is $4-\sqrt{y}$.
$$
\int_0^4\left(\vphantom{\left(\sqrt{y}\right)^2}\right.\!\overbrace{\pi\left(4+\sqrt{y}\right)^2}^{\substack{\text{area of disk}}}-\overbrace{\pi\left(4-\sqrt{y}\right)^2}^{\substack{\text{area of hole}}}\left.\vphantom{\left(\sqrt{y}\right)^2}\!\right)\overbrace{\ \quad\mathrm{d}y\ \quad\vphantom{\left(\sqrt{y}\right)^2}}^{\substack{\text{thickness}\\\text{of the disk}}}\tag3
$$

Both of the integrals equal $\frac{256\pi}3$.
