How do i solve this integration question using the washer and shell method?

Question

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:

$$ 2 \pi \int_{-1}^3 {(x + 3)(4 - \sqrt{x} + 1)^2} \ \ dx $$

Answer 1

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

$$ V_s = 2 \pi \int_{-1}^3 {r(x)h(x) \ dx} $$ $$ = 2 \pi \int_{-1}^3 {(x + 3)(4 - (x - 1)^2) \ dx} $$ $$ = \frac{256\pi}{3} . $$

Answer 2

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 $\frac{256\pi}3$.