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:

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

• You will get better feedback if you show us what you did. Otherwise, how can we know where you made a mistake? Jan 22, 2021 at 1:35
• Alright I added my set ups for each! Thank you! Jan 22, 2021 at 1:41

2 Answers

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

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

• Thank you so much!!! Jan 22, 2021 at 2:34

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 $$\frac{256\pi}3$$.

• Can tell which software you used for the graphics?
– user870492
Feb 1, 2021 at 4:11
• Mathematica 11.3
– robjohn
Feb 1, 2021 at 4:36