# Cylindrical shells (parts in details)?

## $f \left(x\right) = {x}^{2}$, $y = 0$, $x = 1$ Set up a volume (shell) derivative based on the following axes of rotation: (A) around $x = - 2$ (B) around $x = 2$ (C) around $y = - 2$ (D) around $y = 2$ Can you use washers for (A) and (B) in the previous question? Why or why not? "Pikachu claims that no matter what kind of problem, you can always use disks/washers and shells. Is this true? Explain."

Feb 21, 2018

I have some parts of this question answered; feel free to check/change where needed

#### Explanation:

Here are my answers for number 1
$a = 0$ and $b = 1$;
(A)
$\setminus {\int}_{0}^{1} \left[2 \setminus \pi \left(x - \left(- 2\right)\right) \left({x}^{2}\right)\right] \mathrm{dx} = 2 \setminus \pi \setminus {\int}_{0}^{1} \left[\left(x + 2\right) \left({x}^{2}\right)\right] \mathrm{dx}$

(B)
$\setminus {\int}_{0}^{1} \left[2 \setminus \pi \left(2 - x\right) \left({x}^{2}\right)\right] \mathrm{dx} = 2 \setminus \pi \setminus {\int}_{0}^{1} \left[\left(2 - x\right) \left({x}^{2}\right)\right] \mathrm{dx}$

(C)
$\setminus {\int}_{0}^{1} \left[2 \setminus \pi \left(y - \left(- 2\right)\right) \left(1 - \setminus \sqrt{y}\right)\right] \mathrm{dy} = 2 \setminus \pi \setminus {\int}_{0}^{1} \left[\left(y + 2\right) \left(1 - \setminus \sqrt{y}\right)\right] \mathrm{dy}$

(D)
$\setminus {\int}_{0}^{1} \left[2 \setminus \pi \left(2 - y\right) \left(1 - \setminus \sqrt{y}\right)\right] \mathrm{dy} = 2 \setminus \pi \setminus {\int}_{0}^{1} \left[\left(2 - y\right) \left(1 - y\right)\right] \mathrm{dy}$

Feb 22, 2018

For attempted answers to questions 2 and 3. please see below.

#### Explanation:

1. (A)
Shells: $V = 2 \pi {\int}_{0}^{1} \left(x + 2\right) {x}^{2} \mathrm{dx} = \frac{11 \pi}{6}$

Washers: $V = \pi {\int}_{0}^{1} \left(9 - {\left(\sqrt{y} + 2\right)}^{2}\right) \mathrm{dy} = \frac{11 \pi}{6}$

(B)
Shells: ${\int}_{0}^{1} \left(2 - x\right) {x}^{2} \mathrm{dx} = \frac{5}{6} \pi$

Washers: $\pi {\int}_{0}^{1} \left({\left(2 - \sqrt{y}\right)}^{2} - 1\right) \mathrm{dy} = \frac{5}{6} \pi$

3.
Obviously Pikachu is not correct. You cannot use discs/washers or shells to solve a related rates problem.

In theory, for any solid of revolution, we can use either. But consider the following problem.

Find the volume that results if the region bounded by the curve $y = x \sin x$ and the $x$-axis from the origin to $x = \pi$ is rotated about the $y$-axis..

graph{xsinx [-1.456, 4.702, -0.76, 2.318]}

Using shells, this is $\pi {\int}_{0}^{\pi} {x}^{2} \sin x \mathrm{dx} = {\pi}^{2} - 4$

To use washers we need expressions for the $x$ values on the left and right in terms of $y$.

Since $y = x \sin x$ cannot be solved using standard mathematical functions, our best hope would probably be to find a series approximation to the inverse function. (I think that one exists.)
So we could do it using washers (maybe), but we sure wouldn't want to.