1. ## Shell method

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis.

x + y^2 = 16

I'm not sure what to set h(y) equal to. Is it sqrt(16-x)? It can't be that because I need to have a y instead of an x, right? I understand how to find the volume and integrate from here.

Thank you.

2. In this case we're integrating along the $y$-axis, not the $x$-axis as usual. For the volume of the solid obtained by revolving the $x \ge 0$ portion of the curve about the $x$-axis, we can use the shell method, our function of height being

$x = 16 - y^2$

and the volume of each hollow cylinder being

$2\pi \cdot (16 - y^2) \cdot dy.$ Edit: This is incorrect.

Since $x = 0$ at $y = \pm 4$, the total volume equals

$\int_0^4 2\pi (16 - y^2) \, dy.$

3. So, could the answer really be (-7808pie)/(3)?

4. nvm. THank you very much. I made an incorrect calculation.

5. Sorry, I made a mistake.

In calculating the volume of a shell, we use not only the radius of the shell but also its height. My previous formula is therefore wrong.

The volume of one shell is

$2\pi xh(y)\,dy,$

not

$2\pi h(y)\,dy.$

Integrating, we have

$\int_0^4 2\pi(16y - y^3)\,dy.$

I apologize for the confusion.

6. So, you just multiply in an extra y?

7. Correct.

The $y$ is for the radius of the shell, and $x = 16 - y^2$ is for the height.

This was my first time doing a shell problem in a long time, and I momentarily forgot.