# (repost since typos) Volume of revolution using shell method

1. Nov 1, 2009

### Call my name

(EDITED)
1. Use the shell method to find the volume of the solid generated by revolving about the y-axis. $$x=y^2, x=y+2$$

2. same as #1, except change y and x for the two equations and revolve about x-axis.

I tried doing $$2pi\int_{x=0}^4(x)(\sqrt{x}-x+2)dx$$ but the answer is off for #1.

I tried doing $$2pi\int_{y=-1}^4(y)(\sqrt{y}-y+2)dy$$, well it's wrong.

Is there a problem with how I established the shell height?

the answer for these two questions are supposed to give me 72pi/5.

2. Nov 1, 2009

### tiny-tim

Hi Call my name!

(have a pi: π and a square-root: √ and try using the X2 tag just above the Reply box )
Yeees … it's the shell width, isn't it?
Are you sure that's the question? If you swap everything, isn't the volume the same?

3. Nov 1, 2009

### Call my name

Yeah, I meant that those two questions' answers are the same.

So is what I have established wrong?

4. Nov 1, 2009

### tiny-tim

You've used height instead of width

(or you've revolved around the wrong axis).

5. Nov 1, 2009

### Call my name

What do you mean? I'm a newbie so please specify and clarify

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