(repost since typos) Volume of revolution using shell method

Call my name
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(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.
 
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Hi Call my name! :smile:

(have a pi: π and a square-root: √ and try using the X2 tag just above the Reply box :wink:)
Call my name said:
(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

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

Yeees … it's the shell width, isn't it? :wink:
2. same as #1, except change y and x for the two equations and revolve about x-axis.

Are you sure that's the question? If you swap everything, isn't the volume the same?:smile:
 
tiny-tim said:
Hi Call my name! :smile:

(have a pi: π and a square-root: √ and try using the X2 tag just above the Reply box :wink:)


Yeees … it's the shell width, isn't it? :wink:


Are you sure that's the question? If you swap everything, isn't the volume the same?:smile:


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

So is what I have established wrong?
 
Call my name said:
So is what I have established wrong?

You've used height instead of width

(or you've revolved around the wrong axis).
 
tiny-tim said:
You've used height instead of width

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

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