(repost since typos) Volume of revolution using shell method

[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.

 

[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 )

(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?

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?

 

[Call my name]

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?

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

So is what I have established wrong?

 

[tiny-tim]

So is what I have established wrong?

You've used height instead of width

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

 

[Call my name]

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