Efficiently Solve a Challenging Rotation Problem | Homework with Shell Method

[iRaid]

Homework Statement

Homework Equations

Shell method

The Attempt at a Solution

Not sure if this is right, but the integral I set up is:
[tex]2\pi \int_0^1 (4-y)(\sqrt{siny})dy[/tex]


Finding the radius is my big problem with these problems, I can't visualize it very well.

Any help is appreciated.

 

[SammyS]

Homework Statement

Homework Equations

Shell method

The Attempt at a Solution

Not sure if this is right, but the integral I set up is:
[tex]2\pi \int_0^1 (4-y)(\sqrt{siny})dy[/tex]
Finding the radius is my big problem with these problems, I can't visualize it very well.

Any help is appreciated.

That looks fine except for the limits of integration.

How did you get those limits?

 

[iRaid]

graphing [tex]\sqrt{siny}[/tex] from 0 to pi gives me a curve that goes to 1 and back to 0. Since I'm integrating wrt y, I figured it's 0 to 1?

 

[SammyS]

graphing [tex]\sqrt{siny}[/tex] from 0 to pi gives me a curve that goes to 1 and back to 0. Since I'm integrating wrt y, I figured it's 0 to 1?

You don't have [itex]\ \ y=\sqrt{\sin(x)}\ \ [/itex] with x going from 0 to π.

You have [itex]\ \ x=\sqrt{\sin(y)}\ \ [/itex] with y going from 0 to π --- and your integration is w.r.t. y.

 

[iRaid]

You don't have [itex]\ \ y=\sqrt{\sin(x)}\ \ [/itex] with x going from 0 to π.

You have [itex]\ \ x=\sqrt{\sin(y)}\ \ [/itex] with y going from 0 to π --- and your integration is w.r.t. y.

Thanks, probably shouldn't agree with wolfram straight after I graph...