Using the shell method to find a volume of a solid

[deekin]

Homework Statement

Use the shell method to find the volume of the solid generated by revolving the shaded region about the indicated axes. Now, I don't know how to put the graph on here, but the equations are below. The shape revolves around the line y = 2 to produce the solid.

Homework Equations

x = (1/4)(y^4)-(1/2)(y^2)
x = (1/2)(y^2)

The graphs form this leaf shape, on 0≤y≤2.

The Attempt at a Solution

For the following work, please read the integral sign as the definite integral from 0 to 2. I don't know how to include the upper and lower bounds.

V = ∫2π(2-y)((y^2)/2-((y^4)/4-(y^2)/2)))dy

V = ∫2π(2-y)(y^2-(y^4)/4)dy

V= 2π∫(2y^2-(y^4)/2-y^3+(y^5)/4)dy = 2π(2y^3-(y^5)/10-(y^4)/4+(y^6)/24)

Then I evaluated that for y = 2, which gave me a volume of 8π/5 units cubed.

Hopefully I expressed everything clearly in this post. This is my first time posting equations and my work online, so please let me know if anything needs to be clarified. The general method that I used was to integrate 2π(shell radius)(shell height) with respect to y. For the shell radius, I used (2-y), and this is what I'm not sure on. Thank you.

 

[DryRun]

Now, I don't know how to put the graph on here

You can scan the section or just take a clear picture with a camera, and then directly attach the picture to your post.

 

[SammyS]

Homework Statement

Use the shell method to find the volume of the solid generated by revolving the shaded region about the indicated axes. Now, I don't know how to put the graph on here, but the equations are below. The shape revolves around the line y = 2 to produce the solid.

Homework Equations

x = (1/4)(y^4)-(1/2)(y^2)
x = (1/2)(y^2)

The graphs form this leaf shape, on 0≤y≤2.

The Attempt at a Solution

For the following work, please read the integral sign as the definite integral from 0 to 2. I don't know how to include the upper and lower bounds.

V = ∫2π(2-y)((y^2)/2-((y^4)/4-(y^2)/2)))dy

V = ∫2π(2-y)(y^2-(y^4)/4)dy

V= 2π∫(2y^2-(y^4)/2-y^3+(y^5)/4)dy = 2π(2y^3-(y^5)/10-(y^4)/4+(y^6)/24)

Then I evaluated that for y = 2, which gave me a volume of 8π/5 units cubed.

Hopefully I expressed everything clearly in this post. This is my first time posting equations and my work online, so please let me know if anything needs to be clarified. The general method that I used was to integrate 2π(shell radius)(shell height) with respect to y. For the shell radius, I used (2-y), and this is what I'm not sure on. Thank you.

That all looks fine to me !

Yes, 2-y is the radius of the shell(s).

 

[deekin]

Awesome, thank you for checking it out for me.