How do I find the volume of a solid of revolution using the Shell method?

[Heroesrule99]

Homework Statement

Find the volume of the solid obtained by rotating the region bounded by y=2x²-x³ and y=0 about the y-axis

Homework Equations

There no required method (between Disk, Washer, Shell). In my attempt below, I used the Shell method, I believe. 2π(shell radius)(shell height)dx

The Attempt at a Solution

I graphed the equation and ended up with V = ∫²₀ 2π(x)(2x²-x³)dx

V =2π*limit 0 to 2* (2x³-x⁴)dx

V = 2π *limit 0 to 2* x⁴/2 - x⁵/5

V =$$2π (1/2 - 1/5)$$
V= 3π/5

Am I on the right track or did I mess up horribly in the setup? I feel like the area where I might have made a mistake was the limits I set... and still getting used to the functions on here lol sorry

 

[Dick]

Homework Statement

Find the volume of the solid obtained by rotating the region bounded by y=2x²-x³ and y=0 about the y-axis

Homework Equations

There no required method (between Disk, Washer, Shell). In my attempt below, I used the Shell method, I believe. 2π(shell radius)(shell height)dx

The Attempt at a Solution

I graphed the equation and ended up with V = ∫²₀ 2π(x)(2x²-x³)dx

V =2π*limit 0 to 2* (2x³-x⁴)dx

V = 2π *limit 0 to 2* x⁴/2 - x⁵/5

V =$$2π (1/2 - 1/5)$$
V= 3π/5

Am I on the right track or did I mess up horribly in the setup? I feel like the area where I might have made a mistake was the limits I set... and still getting used to the functions on here lol sorry

The setup looks ok to me but it looks like you are putting x=1 for the upper limit of your integral instead of x=2 when you evaluate it. Why would you do that?

 

[Heroesrule99]

Ah, just a silly error. So when I plug in 2 for x rather than one. I get 8 - 6.4 = 3.2pi. But other than that it looks alright? I was worried about how I choose to set up my limits since I found the x intercept for the upper limit...but kinda sort of just guessed that the other was 0. There has to be a more technical way to go about that. @_@

 

[Dick]

Ah, just a silly error. So when I plug in 2 for x rather than one. I get 8 - 6.4 = 3.2pi. But other than that it looks alright? I was worried about how I choose to set up my limits since I found the x intercept for the upper limit...but kinda sort of just guessed that the other was 0. There has to be a more technical way to go about that. @_@

Yes, that's fine. There are two x intercepts. Factor. 2x^2-x^3=x^2*(2-x)=0. So either x=0 or x=2.

 

[Heroesrule99]

Thanks so much!