Find the volume formed by rotating the region enclosed by:
x = 10 y and x = y^3 with y≥0 about the y-axis

Question

Find the volume formed by rotating the region enclosed by:
x = 10 y and x = y^3 with y≥0 about the y-axis

callisto · Answer

I guess it's similar to the previous one...
1. Put x=10y into x=y^3
    => y(y^2 -10) =0  =>  y=0, y=-sqrt10   or   y=sqrt10
2. $$Volume = 2\pi \int_{0}^{\sqrt{10}} y(y^3) -y(10y) dy $$
Can you do it from here?

anonymous · Answer

I'm getting a negative value :s

anonymous · Answer

oops nevermind I forgot to put - sqrt10

callisto · Answer

It's weird.... Let me think about it... Sorry :(

anonymous · Answer

I don't think it's the right integral, I got 264.922 and its not working

callisto · Answer

Oh.... My fault :(
$$volume = \pi \int_{0}^{\sqrt{10}}(y^3)^2 -(10y)^2dy$$
Let see if it works this time :S

callisto · Answer

Something wrong with the limit...

anonymous · Answer

That one worked! using -sqrt 10 :-)

callisto · Answer

I see what's wrong.
it should be 
$$volume = pi \int_{0}^{\sqrt{10}} (10y)^2 - (y^3)^2 dy$$
This is the correct one :S

callisto · Answer

since x=10y has a longer distance to y-axis that x=y^3, so it should be (10y)^2 - (y^3)^2

anonymous · Answer

Sweet, thanks for your help!

callisto · Answer

Welcome :)