R(x)=x^3 bounded by x=0, x=2 and y=1.(adsbygoogle = window.adsbygoogle || []).push({});

a. revolved around x=2

b. revolved around x=10

my pathetic attempt:

a. v=pi[(integral from 0 to 1)(2-y^1/3)^2]dy

so =pi[4y-3y^(4/3)+(3/5)y^(5/3)]evaluated from 0 to 1

=(8/5)pi

b. v=pi[(integral from 0 to 2)(10-x^3)^2]dy

I must admit that I don't think that equation is right. But I just seem to get how to set this up.

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Volumes of revolution

**Physics Forums | Science Articles, Homework Help, Discussion**