- #1
BoldKnight399
- 79
- 0
R(x)=x^3 bounded by x=0, x=2 and y=1.
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.
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.