How Do Volumes Change When Revolving R(x)=x^3 Around Different Axes?

[BoldKnight399]

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.

 

[Mark44]

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

Are you sure you have copied the description of the region that is being revolved correctly? Could it be the region bounded by the graph of y = x^3, y = 0, x = 1, and y = 1? The region as you wrote it doesn't have a boundary on the bottom.

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.

 

[BoldKnight399]

o, ummm. No there is no lower bound. I assume that it is y=0. And for b, I meant to say that I just can't seem to get the set up.

 

[Mark44]

For a) are you sure it doesn't say the region bounded by the graph of y = x³, y = 0, x = 2, and y = 1? That would make more sense. In problems like these, it's very rare that you have to assume how the region is defined.

 

[BoldKnight399]

Consider the region R bounded by   3f x  x , y =1, and x = 2 that is what it says.

 

[BoldKnight399]

oops...copy pasting didnt work, but that is the region

 

[Mark44]

Consider the region R bounded by   3f x  x , y =1, and x = 2 that is what it says.

What you entered is showing up as squares. What does it say?

 

[BoldKnight399]

f(x)=x^3, y=1, and x=2. I just got a response back from my teacher and she said that "The limits of integration are determined by the region described. You have to figure them out." So now I am confuzed more than before.

 

[Mark44]

OK, that's better. In your first post, you also had x = 0 as one of the bounds. The region being revolve has a somewhat triangular shape. It's above the line y = 1, to the right of the curve, and to the left of the line x = 2.

For a, your integrand is correct, but your limits of integration aren't. Your values of dy are running from what to what? Here's the integral. All you need to do are to get the right limits of integration, find the antiderivative, and then evaluate the antiderivative at the two limits of integration.
$$\pi \int_?^? (2 - y^{1/3})^2 dy$$