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

In summary: For b, your limits of integration are correct, but your integrand isn't. You have y being used in the integral, but x is the variable. The integrand should be in terms of x. Here's the integral. All you need to do is to get the integrand correct, and then to find the antiderivative and evaluate it at the two limits of integration.\pi \int_0^2 (10 - x^3)^2 dxIn summary, we are given a region R(x)=x^3 bounded by x=0, x=2 and y=1. We are asked to find the volume of this region when revolved around the lines x=2 and x=10. For
  • #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.
 
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  • #2
BoldKnight399 said:
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.
BoldKnight399 said:
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
Consider the region R bounded by   3f x  x , y =1, and x = 2 that is what it says.
 
  • #6
oops...copy pasting didnt work, but that is the region
 
  • #7
BoldKnight399 said:
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?
 
  • #8
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.
 
  • #9
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.
[tex]\pi \int_?^? (2 - y^{1/3})^2 dy[/tex]
 

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

1. What is a volume of revolution?

A volume of revolution is a three-dimensional figure created by rotating a two-dimensional shape around an axis. This process is also known as rotational symmetry.

2. What are the different methods for calculating a volume of revolution?

There are two main methods for calculating a volume of revolution: the disk method and the shell method. The disk method involves using circular cross-sections to determine the volume, while the shell method involves using cylindrical shells to calculate the volume.

3. When do we use the disk method and when do we use the shell method?

The disk method is typically used when the axis of rotation is perpendicular to the base of the figure, while the shell method is used when the axis of rotation is parallel to the base of the figure. However, either method can be used in certain situations, so it is important to understand the concept and choose the appropriate method.

4. How do we set up the integral for finding the volume of a figure using the disk method?

To set up the integral for the disk method, we need to use the formula V = πr^2h, where r is the radius of the circular cross-section and h is the height of the figure. The limits of integration will depend on the bounds of the function being rotated and the axis of rotation.

5. Can volumes of revolution be negative?

No, volumes of revolution cannot be negative. Since volume is a measure of space, it cannot have a negative value. If the integral for a figure using the disk or shell method results in a negative value, it means that the figure is below the x-axis and the absolute value of the integral should be taken to get the correct volume.

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