Calculus Volume Integration

In summary, the conversation discusses the attempt to solve for the volume of a solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axis. The solution involves integrating the equation pi[80x - (4x^3)/3] and considering the boundaries x=-2 and x=2. The final answer given is 896(pi)/3, but there may be an error in the algebra when expanding the squares inside the integration.
  • #1
kira137
16
0
I've been trying this equation for few hours now and I can't seem to get the right answer..
answer is suppose to be 640(pi)/3

Homework Statement

Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axisThe attempt at a solution
``2
Pi S (9-x^2)^2 - (x^2+1)^2 dx
``-2

``2
Pi S (81-2x^2+x^4-x^4-2x^2-1) dx
``-2

Pi[80x - (4x^3)/3] (x=-2 to 2)

= (448(pi)/3) - (-448(pi)/3) = 896(pi)/3

that's what I keep getting..
and if that isn't the correct answer for rotating around x=0, isn't it suppose to be the answer for when rotating around y=1?

thank you in advance
 
Physics news on Phys.org
  • #2
kira137 said:
I've been trying this equation for few hours now and I can't seem to get the right answer..
answer is suppose to be 640(pi)/3

Homework Statement

Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axis


The attempt at a solution
``2
Pi S (9-x^2)^2 - (x^2+1)^2 dx
``-2

``2
Pi S (81-2x^2+x^4-x^4-2x^2-1) dx
``-2

Pi[80x - (4x^3)/3] (x=-2 to 2)

= (448(pi)/3) - (-448(pi)/3) = 896(pi)/3

that's what I keep getting..
and if that isn't the correct answer for rotating around x=0, isn't it suppose to be the answer for when rotating around y=1?

thank you in advance

Check again your algebra when you expand the squares inside the integration.

Cheers -- sylas
 

1. What is volume integration in calculus?

Volume integration in calculus is a mathematical technique used to calculate the volume of a three-dimensional object or shape. It involves breaking down the object into infinitesimally small pieces, finding the volume of each piece, and then summing them up to get the total volume.

2. What is the difference between volume integration and area integration?

Volume integration is used to find the volume of a three-dimensional object, while area integration is used to find the area of a two-dimensional shape. In volume integration, we integrate over three variables (x, y, and z), while in area integration, we integrate over two variables (x and y).

3. How do you set up a volume integration problem?

To set up a volume integration problem, you first need to identify the shape or object whose volume you want to find. Then, you will need to determine the limits of integration for each variable (x, y, and z). Next, you will need to choose the appropriate integration method (single, double, or triple) and the integrand (function) to be integrated. Finally, you can solve the integral to find the volume.

4. What is the importance of volume integration in real life?

Volume integration is used in various fields such as engineering, physics, and architecture to find the volume of complex shapes and objects. It is essential for calculating quantities like the volume of a liquid in a container, the displacement of a vehicle, or the mass of an irregularly shaped object. It also has applications in real-life scenarios such as calculating the volume of a reservoir or determining the amount of material needed for construction projects.

5. Can volume integration be applied to non-regular shapes?

Yes, volume integration can be applied to non-regular shapes. In fact, it is often used to find the volume of irregular and complex shapes that cannot be easily calculated by other methods. By breaking down the shape into smaller, more manageable pieces, we can use volume integration to find the volume of each piece and then add them together to get the total volume of the shape.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
269
  • Calculus and Beyond Homework Help
Replies
10
Views
286
  • Calculus and Beyond Homework Help
Replies
5
Views
619
  • Calculus and Beyond Homework Help
Replies
34
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
479
  • Calculus and Beyond Homework Help
Replies
8
Views
827
  • Calculus and Beyond Homework Help
Replies
9
Views
698
  • Calculus and Beyond Homework Help
Replies
5
Views
668
  • Calculus and Beyond Homework Help
Replies
9
Views
921
  • Calculus and Beyond Homework Help
Replies
12
Views
936
Back
Top