Need help with computing the volume of a solid by revolving lines

In summary, the task is to compute the volume of a solid formed by revolving a given region about a given line. Using integration, the volume can be calculated as the integral of the cross sectional area with respect to the axis of rotation. For part (a), the cross section is a disk and the volume is (2/3) pi. For part (b), the cross section is a washer and the volume is (4/3) pi.
  • #1
db2dz
7
0

Homework Statement




the directions are:

compute the volume of the solid formed by revolving the given region about the line

region bounded by y=2x, y=2 and x=0 about (a) the y-axis; (b) x=1



Homework Equations



V=∫A(x)dx,a,b


The Attempt at a Solution



From what my teacher showed in class this is how i think it is worked out
(a)
V=∫A(x)dx,0,2

=∫(pi)[r(x)]^2,0,2

=∫(pi)[2x-2]^2,0,2

=∫(pi)[4x^2-a],0,2

=(8(pi))/3

so i am not sure if this is right or if i have to times(8(pi))/3 by 1/3 because the volume of a cone is (1/3)(pi)(r^2)(h)

and i don't know where to begin on (b), i think when it is revolved it will be Cylinder with a cut out. please help
 
Physics news on Phys.org
  • #2
You want to think about what a cross section of solid perpendicular to the rotation axis. In this case, it's easier to integrate dy than dx, since that's the rotation axis. The volume is then integral A(y)*dy where A(y) is the cross sectional area at y. In a) the cross section is a disk? So yes, the area is pi*R(y)^2. What's R(y) as a function of y? In b) the cross section is a disk with a hole in the middle (a 'washer'), but it's the same general idea.
 
  • #3
so i used dy to integrate and for a i got 2(pi)/3 which i think is right because i checked it with V = (1/3) π r² h. But i am not sure how to get b. i know its a cylinder with the cone removed and i think its volume is 4(pi)/3 but i still have no idea.
 
  • #4
You are right. To get b) using integration the cross sectional area is area of an outer circle minus the area of the inner circle that was removed. So it's pi*(outer radius)^2-pi*(inner radius)^2. Integrate that dy.
 

1. What is the formula for computing the volume of a solid by revolving lines?

The formula for computing the volume of a solid by revolving lines is V = ∫2πy(x) dx, where y(x) represents the cross-sectional area of the solid at a given x-value.

2. Can you provide an example of computing the volume of a solid by revolving lines?

Yes, an example would be finding the volume of a cone by revolving a line segment with length x and base radius r around the x-axis. The cross-sectional area in this case would be given by y(x) = π(x/r)^2, and the integral would be V = ∫2ππ(x/r)^2 dx, which simplifies to V = 2π^2r^2x.

3. What information do I need in order to compute the volume of a solid by revolving lines?

In order to compute the volume of a solid by revolving lines, you will need to know the function for the cross-sectional area, the limits of integration, and the axis of rotation.

4. Are there any limitations to computing the volume of a solid by revolving lines?

Yes, this method can only be used for solids with a circular cross-section and a known function for the cross-sectional area. It also assumes that the solid is rotationally symmetric and the axis of rotation is known.

5. How is this method different from other methods of computing volume?

This method is different from other methods of computing volume, such as using the formula for the volume of a cylinder or the volume of a cone, because it allows for more complex and irregularly shaped solids to be calculated. It is also a more general method that can be applied to various shapes and dimensions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
857
  • Calculus and Beyond Homework Help
Replies
2
Views
852
  • Calculus and Beyond Homework Help
Replies
5
Views
683
  • Calculus and Beyond Homework Help
Replies
3
Views
943
  • Calculus and Beyond Homework Help
Replies
2
Views
434
  • Calculus and Beyond Homework Help
Replies
4
Views
929
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
903
  • Calculus and Beyond Homework Help
Replies
10
Views
389
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
Back
Top