Solid generated by revolving region, find the diameter of the hole

In summary, a solid is generated by revolving a region bounded by y=(1/2)x^2 and y=12 about the y axis. A hole is drilled through the solid, removing 1/4 of the total volume. Using the shell method integral, the boundaries x=0 to x=12 are found and the total volume is calculated to be 864*pi. Dividing by 4, the area to be removed is determined to be 216*pi. To find the diameter of the hole, a sketch of the solid and the hole is suggested, with the intersection of the solid and the hole in the x,y-plane as a starting point for integration. It is recommended to give the hole a radius of r and then
  • #1
isukatphysics69
453
8

Homework Statement


A solid is generated by revolving region bounded by y=(1/2)x^2 and y=12 about the y axis. A hole centered along the axis of revolution is drilled through this solid so that 1/4 of the volume is removed. find the diameter of the hole.

Homework Equations


y=(1/12)x^2 y = 12
shell method integral

The Attempt at a Solution



i figured out the boundries are x=0 to x=12 and integrated using the shell method and got the answer of the total volume as 864*pi. I divided that by 4 and got 216*pi as the area that needs to be removed but now i am stuck and don't know how to figure out the diameter of the hole! [/B]
 
Physics news on Phys.org
  • #2
isukatphysics69 said:

Homework Statement


A solid is generated by revolving region bounded by y=(1/2)x^2 and y=12 about the y axis. A hole centered along the axis of revolution is drilled through this solid so that 1/4 of the volume is removed. find the diameter of the hole.

Homework Equations


y=(1/12)x^2 y = 12
shell method integral

The Attempt at a Solution



i figured out the boundries are x=0 to x=12 and integrated using the shell method and got the answer of the total volume as 864*pi. I divided that by 4 and got 216*pi as the area that needs to be removed but now i am stuck and don't know how to figure out the diameter of the hole! [/B]
Have you made a sketch of the solid and the hole? That might help you figure it out. Actually, just start with the intersection of the solid and the hole with the x,y-plane, then it should be pretty clear how to integrate the solid of revolution for the hole. Give the hole a radius of r, do the integration, and then figure out what r has to be to give you the right volume for the hole.
 
Last edited:

Related to Solid generated by revolving region, find the diameter of the hole

1) What is a solid generated by revolving region?

A solid generated by revolving region is a 3-dimensional shape formed by rotating a 2-dimensional region (such as a circle or rectangle) around a given axis.

2) How do you find the diameter of the hole in a solid generated by revolving region?

To find the diameter of the hole, you first need to determine the shape of the 2-dimensional region and the axis of rotation. Then, you can use mathematical formulas (such as the Pythagorean Theorem or the formula for circumference) to calculate the diameter.

3) Can the diameter of the hole vary in a solid generated by revolving region?

Yes, the diameter of the hole can vary depending on the shape and size of the 2-dimensional region and the axis of rotation used. It can also change if the shape is rotated at different angles or if the axis of rotation is shifted.

4) What are some real-life examples of solids generated by revolving region?

Some examples include a cup (formed by revolving a circle around a central axis), a vase (formed by revolving a rectangle around a central axis), and a donut (formed by revolving a circle around a smaller central circle).

5) How is the diameter of the hole important in a solid generated by revolving region?

The diameter of the hole is important because it determines the size and shape of the resulting solid. It can also affect the stability and functionality of the solid, as well as its aesthetic appearance.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
973
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
902
  • Calculus and Beyond Homework Help
Replies
8
Views
3K
Back
Top