# Thread: Calculus - Integration - Volume of Revolution

1. ## Calculus - Integration - Volume of Revolution

A manufacturer drills a hole through the centre of a metal sphere of radius 5cm. The hole has a radius of 3cm. What is the volume of the resulting metal ring?

I would like to have full solutions to this problem and if possible, directions to how I can draw a diagram. Thank you. Please solve this problem using volumes of revolution.

2. Originally Posted by bhuang
A manufacturer drills a hole through the centre of a metal sphere of radius 5cm. The hole has a radius of 3cm. What is the volume of the resulting metal ring?

I would like to have full solutions to this problem and if possible, directions to how I can draw a diagram. Thank you. Please solve this problem using volumes of revolution.
In general we don't like to give out full solutions just like that. Here's how to approach this problem. Consider the region bounded by the curves $\displaystyle y = \sqrt{25 - x^2}$ (the upper half of the circle of radius 5) and $\displaystyle y = 3$. The desired volume can be obtained by revolving this region about the x-axis. I'd use the disk (washer) method for this. Now try to do the problem

3. What do you mean when you say the upper half is radius 5 and y=3? Are 5 and 3 the limits of integration?

4. Originally Posted by bhuang
What do you mean when you say the upper half is radius 5 and y=3? Are 5 and 3 the limits of integration?
i said the upper half of the circle of radisu 5. I am just describing a semi-circle of radius 5 centered at the origin.

First step: draw the region i'm talking about.

and to find the limits of integration, you would set the two functions equal to each other and solve for x.

5. is the region √)25-x^2)?
are the two functions √(25-x^2) and -√(25-x^2)?

6. Originally Posted by bhuang
is the region √)25-x^2)?
are the two functions √(25-x^2) and -√(25-x^2)?
I gave you the exact two functions you are to use. please read my first post.

and I have tried doing this question without integration, just simple volume of sphere and cylinder but I don't get the same answer. Am I not supposed to do it with volume of sphere and cylinder?

8. Originally Posted by bhuang
and I have tried doing this question without integration, just simple volume of sphere and cylinder but I don't get the same answer. Am I not supposed to do it with volume of sphere and cylinder?
yup, that's what i got. and no, you would not get the same answer by subtracting the volume of a cylinder from that of a sphere. you'd be leaving out the spherical caps. draw a diagram and you will see what i'm talking about. there would be round edges on the cylinder...

9. ya I know what you're talking about, I subtracted those too. What I had was the volume of the whole sphere (4/3)(pi)5^3) subtract the volume of the spherical caps (4/3)(pi)3^3) subtract the volume of the cylinder (pi)(3^2)(4).

Since the radius of the hole is 3 cm, I assume the radius of the spherical caps to be 3 cm. And if they're 3 cm, then the height of the cylinder becomes 10-6 = 4.

10. Originally Posted by bhuang
ya I know what you're talking about, I subtracted those too. What I had was the volume of the whole sphere (4/3)(pi)5^3) subtract the volume of the spherical caps (4/3)(pi)3^3) subtract the volume of the cylinder (pi)(3^2)(4).

Since the radius of the hole is 3 cm, I assume the radius of the spherical caps to be 3 cm. And if they're 3 cm, then the height of the cylinder becomes 10-6 = 4.
actually, the length of the cylinder would be 8. moreover, the volume for a spherical cap can be found here. it doesn't seem like you used that formula