Volume: The Disk Method

1. Aug 7, 2008

carlodelmundo

1. The problem statement, all variables and given/known data

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

Solution:

You can imagine the ring to be generated by a segment of the circle who equation is x^2 + y^2 = 25. Because the radius of the hole is 3 inches, you can let y = 3 and solve the equation x^2 + y^2 = 25 to determine that the limits of integration are x = +- 4 . So, the inner and outer radii are r(x) = 3 and R(x) = sqrt(25-x^2).

2. Relevant equations

x^2 + y^2 = r^2 (Standard Form of A Circle)

3. The attempt at a solution

This is an example problem in my textbook, Calculus 8th Edition by Larson (pg. 461 Section 7.2)

My fundamental question: How did they find the limits of integration by substituting 3 with y?

I just don't see how plugging in the inner radius (y = 3) into the equation of a circle would find the limits of integration? I can see how in a 2D circle, x has solutions when it is -4 and 4.....but why couldn't we plug in radius (y = 5) in instead of 3?

Thanks

2. Aug 7, 2008

moemoney

In this case, take the center corss section of the sphere and look at the top half of it. It will look like a semi-circle with radius 5 correct? (y = sqrt(25-x^2))

Now the sphere was drilled and the resulting hole has a radius of 3. So now the bottom half of the semi-circle is now removed and the semi-cricle is sitting 3 inches away form the x-axis. Can you visualize this? I wish I could show you a diagram...

By substituting y = 3 into the circle equation, you will find where the line intersects with the semi-circle. This will give you the end points of the semi-circle and thse end points define the limits of integration.

3. Aug 7, 2008

Defennder

You have draw it out to understand the solution. Draw the equation of the circle and the line y=3. Without the line you can see that the volume of the sphere can be generated by revolving the semicircle about the x-axis. You should see that the volume of solid generated by rotating the graph from the top of the circle to the line y=3 is the remaining solid.

Plugging in y=5 would mean that the entire sphere was drilled (destroyed).

4. Aug 7, 2008

carlodelmundo

Thank you both!

Before, what I was doing was:

1.) Drawing a circle whose center is on (0,5) instead of the origin (0,0). I thought the only way to create a sphere is if the whole circle was above the x-axis (though, I will learn later that a semi-circle is enough to make the sphere with radius 5.

Basically, I followed your methods: the top semicircle is defined by y = sqrt(25-x^2) and the "hole" will be drilled at y = 3 (and y = -3... radius is still equal to 3).

Anyway, focusing on one semi circle...to find the limits of integrations, you must equate the two equations (y = sqrt(25-x^2) and y = 3).... which the solution is x = +- 4.

Thank you. The book's interpretation was far more complicated than you two's.