Question

I am working on the test question #10 and am not sure if I am taking the right direction. It is as follows:

Answer 1

10. The disk bounded by the circle x^2+ y^2 = a^2
is revolved about the y-axis to make a sphere. Then a hole of diameter a is bored through the sphere along the y-axis (from north to south pole, like a cored apple). Find the volume of the resulting “cored” sphere. (Hint: Draw a picture of the two-dimensional region to be revolved, and label parts of your picture to help set up the integration.)


So I drew the picture on my piece of paper. (also I am assuming that the "a" variable in the sphere is different from the a in the hole that's bored through, is this a correct assumption?)

The integral I get for calculating the volume is this:
\[\int\limits_{1/2a}^{b} 2\pi(\sqrt(b^2+x^2)(x-1/2a)dx\]

So my question is: Is this the correct integral to calculate the volume of the bored sphere?

Thanks in advance and hopefully someone else is also working on this :D

Answer 2

Also this is the top half calculation and I would take the final answer times 2.

Answer 3

I'm guessing the "a" is the same. No reason why not. I'm guessing you are finding the volume for the special case where the drill has a diameter half that of the sphere.This one took so long but was interesting. I did the following:
Drew my diagram or a circle in quadrant 1 with the drilled part. I did cylindrical shell method:
\[1/2V=\int\limits_{0.5a}^{a}x \sqrt{a^2-x^2}dx\]

Answer 4

I'm going to go back hopefully tomorrow and revisit this problem however I have a question. If you revolve a circle that is in quadrant 1 about the y axis you will get a torus(or doughnut) not a sphere right? And this problem is asking us to make a sphere and then drill a hole through the center.

Answer 5

P.S. I will try and post my handwritten notes of my rework.

Answer 6

Err I didn't explain it well, sorry. The equation is also flawed =/
Its the same drawing of a semicircle in the first two quadrants and you probably have. The middle is drilled off. So from -a/2 to a/2 will be empty space. On the right (first quadrant) i am left with this sort of quarter circle which if i revolve around the y-axis using shells i will get the area of the top part of the drilled cylinder. I multiply this by 2 to include the area of the bottom.
The equation should be:

\[1/2V=2\pi \int\limits_{a/2}^{a}x \sqrt{a^2-x^2}dx\]

Answer 7

drilled sphere i mean. sigh

Answer 8

Ok, yes, I got the same equation as you :D. I had to solve this one tonight. I double checked my answer by also finding the drilled portion and adding it to my answer which gave a total volume exactly equal to the volume of a sphere of the same radius. I have posted my final notes for anyone else wanting to see the solution (after they attempted of course :D).

P.S. I appreciate the dialog Xavier as it has been helping me when I get stuck!

Answer 9

Glad to help. Also a u substitution with u=a^2-x^2 would have made it a bit less work.

Answer 10

Heh, yeah, I have been doing too many trig subs lately.