anonymous 5 years ago I am working on the test question #10 and am not sure if I am taking the right direction. It is as follows:

1. anonymous

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

2. anonymous

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

3. anonymous

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: $1/2V=\int\limits_{0.5a}^{a}x \sqrt{a^2-x^2}dx$

4. anonymous

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.

5. anonymous

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

6. anonymous

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

7. anonymous

drilled sphere i mean. sigh

8. anonymous

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!

9. anonymous

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

10. anonymous

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