Question

"A cylindrical hole of radius 1 is bored through a solid ball of radius 2. Find the volume and outer surface area." I calculated the volume fine, which came out to be \(4\pi\sqrt{3}\), however, when I tried calculating the outer surface area, I got the same numerical result, though through entirely different means. Can anyone verify this for me if the surface area is in fact \(4\pi \sqrt{3}\)?

Answer 1

Kidding...I recalculated the surface area as \(8 \pi \sqrt 3\). Can anyone verify this for me? Thank you.

Answer 2

I got the volume of the sphere to be 32 pi / 3.
( V = 4/3 pi (2)^3 )
The volume of the cylinder is 4 pi.
(V = pi* (1)^2 (4) )
So the volume of the sphere with hole = 32pi/3 - 4 pi = 20 pi/3.
Please check.

Answer 3

For surface area of the sphere with hole, I got 24 pi.
SA = area of sphere + lateral area of cylinder
SA = 4 pi (2)^2 + 2 pi(1) (4)
SA = 16 pi + 8 pi = 24 pi
Please check.

Answer 4

Sorry, Directrix, your answers are wrong. The portion of the ball that is drilled out is NOT in fact cylindrical, so you cannot simply subtract out a cylindrical volume from the entire volume of the ball. For your surface area calculation, you are not considering that the top and bottom caps of the sphere are removed when we drill the hole in the ball. You can't simply use the sphere surface area formula here. Additionally, it only asked for the outer surface area, so we don't need to deal with the cylindrical surface area formula at all.

You need to do multiple integration with coordinate parametrization for the surface area calculation, and you need to apply Cavalieri's principle for the volume--that I am sure of. I just need a check on my answers.

Answer 5

But it says cylindrical hole of radius 1, what's it for?

Answer 6

The remaining lateral area is:\[8 \sqrt{3} \pi \]A solution is attached.

Answer 7

Ishaan94 - It's a cylindrical hole, but the portion of the ball that is removed is in fact not cylindrical (it has rounded caps). Think about it.

Robtobey - Thank you so much! I found my algebraic error which lead to the missing factor of 2. Your solution was very detailed and helpful.

Answer 8

Robtobey - This was a problem in a Calc III textbook, so the idea was to use integration to solve this instead of canned formulas, just so you know, but for the purposes here, I just needed a check with the answer. Thanks!

Answer 9

@yakeyglee

Thank you for the medal(s).

Sorry about the non-calculus solution. My approach was to solve the problem in the simplest way. Discovered the cap volume and area formula with a google search. These expressions were probably obtained with the use of the calculus in the first place.