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Homework Statement
A hole of radius r is bored through the center of a sphere of radius R. Find the volume V of the remaining portion of the sphere.
Homework Equations
The Attempt at a Solution
Wouldn't is be (4/3)∏(R^3-r^3)?
No. They're drilling a cylindrical hole through the sphere, not removing a sphere of radius r from it.Homework Statement
A hole of radius r is bored through the center of a sphere of radius R. Find the volume V of the remaining portion of the sphere.
Homework Equations
The Attempt at a Solution
Wouldn't is be (4/3)∏(R^3-r^3)?
You can use washers to compute the volume of the sphere (with a hole drilled in it) or you can use shells - your call.Right, right. Should I use the method of washers as the cross sectional area?
Which means it's a different question. The question in the OP is what it is -- I don't see anything wrong with the problem statement.Just wanted to check the question is written correctly. A hole 'of radius r' clearly means that the distance from the cylindrical axis of the hole to the edge of the hole is r. However I've seen this question before where the LENGTH of the hole is given.
davidmoore63@y said:Then it turns out that the volume remaining in the sphere is independent of R, which is interesting.
If the length of the hole is given, then for a sphere of radius R, the radius of the hole is fixed. Conversely, if the radius of the hole is given then the length is fixed so these are the same problem. Draw a line from the center of the sphere to the edge of the hole and a line from the center of the sphere perpendicular to the side of the hole. That gives a right triangle with hypotenuse of length R, one leg of length r, and the other leg of length h/2 so we must have [itex](h/2)^2+ r^2= R^2[/itex].Just wanted to check the question is written correctly. A hole 'of radius r' clearly means that the distance from the cylindrical axis of the hole to the edge of the hole is r. However I've seen this question before where the LENGTH of the hole is given. Then it turns out that the volume remaining in the sphere is independent of R, which is interesting.