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Find Volume of sphere

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  • #1
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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)?
 

Answers and Replies

  • #2
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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.
 
  • #3
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Right, right. Should I use the method of washers as the cross sectional area?
 
  • #4
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Right, right. Should I use the method of washers as the cross sectional area?
You can use washers to compute the volume of the sphere (with a hole drilled in it) or you can use shells - your call.

I would start with a drawing of a circle of radius R, centered at the origin. Remove a central rectangle of width 2r. That will be what a cross section of the solid looks like. Then use washers or shells to get your integral for the volume.
 
  • #5
HallsofIvy
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Start by drawing a picture: A circle of radius R (representing a two-dimensional view of the sphere) and rectangle, of width r (representing the cylinder) inscribed in the sphere.

Notice that there are three volumes you need to compute. The rectangle has width r and height [itex]2\sqrt{R^2- r^2}[/itex] (use the Pythagorean theorem to get that) so represents a cylinder of radius r and height [itex]2\sqrt{R^2- r^2}[/itex] so the cylinder has volume [itex]\pi r^2h= \pi r^2\sqrt{R^2- r^2}[/itex].

The other two parts are the "caps" on each end of the cylinder. They are portions of the sphere from radius r to R and represented on your picture by the portions of the circle from radius r to R. The two volumes are the same, of course, and can be calculated as rotating the region between [itex]y= \sqrt{R^2- x^2}[/itex] and y= R around the x-axis.

Of course, to find the volume remaining after that volume is removed, subtract that result from the volume of the sphere.
 
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  • #6
vanhees71
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My guess is that the most direct way to calculate the volume of the cylinder-like hole (note that the caps are not plain but parts of the sphere!) is to introduce cylinder coordinates and then do the integral!
 
  • #7
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.
 
  • #8
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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.
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.
davidmoore63@y said:
Then it turns out that the volume remaining in the sphere is independent of R, which is interesting.
 
  • #9
HallsofIvy
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lf You drill a hole through the center of a sphere of radius R, the radius of the cylindrical hole is fixed. Draw a line from the center of the hole to one
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.
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].

The fact that the answer is independent of R is very interesting- and allows for a very easy solution to this problem. Given h, the length of the hole, imagine a sphere of radius R such that the radius of the hole is 0. That is [itex](h/2)^2= R^2[/itex] so that R= h/2. The volume removed from the sphere is 0 so the "volume remaining" is [itex](4/3)\pi (h/2)^3= \pi h^3/6[/itex].
 

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