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 January 30th, 2007, 02:55 AM #1 Senior Member   Joined: Dec 2006 From: New Jersey Posts: 378 Thanks: 1 Cylinder in a Sphere Inscribed a right circular cylinder of height h and radius r in a sphere of fixed radius R. Express the volume V of the cylinder as a function of h. HINTS GIVEN: (1) V = pi(r^2)(h). (2) R = the hypotenuse and r = one of the legs of the right triangle in the cylinder.
January 30th, 2007, 03:13 AM   #2
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I don't get the point of this problem. It looks like a complicated trick question to me.

Quote:
 Inscribed a right circular cylinder of height h and radius r in a sphere of fixed radius R. Express the volume V of the cylinder as a function of h.
Well, it doesn't matter where the cylinder is, how it is oriented, whether it is inscribed, or whatever.
A cylinder's volume is always related to its height by the equation:
V=pi(r^2)(h), so they actually already gave you the answer to the question in their hints.

 January 30th, 2007, 03:20 AM #3 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Oh, OK, I see now what they're getting at. They want to get rid of the 'r' variable and express the whole thing in term of h only. So, all we must do is find an expression that relates r and h.
 January 30th, 2007, 01:09 PM #4 Senior Member   Joined: Dec 2006 From: New Jersey Posts: 378 Thanks: 1 ok How do we find an expression that relates r and h?
 January 30th, 2007, 02:31 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,472 Thanks: 2039 Follow the second hint - the unmentioned third side of the triangle is of length h/2. Pythagoras's theorem gives the relationship needed.
 February 2nd, 2007, 03:52 PM #6 Senior Member   Joined: Dec 2006 From: New Jersey Posts: 378 Thanks: 1 ok Can you show me how?
 February 2nd, 2007, 05:14 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,472 Thanks: 2039 Sketch the situation. You should find it easy to spot a triangle containing a right angle, with sides are of lengths r, h/2 and R (for the hypotenuse).
 February 3rd, 2007, 08:35 AM #8 Senior Member   Joined: Dec 2006 From: New Jersey Posts: 378 Thanks: 1 ok We have: r, h/2 and R = hypotenuse, right? I get that part but what am I solving for? Do I use the Pythagorean Theorem to solve for r, h/2 or R?
 February 3rd, 2007, 12:03 PM #9 Global Moderator   Joined: Dec 2006 Posts: 20,472 Thanks: 2039 In order, "yes", "r" and "yes, r".
 February 3rd, 2007, 01:45 PM #10 Senior Member   Joined: Dec 2006 From: New Jersey Posts: 378 Thanks: 1 ok You are saying to solve for r, h/2 and R INDIVIDUALLY IN ORDER?

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### Inscribe a right circular cylinder of height hand radius rin a sphere of fixed radius R.See the illustration.Express the volume Vof the cylinder as a function of h.

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