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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  
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
I don't get the point of this problem. It looks like a complicated trick question to me. Quote:
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: 20,931 Thanks: 2205 
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: 20,931 Thanks: 2205 
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: 20,931 Thanks: 2205 
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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