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        riddles >> putnam exam (pure math) >> Find the Volume and Surface Area
        
(Message started by: THUDandBLUNDER on Jan 11^th, 2007, 9:52am)

Title: Find the Volume and Surface Area
Post by THUDandBLUNDER on Jan 11^th, 2007, 9:52am

I first saw the wedge-shaped solid below in one of Martin Gardner's books. From the front it looks like a square with side length 2 units, from the side it looks like an isosceles triangle with base and altitude both 2 units, and from the top it looks like a circle with diameter 2 units.

What is its volume and surface area?

Title: Re: Find the Volume and Surface Area
Post by balakrishnan on Jan 11^th, 2007, 10:09am

[hide]volume is pi-4/3[/hide]

Title: Re: Find the Volume and Surface Area
Post by towr on Jan 11^th, 2007, 10:19am

Hmmm.. That's not what I got.. I think I'll check my calculations again..

Title: Re: Find the Volume and Surface Area
Post by towr on Jan 11^th, 2007, 10:24am

Ok, i shouldn't have read isosceles as equilateral, but I still get twice that..

Title: Re: Find the Volume and Surface Area
Post by balakrishnan on Jan 11^th, 2007, 10:30am

[hide]
It is indeed twice.
Seeing the figure,I though it is just 1 half of what it is actually.
First let us compute the volume:
z=2(1-x)
So the volume is
2*int[int[2(1-x)] dy_{y=-sqrt(1-x^2)}^{y=sqrt(1-x^2)} dx_{x=0 to 1}
which gives 2*pi-2/3

For the surface area
the curved surface area is
int[4*asin[1-z/2]}_{z=0 to 2}
adding this to [pi+sqrt(5)*pi] gives
pi*(5+sqrt(5))-8[/hide]

Title: Re: Find the Volume and Surface Area
Post by towr on Jan 11^th, 2007, 10:33am

Since you integrate from 0 to 1, shouldn't you double it to get the other half?
I did the integration along the line of the square view. All the rectangles make it simple, imo.

2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif¹₀ 2(1 - x) 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - x²) dx