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Topic: Find the Volume and Surface Area (Read 883 times) |
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ThudnBlunder
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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?
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balakrishnan
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Re: Find the Volume and Surface Area
« Reply #1 on: Jan 11th, 2007, 10:09am » |
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volume is pi-4/3
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towr
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Re: Find the Volume and Surface Area
« Reply #2 on: Jan 11th, 2007, 10:19am » |
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Hmmm.. That's not what I got.. I think I'll check my calculations again..
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towr
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Re: Find the Volume and Surface Area
« Reply #3 on: Jan 11th, 2007, 10:24am » |
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Ok, i shouldn't have read isosceles as equilateral, but I still get twice that..
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balakrishnan
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Re: Find the Volume and Surface Area
« Reply #4 on: Jan 11th, 2007, 10:30am » |
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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
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| « Last Edit: Jan 11th, 2007, 10:50am by balakrishnan » |
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towr
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Re: Find the Volume and Surface Area
« Reply #5 on: Jan 11th, 2007, 10:33am » |
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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  10 2(1 - x) 2 (1 - x2) dx
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| « Last Edit: Jan 11th, 2007, 10:34am by towr » |
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