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Wednesday, March 30, 2011

Integration trick

I was working on this problem, long story short it didn't amount to much. But I thought the technique I used trying to solve it was interesting in itself. If you have a certain graph in a cube like so:
x^2 in 3 dimensions
  But you don't want to find the volume of the whole rectangular area looking form the top, you just want it to look like this when you look down on it from the top (you just want between the black lines):
topview cutout
  SO that an artists poor rendering might be this:
You want the volume of this triangular wedge thingie
So the technique I use is to make two functions, one that goes from 4 to 10  and one that goes from 0 to -8 in the time it takes y to go from 0 to 1. Like so:
b = 4*(1-y)+10*y
a = 0*(1-y)-8*y  
If you mess around with these you can see b = 4 and a=0 when y=1 and a=-8 and b = 10 when y=1 and they go in lines between the points in between. 

So I set up my integral like this:

Then integrate with y going from 0 to 1 and x going from 0 to 10...
That gives the wedge shaped thing's volume to be:
V=148

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