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Using a triple integral to find volume

  • Thread starter Jgalt
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I'm supposed to find the volume of a solid bound by co-ordinate planes x=0,y=0, z=0 & surface z=1-y-x^2 and am having a lot of difficulty doing so. f(x,y,z) is not given so I am assuming it is one. I figure I should then take the triple integral dzdydx. Then, I made a 2D sketch for the xy plane to determine limits of integration. I took the ∫dz from 0 to 1-y-x^2 then ∫dy from 0 to 1-x, then ∫dx from 0 to 1, but I keep getting a negative volume so I must be doing something wrong... Please help!
 

Answers and Replies

olgranpappy
Homework Helper
1,271
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you want the volume under the surface f(x,y)=1-y-x^2 so find the limits in the x,y plane and compute the integral
[tex]
\int dx dy f(x,y)
[/tex]
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
I'm supposed to find the volume of a solid bound by co-ordinate planes x=0,y=0, z=0 & surface z=1-y-x^2 and am having a lot of difficulty doing so. f(x,y,z) is not given so I am assuming it is one. I figure I should then take the triple integral dzdydx. Then, I made a 2D sketch for the xy plane to determine limits of integration. I took the ∫dz from 0 to 1-y-x^2 then ∫dy from 0 to 1-x, then ∫dx from 0 to 1, but I keep getting a negative volume so I must be doing something wrong... Please help!
And what did you see when you made a 2D sketch for the xy plane to determine the limits of integration? The plane z= 1- y- x^2 crosses the plane z= 0, when 0= 1- y- x2 or y= 1- x2. Together with with x= 0, y= 0, that is half of a parabola. For each x, y goes from 0 to 1- x2, not 1-x.
 

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