Using a triple integral to find volume

[Jgalt]

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!

 

[olgranpappy]

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
<br /> \int dx dy f(x,y)<br />

 

[HallsofIvy]

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.