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I'm confused with this problem.

Determine the volume of the solid limited for $x = 1-y$, $x = 3-y$, $y = 0$, $z = 0$ and $z = 1-y^2$.

What I tried to do: well, first I suppose that the function I have to integrate is $f(x,y,z) = z+y^2-1$$, and my triple integral is

$$ \iint_D\int_0^{1-y^2}\,(z+y^2-1)\,dz\,dA $$

Where D is the region limited by the lines $x+y = 1$ and $x+y = 3$, so my integral is

$$ V = \int_1^3\int_{1-x}^{3-x}\int_0^{1-y^2}\,(z+y^2-1)\,dz\,dy\,dx $$

I'm not sure if I'm right, so I appreciate all your ideas.

Alexei0709
  • 1,184
  • If my sketch is correct, then there are two solids, one with $y \gt 0$ and the other with $y \le 0$. And for computing a volume via integrals one starts usually with the integrand "1". – andre Oct 13 '14 at 21:30
  • So the function I have to integrate is not $f(x,y,z) = y^2+z-1$? – Alexei0709 Oct 14 '14 at 02:17
  • I make an example for a $3$-simplex $S$ given by the $4$ vertices $e_0=(0,0,0)$, $e_1=(1,0,0)$, $e_2=(0,1,0)$, $e_3=(0,0,1)$. The volume of $S$ is the volume of the convex hull of the 4 vertices. Which gives $$\int\limits_{0}^{1}\int\limits_{0}^{1-x}\int\limits_{0}^{1-x-y} 1 : dz:dy:dx = 1/6.$$ For a $3$-dimensional volume you have to find the correct bounds for each of the three integrals and integrate the function $f(x,y,z)=1$. – andre Oct 14 '14 at 07:56

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