Volume computed by a double integral We are asked to find the volume of the following solid:
$$x=0, y=0, z=0, x+y+z=2, y^2=1-z, y>0$$
We should compute $\displaystyle \iint_{T}(x+y-y^2-1)dxdy$, but I cannot figure out which is the orthogonal $T$, in which we can integrate. I have tried to transform the above expressions into polar coordinated, but this just made things worse, so I was trying to consider the zyx coordinates system instead. I know that the result is $\frac{49}{60}$.
Any advice on this problem would be appreciated.
 A: The region is parabolic cylinder $y^2 = 1-z~$ in the first octant and bound above by $x = 2 - y - z$. If you take projection of the region in a coordinate plane other than yz-plane, you will be dealing with split integral. So it is easier to integrate in the order $dx, ~dz $ and finally $dy$.
The bounds of the integral is,
$0 \leq x \leq 2 - y -z, 0 \leq z \leq 1 - y^2, 0 \leq y \leq 1$
And $ ~\displaystyle \frac{49}{60}~$ is indeed the volume of the region.
A: The integration volume is in the picture below. $x$ axis is in red, $y$ is in green, and $z$ is in blue.

You have a pyramid with vertices $(0,0,0), (2,0,0), (0,2,0), (0,0,2)$ from the first four equations. Then you have the parabolic cylinder surface. You arte interested in the volume inside the intersection of the two. So if your integration region is in the $xy$ plane, you need to split your integral into two parts, where the upper limit of $z$ is given by the pyramid, and the region where the upper limit of $z$ is given by the parabola.
If I were you, I would choose the integration domain to be in the $yz$ plane ($x=0$), the region where it is inside of the parabola. Then the "height" of the function is the distance from $x=0$ plane to the $x+y+z=2$ plane.
A: The problem becomes considerablty easier if you choose the right order to integrate the coordinates. In this case, we want to start with $x$. This is because in the $yz$ plane, the height of the surface is always $2-y-z$, which means we don't need to do any splitting of the region. Once we have $y$ and $z$ left, the region to integrate over is $0 < z < 1-y^2$. We could also express it as $0 < y < \sqrt{1-z}$, but that's not as simple. Finally, we want $y$ to go from $0$ to $1$ to cover the whole region, so the integral will be
$$
\int_0^1\int_0^{1-y^2}\int_0^{2-y-z}dxdzdy.
$$
Standard integration methods will give the result $49/60$.
