Find a volume using a triple integral.. This is the problem:
Find the volumen of de solid bounded by $x^2+y^2=2$, $z=0$ and $x+2y+2z=2$.
I have set the parameters to:
$0 \leq z \leq \dfrac{2-x-y}{2}$
$ 0\leq y\leq 2-x$
$0 \leq x \leq 2$
and evaluated:
$
  \displaystyle{  \int\limits_{0}^{2} \int\limits_{0}^{2-x} \int\limits_{0}^{\dfrac{2-x-y}{2}}  dz\,dy\,dz}=\dfrac{4}{3}
$
Is there anything wrong?? Please help me... thank you so much...
 A: There is no way in which a 2d-circle and a plane may bound a solid, so I assume you are trying to find the volume of the intersection between the cylinder $\{x^2+y^2\leq 2,z\geq 0\}$ and the half-space $x+2y+2z\leq 2$. It is worth to apply a rotation around the $z$-axis, by setting:
$$ u=\frac{x+2y}{\sqrt{5}},v=\frac{2x-y}{\sqrt{5}} $$
in order that the problem can be stated as: find the volume of the intersection between the cylinder $\{u^2+v^2\leq 2,z\geq 0\}$ and the half-space $\frac{\sqrt{5}}{2}u+z\leq 1$. Now the solid is symmetric with respect to the $uz$-plane.
Since $-\sqrt{2}\leq u\leq\sqrt{2}$, $z$ ranges between $0$ and $1+\frac{1}{2}\sqrt{10}$. Let us compute the area of a section $A_z$ according to the value of $z$ in the previous range:
$$ A_z = 2\int_{-\sqrt{2}}^{\frac{2}{\sqrt{5}}(1-z)}\sqrt{2-u^2}\,du =2\left.\left(w\sqrt{1-w^2}+\arcsin w\right)\right|_{-1}^{\sqrt{\frac{2}{5}}(1-z)}$$
giving:
$$ A_z = \pi + 2\left(\sqrt{\frac{2}{5}}(1-z)\sqrt{1-\frac{2}{5}(1-z)^2}+\arcsin\left(\sqrt{\frac{2}{5}}(1-z)\right)\right).$$
Now the volume is given by:
$$ V = \int_{0}^{1+\frac{1}{2}\sqrt{10}}A_z\,dz = \color{red}{\pi+\frac{4}{5}\sqrt{6}+2\arcsin\sqrt{\frac{2}{5}}}.$$
A: $\int_0^2\int_0^{2-x}\int_0^{(2-x-y)/2}dzdydx=\int_0^2\int_0^{2-x}\frac{2-x-y}{2}=\int_{0}^2(0-1)=2$
