Evaluate the triple integral

Question

Evaluate triple integral bounded by a solid $S$ which is a tetrahedron defined by vertices $(0,0,0)$, $(0,0,1)$, $(1,1,0)$ and $(-1,1,0)$.

$$\iiint y^2\, dV$$

So, I tried to find the plane that contains the tree vertices except $(0,0,0)$ and with that I was planning to integrate the solid as I was doing with regions like that. But the plane doesn't have a coordinate x and so I'm in trouble at the integration. In the end, I have found the one set of limits:

$ S= {{(x,y,z): 0<z<1 , 0< z <2-y , -y<x<-y}} $

Answer 1

The plane that contains all the vertices except $0,0,1)$ is easy to find; it is the XY plane. Now if we slice the tetrahedron by a plane at $z = z_0$ where $0 \leq z_0 \leq 1$ that slice is a right triangle with vertices $$(0,0,0), (1-z_0,1-z_0,0), (z_0-1,1-z_0,0) $$ so the limits of integration to use to deal with such a plane is $$ \int_{y=0}^{1-z} \int_{x=-y}^y dx \,dy $$ And the volume integral you want is $$ \int_{z=0}^1\int_{y=0}^{1-z} \int_{x=-y}^y y^2\,dx \,dy \, dz $$ SPOILER ALERT Don't read on if you want to do the integral yourself. $$ \int_{z=0}^1\int_{y=0}^{1-z} \int_{x=-y}^y y^2\,dx \,dy \, dz = \int_{z=0}^1\int_{y=0}^{1-z} \left[ xy^2 \right]_{x=-y}^{x=y} \,dy \, dz \\ = \int_{z=0}^1\int_{y=0}^{1-z} 2y^3\,dy \, dz = \int_{z=0}^1 \left[ \frac{y^4}{2}\right]_{y=0}^{1-z}\\ = \frac12\int_{z=0}^1 \left( 1-z \right)^4 \, dz=\frac{1}{10} $$