I have this question which states:
Let $T$ be the tetrahedron bounded by the planes $x = 0, y = 0, z =0$ and $x+y+z = 2$
Integrate $$J = \iiint_{T} y^2 \,dx\,dy\,dz$$ Using the above integral, or otherwise, evaluate $$K = \iiint_{P}(x^2+\sin(xy^2)+ y^2) \,dx\,dy\,dz$$
where $$P = \left \{(x,y,z) : 0 \leq z \leq 2 -|x| -|y|\right \}$$
My thoughts are as follows I think the first part of the question is understandable just use the limits and follow through with the solution $J = 0$ but the second part is tougher.
I'm guessing we realise (if we let) $f(x,y) = x^2y$ this is odd w.r.t. y and domain P is symmetric about y-axis. And the same for $g(x,y) = \sin(xy^2)$ which is odd w.r.t. x and P is symmetric about x-axis. We can say their integral over P is $0$
So the final solution is $$K = J = 0$$
I hope someone could either correct me and tell me how to proceed if this is wrong thank you!
I'm going to give you an hint only for the triple integral $J$ and hope I will not do any stupid error.
The region you are considering is $x+y+z = 2$, $x\ge0$, $y\ge0$ and $z\ge0$. Then you can get for example the following parameterization: $$y\in [0,2], \quad z \in [0,2-y], \quad x \in [0,2-y-z]$$
Once you got this you can write the integral as follows: $$\int_0^2 \int_0^{2-y} \int_0^{2-y-z}y^2 \ \ dx\ dz\ dy$$
Now you just have to perform the integrations correctly and you should get $J = \frac{8}{15}$