The problem is:
Calculate
$$\int_{R}\sin(\pi y^{3}) \ dzdydx$$
Where $R$ is the pyramid having vertices at $B= (0,0,0),D=(0,1,0),A=(1,1,0),E=(1,1,1)$ and $C=(0,1,1)$.
I approached the problem by setting out the points in Geogebra 3D. I then concluded that one can enclose the pyramid with base parallel to the $xz$-plane by $x=0,z=0,z-y=0,x-y=0$ and $y=1$. My question is if it is then correct to say that
$$\int_{R}\sin(\pi y^{3}) \ dzdydx =\int_{0}^{1}\int_{0}^{y}\int_{0}^{z} \sin(\pi y^{3})\ dxdzdy+\int_{0}^{1}\int_{0}^{y}\int_{0}^{x} \sin(\pi y^{3})\ dzdxdy$$
Added:
Best Answer
I would calculate your integral on this way
$$\int_{R}\sin(\pi y^{3}) \ dzdydx=\int_0^1\int_0^y\int_0^y\sin \pi\,y^3\,dz\,dx\,dy=\frac{2}{3\,\pi}$$