Triple integral (mass) - setting up region between planes and parabolic cylinder

Question

I am trying to set up the following triple integral using the xy plane.

$E$ is bounded by the parabolic cylinder $z=1-y^2$ and the planes $x+6z=6, x=0$, and $z=0 ; \quad \rho(x, y, z)=8$.

I set up these ok:

$m=\iiint_E \rho d x d z d y$ and $m=\iiint_E \rho d y d z d x$

but also wanted to try to set up $d z d x d y$.

Where am I going wrong?

$\int_{y=-1}^1 \int_{x=6 y^2}^6 \int_{z=0}^{1-y^2} 8 d z d x d y$

$\begin{aligned} & x=6-2\left(1-y^2\right) \\ & x=6-6\left(1-y^2\right) \\ & x=6 y^2\end{aligned}$

Thanks.

Answer 1

I think the problem here is that the way you write the triple integral is actually depicting a dydzdx-view region. If you want to set up dzdxdy, you need to depict the region in a dzdxdy-view.

If you try to set up dzdxdy , actually we are looking from the direction of the positive y-axis, the region is a cylinder (the graph of this region). If we decide the positive direction of y-axis is pointing to the top, then the side of curved-topeed cylinder is a closed-curve on xz plane (composed of line z=0, z=1, x=0 and x=6-6z),the upper surface is $y=\sqrt{z-1}$ and the lower surface is $y=-\sqrt{z-1}$. The triple integral is $8\int_{z=0}^{z=1}dz\int_{x=0}^{x=6-6z}dx\int_{y=-\sqrt{1-z}}^{y=\sqrt{1-z}}dy$. Or you can multiply $8\int_{z=0}^{z=1}dz\int_{x=0}^{x=6-6z}dx\int_{y=0}^{y=\sqrt{1-z}}dy$ by 2 and write it as $16\int_{z=0}^{z=1}dz\int_{x=0}^{x=6-6z}dx\int_{y=0}^{y=\sqrt{1-z}}dy$.