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So, I am having trouble (again) with the domain for a triple integral of a function, bounded by the paraboloid $2y^2=x$ and the $x+2y+z=4$ and $z=0$ planes

I have tried to guess the bounds for x,y and z in cartesian coordinates with no luck, and it's somewhat apparent that this needs to be done in cylindrical polar coordinates. The issue here is that I haven't been able to see any clear values of the bounds for the integrals for either $\theta, r$, or $z$ ,so I am not sure if I am missing something else.

So far, my polar conversion is:

$$\begin{aligned} & 2r^2 \sin^2 \theta = r \cos \theta \\ & r \cos \theta + 2 r \sin \theta+z=4 \\ & 0 \leq z \leq 4 - r \cos \theta - 2 r \sin \theta \\ & 0 \leq r \leq \text{(?)} \\ & 0 \leq \theta \leq \text{(?)} \end{aligned}$$

Any help is welcome.

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    $\begingroup$ Try it in simple Cartesian coordinates $\endgroup$ – Tojrah Apr 23 at 16:59
  • $\begingroup$ What makes you think that cylindrical coordinates are called for here? $\endgroup$ – amd Apr 23 at 17:41
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We better keep the given coordinates $(x,y,z)$. The desired finite domain $D$ can be described as follows: The infinite parabolic cylinder $P:=\bigl\{(x,y,z)\bigm| x\geq 2y^2, \ -\infty<z<\infty\bigr\}$ is cut below by the plane $z=0$ and on top by the plane $z=4-x-2y$. These two planes intersect in the line $$\ell: \ x+2y=4\ \wedge\ z=0\ ,$$ creating an infinite wedge $W$ above $z=0$. Our $D$ then is the intersection $P\cap W$. The line $\ell$ intersects the parabola $x=2y^2$ in the points $(8,-2)$ and $(2,1)$. We therefore can write $D$ in the form $$D:=\bigl\{(x,y,z)\bigm| -2\leq y\leq 1, \>2y^2\leq x\leq4-2y, \ 0\leq z\leq 4-x-2y\bigr\}\ .$$ This indicates that we can write an integral over $D$ in the form $$\int_D f(x,y,z)\>{\rm d}(x,y,z)=\int_{-2}^1\int_{2y^2}^{4-2y}\int_0^{4-x-2y} f(x,y,z)\>dz\>dx\>dy\ .$$

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not a full answer... so don't downvote.

First, just plot the region:

enter image description here

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