# Understanding the domain of the triple integral for $f(x,y,z)=x+2y+z^2$

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.

• Try it in simple Cartesian coordinates – Tojrah Apr 23 at 16:59
• What makes you think that cylindrical coordinates are called for here? – amd Apr 23 at 17:41

We better keep the given coordinates $$(x,y,z)$$. The desired finite domain $$D$$ can be described as follows: The infinite parabolic cylinder $$P:=\bigl\{(x,y,z)\bigm| x\geq 2y^2, \ -\infty is cut below by the plane $$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\ .$$ 