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The triple integral is bounded by

$$z=4-y^2$$ $$y=2x$$ $$z=0$$ $$x=0$$

I have to rewrite it so that the plane is on the $xz$ plane, and therefore I fix $y$.

First I notice that $z>=0$, and so $-2\leq y\leq 2x$

Now on the $xz$ plane, I have the following:

enter image description here

The red line is $x=0$, the yellow line is $z=0$, and the blue line is $z=4-y^2$, but since $y=0$, we just have $z=4$

I see no "bounded area". What is my mistake here? I know that this integral is supposed to converge. I also took into account all the functions.

enter image description here

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Your bounded region doesn’t intersect the $xz$-plane except along the line $x=0$. You want the region with $2x<y<\sqrt{4-z}$. You can see nice cross-sections of this region in the $xy$-plane and in the $yz$-plane. You can also get non-trivial pictures by setting $y=k$ with $0<k<2$.


Edit

To project this region onto the $xz$-plane, it helps a lot to visualize the region in $3$ dimensions first. At any rate, the trick is to look at the intersection curves of the defining surfaces, and then we project those curves by eliminating $y$ algebraically. Thus, the surface $y=2x$ and the surface $z=4-y^2$ project onto the curve $z=4-(2x)^2$, or $z=4-4x^2$. That curve, together with $z=0$ and $x=0$ define a region in the $xz$-plane, which you can use to set up an integral with order of integration $dy\,dz\,dx$. Namely:

$$\int_0^1\int_0^{4-4x^2}\int_{2x}^{\sqrt{4-z}}f(x,y,z)\,dy\,dz\,dx$$

Does that help?

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  • $\begingroup$ The question says to rewrite the integral so that it is projecting on each of the planes seperately. I did $yz$ and $xy$ planes, that's why I need to do $xz$? $\endgroup$ – K Split X Nov 25 '17 at 14:41
  • $\begingroup$ I don't understand what you're asking in this comment. I was simply addressing why you saw no "bounded area" when you looked at the xz plane. In which order are you trying to write the integration? Do you want $dz\,dx\,dy$? $\endgroup$ – G Tony Jacobs Nov 25 '17 at 15:04
  • $\begingroup$ Sorry I wasn't clear enough. Please take a look at the picture i've added in the question. In red it says that we must rewrite the integral so that it projects on each of the planes seperately. I reached the $xz$ plane, and see that they don't intersect the plane anywhere, so is there no integral I can write? $\endgroup$ – K Split X Nov 25 '17 at 15:18
  • $\begingroup$ I see. This is your mistake: There's a difference between projecting onto a coordinate plane and intersecting with that coordinate plane. Setting $y=0$ doesn't give you the projection of your region onto the $xz$ plane, but only the (trivial) intersection with the $xz$-plane. I'll edit my answer to say more. $\endgroup$ – G Tony Jacobs Nov 25 '17 at 15:27
  • $\begingroup$ Thank you your edited answer helps. Basically I needed to replace $y$ with $2x$, to get a correspding intersection $\endgroup$ – K Split X Nov 25 '17 at 15:46

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