# triple integral upper lower bounds

In this problem there is an oblique plane and they are trying to find the volume ABOVE the plane.

I understand how bottom and top limit of this question was found and it also makes sense to me how upper limit of integration was found when integrating with respect to x after it is integrated by Z but I dont get why when integrated with respect to x's lower limit is 0 because isn't the oblique plane changing the lower bound of x?

I am getting really frustrated and hazy when trying to figure out the upper and lower limits for triple integrals with variables for limits....isn't there an easier way to visualize or even come up with these limits?

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Let look at the area, we are working on:

you see that the plane intersects $z=0$ and so makes the red line and makes the following flat area on $xy-$plane:

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my confusion is that if you are integrating x in that 3D diagram doesn't that slanted plane decrease area for x compared to your x-y plane diagram above? What I am trying to say is that your x-y diagram shows the entire region x can be integrated over but in the 3D diagram not all the region is available for x to integrate over since the slanted plane reduces the area...for example, take slices of that 3D diagram at each Z level and see that the x-y diagram will vary for each level of Z, your diagram is true for z=0 but not for z=1 or z=1.2 because area for x reduces for integrat –  Raynos Dec 6 '12 at 7:00
@Raynos: What is we always do in this kind of 3D are is to consider the whole area firstly. Then we search for a proper range for $x$ and then for $y$ secondly. An finally we search for a suitable range for $z$. If your 3D area is as you had shown or as I showed above so we find the intersection of the plane and $z=0$. OK?? –  B. S. Dec 6 '12 at 7:18
@Raynos: The red line shows a part of the line $6-2 x-y=0$. What is the range for $x$? It is $[0,3]$ and what is for $y$? it is $[0,6-2x]$ where $x$ varies in its interval $[0,3]$. This makes you to avoid considering what you had noted completely in above comment. I see what you were trying to tell me, but know that we don't treat the area as you mentioned above. We treat the area as I pointed here. Are you satisfied ? Tell me until we get a common view. :) –  B. S. Dec 6 '12 at 7:42
I think I get what you are saying....So let me try to rephrase what you say. Basically you start off with picking upper and lower limit for z and then you remove Z from the equation making z=0 and then look at the remaining variables' plane and put their max range for integration. Is this what you are saying? –  Raynos Dec 6 '12 at 7:54
@Raynos: Yes. We want to find a proper ranges for $x$ and $y$ and this can't be happen unless you put $z=0$. Always, for finding the range for $y$, draw a thin line perpendicular to $x$ axis. This line is just to show us the range for $y$ and after that we can erase it. I drew it above. You see that it intersect the $x$ axis at $y=0$ and intersect the red line at what? Intersect $x=k$ with $6-2x-y=0$ then you have $y=6-2k$ so if $x$ varies then $y=6-2x$varies. The same way can be said about the $z$. –  B. S. Dec 6 '12 at 8:09