Set up of Triple Integral

KindofSlow

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Assignment is to set up (don't have to solve) triple integral for space above the plane 2x + y + 3z = 6 and below the plane z=2.
Also, to show that order doesn't matter, to set it up as dV = dz dx dy.
For the dz limits, I got the correct answer of - from 2 - 2/3x - y/3 up to 2.
For the dx limits, I got the correct answer of - from 0 up to 3 - y/2.
So good so far.
For the dy limits, I got from 0 up to 6-2x.
The book says this should be from 0 up to 6.
I cannot figure out why the upper dy limit is 6 instead of 6 - 2x.
Any explanation or clarification will be greatly appreciated.
Thank you
 
You are expecting to get a numerical answer, so your outermost integral must only contain numbers as its bounds.

Now assuming you are working in the first octant, draw the region of intersection of your plane with the xy plane. (set z=0) This is how you get the relation x=3-y/2. Now, as x is restricted to values from 0 to 3-y/2 (y>=0), what is the maximum value y achieves? Well, when x and z are both zero, the equation of the plane yields y=6.
 
For the dy limits, I got from 0 up to 6-2x.

If you were integrating dzdydx, then your y limits would be OK.

\(\displaystyle \int_{0}^{3}\int_{0}^{6-2x}\int_{2-\frac{2x}{3}-\frac{y}{3}}^{2}dzdydx\)
 
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