Tri-integral: Help Solving x+y=2, 2y+x=6

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\int \int\int z dx dy dz

where the intergral area is determined by these curves:

y=0,z=0,x+y=2,2y+x=6

and the cylinder: y^2+z^2=4
 
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What have you attempted so far or what difficulties do you have?

PS: since you sent me a PM about LaTeX, here's a hint.
You can create a 'nice' triple integral, with less space in between. Click for the code.

\iiint z dx dy dz
 
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\int_{2}^{6} dx \int_{2}^{2-x} dy \int_ {-2}^{\sqrt {4-y^2}} zdz
 
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Perhaps you should clarify that a bit, what are the limits for x,y,z?
 
x:2<<6
y:2<<2-x
z:-2<< sqrt(4-y^2)

but I'm cofused about x,y limits, I'm not sure and i mix between those too many curves to get the variable limits
 
I wonder why nobody hasen't reply on my post yet! Have I made a mistake or something wrong?
 
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