Triple Integrals: Solving G Bounded by x, 2-x, and y^2

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SUMMARY

The discussion focuses on solving the triple integral \(\int \int \int_{G} (xy + xz) dx dy dz\) over the region \(G\) bounded by the surfaces \(z=x\), \(z=2-x\), and \(z=y^2\). Participants established that the lower bound for \(x\) is 1 and the upper bound is 2. The lower bound for \(z\) is confirmed as 0, while the upper bound is \(y^2\). The lower and upper bounds for \(y\) are determined to be \(\sqrt{x}\) and \(\sqrt{2-x}\), respectively, with suggestions to integrate \(dx\) first while keeping \(z\) constant.

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stunner5000pt
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[tex]\int \int \int_{G} (xy + xz) dx dy dz[/tex]
G bounded by z=x, z= 2-x, and z = y^2

solving the first 2 i get x =1

equating y^2 = z =x and y^2 = 2-x
so x can go from 1 to 2?
not sure how to proceed for the y part, however..

please helppppp
 
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so far i got the x lower boun to be 1 and upper boun to be 2

i (think) that the z lower bound is zero and upper bound is y^2

also the y lower boun would be root x and upper bound would be root (2-x)


is this correct??
Integrating would be a real bugger in this case

please help!
 
Try integrating dx first, with z constant. I've attached a graph to help you figure out the limits. Set your constraints in terms of x= and y=

http://www.public.asu.edu/~hyousif/prob2.JPG"

The thin slant is z=2-x, the plane is z=x and the parabola is obviously z=y^2
 
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