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Find the volume of the solid under the surface z=xy and above the triangular region in the xy-plane bounded by the lines y=2x, y=-x+6, and y=0.
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@wio @radar @robtobey @Directrix
draw it
2x=-x+6 x=2 they intersect at x=2
Okay, I think that we can start with :\[ 0\leq z \leq xy \]
I think \(y=0\) means it is \(y\) simple.
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So what are the limits of integration for x and y?
First solve for \(x\) in terms of \(y\).
We have \[ x=\frac y2\\x=6-y \]We know at \(0\), that the second equation will have a greater \(x\), so we say: \[ \frac y2 \leq x \leq 6-y \]
Finally, we know \(y\) starts at \(0\), and solving for the intersection, we get \(y=4\).\[ 0\leq y \leq 4 \]
Thank you!
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