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Set up the double integral giving the volume of the solid bounded by \(y=x^2,\quad z=3y,\quad z=2+y\)
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this is messing with my head. I am taking R to be \[\left[3y,\,2+y\right]\times\left[ 0, 1\right]\]
sorry that is z=3y to z=2+y and y=0 to y=1
It's been awhile since I have done this haha
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