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Volume by DOUBLE integral. Still struggling with the same double integral problem! Test is in two hours! Can someone check my attempt?
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set up the double integral to evaluate the volume of the solid bounded by \(y=x^2\), \(z=3y\) and \(z=2+y\)
I have set it up like this. does it look correct? \[V=\int^1_0\int^{2+y}_{3y}x^2\,dzdx\]
how did you determine the solid that is bounded?
I plugged \(y=x^2\) into \(z=3y\) and \(z=2+y\). And found \(x=\pm 1\).
I am very confused, all of the text book questions I breezed through and this one has me second guessing...
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