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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 \(x=\pm 1\).

|dw:1352407804481:dw|

|dw:1352408012819:dw|

its been a while so i gotta wonder if im interpreting the shape correctly

|dw:1352408252457:dw|
or does it look more like a wedge of an orange

the wedge looks more "bounded" by the 3 surfaces to me

so, if we use half of this, it being symmetric in shape;
x = 0 to 1
y = x^2 to 1
z = 3y to 2+y

\[\huge \int_{x=0}^{x=1}~\int_{y=x^2}^{y=1}~\int_{z=3y}^{z=2+y}~~dz~dy~dx\]

yeah it has to be a double integral though. that's my huge problem. I'm missing something...

solve the inner integral and it turns into a double :/

I have posted it here before. furthermore I haven't gotten a DI to equal the result of that TI.

yup that's what I might do on the exam haha.

do you have an answer key to check with?

no

the cylinder has vertical sides, so I think this is correct.

anyways thanks a lot!

good luck :)