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i'm having trouble picturing this and figuring out what the limits of my integrals should be.

Calculate the volume integral of $f(\rho,\phi,z) = 2z$ over the wedge shaped cylindrical region, where the cylinders axis is the z-axis, it's radius is $1$, and it is situated in $0 < z < 4$; the angle of the wedge is $\frac{\pi}{3}$

Is this the shape which is being described? (ignoring the method that is being employed in this picture, i'm being asked to perform the triple integral)

Cylindrical wedge shaped region

Is this really as simple as

$V = \int^4_0 \int^{\frac{\pi}{3}}_0 \int^1_0 2z \rho d\rho d\phi dz$ ?

Conversations with my class mates makes me think i'm wrong, but I don't understand why.

Thank you for any help you can give.

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  • $\begingroup$ Definitely wrong. The region of your integral has an horizontal "roof" and can't be the wedge. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 3 '15 at 14:47
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You will only have constants for every bound of integration if all of the boundaries of the volume are parallel to the coordinate planes. The wedge part is not, so it will need something else.

You know that the angle of the wedge is $\phi=\pi/3$. Since $\tan{\phi}=z/\rho$, you can say that $z=\sqrt{3}\rho$. There is your upper bound on z (if you do z first here). Also, the bounds on the angle should be 0 and $\pi$, not $\pi/2$.

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  • $\begingroup$ Ok, it makes sense that my limits would not be constants. $\rho$ is constant, so the limits there should be between 0 and 1. I can see where the limits on the angle come from now after taking another look at the definitions of cyclindrical coordinates. Thanks, i think i can do this now. $\endgroup$ – jm22b Mar 3 '15 at 14:56

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