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If W is the region bounded by cone $z=\sqrt{x^2+y^2}$ and $z=2$, how do I determine the sign of $\int{xyz\space dV}$ over W? Is there a way to do this without actually doing the integral e.g. qualitative analysis of graphs?

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3 Answers 3

Split the volume $W$ as $$xy > 0, z \in [0,2]$$ and $$xy < 0, z \in [0,2]$$ Now make use of symmetry to argue why the integral should be zero.

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The isometry $(x,y,z)\mapsto(-x,y,z)$ sends the function $f:(x,y,z)\mapsto xyz$ to its opposite $-f$ and leaves $W$ unchanged. Hence $\int\limits_Wf=\int\limits_W-f=-\int\limits_Wf$, which implies $\int\limits_Wf=0$.

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I get the answer by calculate the integral exactly.

by Cylindrical Transform:

$$\int_{0}^{2}\int_{0}^{\sqrt{2}z}\int_{0}^{2\pi}r^3z\cos(x)\sin(x)dxdrdz$$

but $\int_{0}^{2\pi}\cos(x)\sin(x)=0$. then the integral $=0$

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