My Math Forum Spherical Challenge....

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 October 15th, 2017, 11:07 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Spherical Challenge.... An object occupies the region in the first octant bounded by the cones $\displaystyle \phi = \frac { \pi }{4}$ and $\displaystyle \phi = arctan 2$, and the sphere $\displaystyle \rho = \sqrt {6}$, and has density proportional to the distance from the origin. Find the mass. is the following correct? $\displaystyle \int_{0}^{ \frac { \pi } {2} } \int_{ \frac { \pi } {4} }^{arctan(2)} \int_{0}^{ \sqrt{6} } \rho ~ \rho^2 sin \phi ~d \rho ~d \phi ~d \theta$
October 15th, 2017, 11:27 AM   #2
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 Originally Posted by zollen An object occupies the region in the first octant bounded by the cones $\displaystyle \phi = \frac { \pi }{4}$ and $\displaystyle \phi = arctan 2$, and the sphere $\displaystyle \rho = \sqrt {6}$, and has density proportional to the distance from the origin. Find the mass. is the following correct? $\displaystyle \int_{0}^{ \frac { \pi } {2} } \int_{ \frac { \pi } {4} }^{arctan(2)} \int_{0}^{ \sqrt{6} } \rho ~ \rho^2 sin \phi ~d \rho ~d \phi ~d \theta$
yep that looks correct, though you should include a proportionality constant for completeness.

 October 15th, 2017, 12:44 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 That would be correct if the density were equal to the distance from the origin. Instead it is "proportional" to that distance. The density is $\kappa \rho$ where $\kappa$ is the "constant of proportionality". Thanks from zollen Last edited by Country Boy; October 15th, 2017 at 12:46 PM.

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