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Question 4: (Correction)

`B`  is a unit ball on the first octant. The first octant is defined by `{x,y,z in RR^3:x,y,z>= 0}` .

The density of `B`   is given by `f = 60xyz` .

i)Remember the Jacobian J of the transformation to polar coordinates! To calculate the mass `M`...

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Question 4: (Correction)

`B`  is a unit ball on the first octant. The first octant is defined by `{x,y,z in RR^3:x,y,z>= 0}` .

The density of `B`   is given by `f = 60xyz` .

i) Remember the Jacobian J of the transformation to polar coordinates! To calculate the mass `M` of `B`   we integrate its density over the region in `RR^3` it covers so that

`M = int_0^sqrt(1-x^2-y^2)int_0^sqrt(1-x^2-z^2)int_0^1 (60xyz) dxdydz`

Now, it is more sensible to integrate over the polar coordinates. For a sphere the required conversion equations are

,  ,   .

`J = (del(x,y,z))/(del(r,phi,theta)) = r^2sin phi`

where `phi` is the angle from the plane `z=0` and `theta` is the angle from the plane `x=0`

Then, we have that

`M = 60 int_0^1int_0^(pi/2)int_0^(pi/2) r^3sin^2phi cos phi sin theta cos theta |J| dr d phi d theta`

`= 60 int_0^1 int_0^(pi/2)_0^(pi/2) r^5 sin^3phicos phi sin theta cos theta dr d phi d theta`

`= 60(r^6/6|_0^1) ((sin^4phi)/4|_0^(pi/2))((sin^2theta)/2|_0^(pi/2))`

`= 60(1/6)(1/4)(1/2) = 5/4`

ii) The average density `mu` is the total mass `M` averaged over the volume of the total region `A` occupied by the object `B` .  `B`   is an 8th of a unit sphere, so its volume is

`(1/8)4/3pi r^3 = pi/6`   , since `r=1` in this case

The average density is then `mu = M/A = (5/4)(6/pi) = 15/(2pi)`

iii) The center of mass `m` is the balance point of the object `B` . Since it is symmetric about the line `x=y=z`, then the center of mass will lie on this line. The balance point is the cut point where the integral of the mass curve along this line is equal either side of the cut.

Along the line `x=y=z` we have in polar coords that `r in (0,1)` , `phi=pi/4` and `theta=pi/4`. The mass curve along the line is given by

`f = 60xyz = 60r^3sin^2 phi cos phi sin theta cos theta `

`= 60r^3(1/sqrt(2))^2(1/sqrt(2))^3` `= 15/sqrt(2)r^3`

The total integral of the mass curve is then `int_0^1 15/sqrt(2) r^3 dr = (15r^4)/sqrt(2)|_0^1 = 15/sqrt(2)`

To have equal mass either side of the cut point `m` we must then have `r_m` such that

`(r_m)^4 = 1/2`  which ` `implies that `r_m = (1/2)^(1/4)`

Translating this back to cartesian coordinates

`x_m=y_m =z_m`  are such that `3x_m^2 = (1/2)^(1/2)`  ie

`x_m=y_m=z_m = sqrt(1/(3sqrt(2)))`

i) mass M = 5/4
ii) average density  `mu` = 15/(2`pi` )

iii) center of mass m is at (`m_x,m_y,m_z` ) where `m_x = sqrt(1/(3sqrt(2)))`

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