Variable Density Problem

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November 6th 2012, 05:09 PM
citcat

Variable Density Problem

Two cones are the same size w/ radius 3in and height 6in. One has a density of 80oz/in^3 at its base and 50oz/in^3 at the vertex. The other has a density of 50oz/in^3 at its base and 80oz/in^3 at its vertex. In both cones, the density varies linearly with the distance from the plane of the base. What are the masses of the cones? How do you solve this? The answers are supposed to be 1350pi and 1035pi, but I cannot get the right answer. One time I got 337.5pi which is 1/4 of 1350pi, but can't see why I would need to multiply by four. Thanks!
November 6th 2012, 09:43 PM
chiro

Re: Variable Density Problem

Hey citcat.

Can you show your integral calculations specifically to find these masses given the density functions (and also the volume regions through the limits)?
November 7th 2012, 10:10 AM
citcat

Re: Variable Density Problem

For cone one:
P=rho=density
V=volume
dM=PdV
P=-5y+80
y=-4x+6 so x=(y-6)/-4
dV=(pi)(r^2)dy=(pi)((y-6)/-4)^2dy
dM=(-5y+80)(pi)((y-6)/-4)^2dy
M=intgeral from 0-6 of (-5y+80)(pi)((y-6)/-4)^2dy=326.5pi
326.5pi is a quarter of what the answer should be which is 1350 pi
November 8th 2012, 08:12 PM
chiro

Re: Variable Density Problem

The volume doesn't look right: shouldn't it be over the region of the cone which is a dV = dxdydz (with the right limits)?
November 9th 2012, 06:26 AM
hollywood

Re: Variable Density Problem

I get:

Mass = density * Volume:

$\int_0^6\rho(x)Adx$

$=\int_0^6\rho\pi{r}^2dx$

$=\int_0^6(80-5x)\pi(3-\frac{1}{2}x)^2dx$

$=1305\pi$

and:

$\int_0^6(50+5x)\pi(3-\frac{1}{2}x)^2dx$

$=1035\pi$

Your expression $r=\frac{y-6}{-4}$ is incorrect. When y=0, r should be 3. Since r is squared, you're off by a factor of 4.

- Hollywood