# Variable Density Problem

• Nov 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!
• Nov 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)?
• Nov 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
• Nov 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)?
• Nov 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