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)?

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

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)?

Re: Variable Density Problem

I get:

Mass = density * Volume:

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

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

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

$\displaystyle =1305\pi$

and:

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

$\displaystyle =1035\pi$

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

- Hollywood