# Finding mass from mass density of lamina

1. Apr 25, 2016

### reminiscent

1. The problem statement, all variables and given/known data

2. Relevant equations
m = ∫∫sρ(x,y,z)dS

3. The attempt at a solution
I used polar coordinates for this (is that necessary here)? I found zx and zy, took the cross product of those, and found it to be √(2). So dS = √2 dA.
√2 02pi25r2drdθ = 78pi√2

2. Apr 25, 2016

### Simon Bridge

Polar coordinates are entirely appropriate: try cylindrical-polar and exploit the symmetry.
Don't know what you mean by $z_x$ and $z_y$ or the role you used $dS$ and $dA$ for.
You have to spell out your reasoning - different courses can use different conventions.

Since it is a lamina - the mass density is probably an area rather than a volume density; so $dm = \rho\;d\!A$ (check).
Here dA is an element of area lying on the surface of the cone and dm comes from: $m=\int_{Area}\;dm$

To select a strategy you can feel confident with, 1st figure out what $\rho$ depends on.
Does it depend on r? z? $\theta$? All of them? None of them? Which?
Then a good choice for the geometry of dA may present itself.

3. Apr 25, 2016

### reminiscent

Doesn't it depend on all of them?

4. Apr 25, 2016

### reminiscent

Okay I'm thinking about dA and would it be dzrdrdθ? The boundaries for z would be from z=2 to z=5, θ would be from 0 to 2pi, but what about r?

5. Apr 25, 2016

### Simon Bridge

Back up a bit.
You have $\rho(x,y,z)=\sqrt{x^2+y^2}$ in post #1.
Notice how that equation does not have a z in it? That means that the value of $\rho$ does not depend on the value of $z$.
Can you write that equation in terms of the cylindrical-polar coordinates?

Do the same for the equation of the cone.

6. Apr 25, 2016

### reminiscent

So it only depends on r and θ, correct?
p(x,y,z) = r
z = 6-r

7. Apr 25, 2016

### Simon Bridge

OK - so you have $\rho = r$ ... look at what you wrote: is there a $\theta$ in that equation?

Note: $\rho(x,y,z)=r$ is a bit misleading because x,y,z belong to rectangular coords while the r is cylindrical. You could write $\rho(r,\theta,z)=r$

8. Apr 25, 2016

### reminiscent

No... so it only depends on r?
I am getting confused on what to use now...

9. Apr 25, 2016

### Simon Bridge

You are doing fine. The density only depends on the distance from the z-axis.
This means you could use a shortcut ... you only need dA to be the area of the surface of the cone between r and r+dr.
Technically you could also make dA the area of the cone between z and z+dz (sketch).

If the density depended on the angle as well, then you'd need dA to be the area of the cone between r and r+dr which is also between $\theta$ and $\theta + d\theta$. You can still do it that way if it helps you think about it.

Last edited: Apr 25, 2016
10. Apr 25, 2016

### reminiscent

I think I see what you are saying... So since this is not dependent on z or theta, would the integrand just be rdr?

11. Apr 25, 2016

### Simon Bridge

No. the integrand is $\rho\; d\!A$
The area on the cone between r and r+dr is circular so you'd expect to see a $\pi$ in there someplace right?
Did you draw a sketch?

12. Apr 25, 2016

### reminiscent

A cone with its center at (0,0,-6), but between the planes z=2 and z=5, is that correct?

13. Apr 25, 2016

### reminiscent

But pdA = rdr? Actually, would dA = dzrdr?

14. Apr 25, 2016

### reminiscent

Can anyone confirm I am thinking correctly? This homework is due tonight. ):

15. Apr 25, 2016

### vela

Staff Emeritus
If I understand what you did, you found $dS = \sqrt{2}\,dx\,dy$ since you differentiated with respect to $x$ and $y$. Now if you want to use polar coordinates, you need to find the equivalent of $dx\,dy$ in terms of polar coordinates. Do you know what that is?

The other thing you need to do is figure out the bounds of integral. I've attached a top-down view of the lamina, so you can see its projection onto the xy-plane.