Find the mass of the density

Question

Find the surface integral of the density $p = k\sqrt{x^2+y^2}$ over the surface $z = 4-2\sqrt{x^2+y^2}, z\in [0,4]$

$\phi(r,t) = (r\cos(t),r\sin(t),4-2r)$

$$ \frac{\partial \phi}{\partial r} = (\cos (t),\sin (t),-2) $$

$$ \frac{\partial \phi}{\partial t} = (-r\sin (t),r\cos (t),0)$$

$\Rightarrow \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} = (2r\cos(t),2r\sin(t),r)$

$\Rightarrow || \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} || = 5r^2 $

$$\therefore \int_S p = \int_0^{2\pi}\int_0^2 r(4-2r)5r^2 dr\, dt$$

$$ ... = 32\pi $$

Can I check if what I have done is correct? (This isn't homework — It's just on a question sheet with no answers.)

Answer 1

I notice some problems. It is not clear what you are doing: in a flux, you have a surface and the density is a vector field, not a scalar. Maybe what you want is the surface integral of the density?

Even without clearing the above issue, there are some obvious problems:

1. The surface has $0\leq r\leq 2$ (and not $ 4$).

2. You seem to be integrating the dot product of the normal vector with the parametrization of the surface. This makes no sense.

3. This is a repetition of 2, but it should be obvious: you are not using the density.

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After the edit: You are still making up stuff inside the integral, and missing at least part of your density. You have $$ \iint_S p\,dS=\iint_A\,p\,\left\| \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} \right\|\,dA =\int_0^2\int_0^{2\pi}kr(5r^2)\,dt\,dr =10k\pi\int_0^2r^3\,dr =40k\pi. $$