Double integral of mass of circular cone

[braindead101]

Find the mass of a right circular cone of base radius r and height h given that the density varies directly with the distance from the vertex


does this mean that density function = K sqrt (x^2 + y^2 + z^2) ?
is it a triple integral problem?

 

[HallsofIvy]

Strictly speaking, no, because you are not given a coordinate system!

However, I understand what you mean. You have set up a "standard" coordinate system so that the vertex of the cone is at (0,0,0) and the central axis of the cone is on the positive z- axis. Yes, in this choice of coordinate system, "density varies directly with distance from the vertex" means that density= k$\sqrt{x^2+y^2+z^2}$.

Yes, this is a triple integral (not the double integral your title implies).

 

[fikus]

you can write it also as double integral
$$\rho \propto \sqrt{r^2 +z^2}$$, and then write the volume element as $$dV=2\pi r dr dz$$

 

[fikus]

you would have problems with limits calculating this integral in x,y,z system.

 

[braindead101]

how do you get dV = 2pi r dr dz
p ~ sqrt (r^2 + z^2)
p ~ sqrt (p^2)
so p ~ p

limit on phi:
z = sqrt (x^2 + y^2) (cone)
z = sqrt (p^2 sin^2 phi)
z = p sin phi
p cos phi = r = p sin phi
cos phi = sin phi
phi = pi/4
so limit of phi is 0 -> pi/4

limit of theta is 0 -> 2pi

$$\int^{2pi}_{0}\int^{pi/4}_{0}\int^{}_{} p dp dphi dtheta$$

is this correct so far?

 

[braindead101]

how do i go about looking for the limit of p?

also it is suppose to say : p dp dphi dtheta in integral

 

[fikus]

you should do it this way.
$$\rho \propto \sqrt {r^2+z^2}~~~~(r^2 = x^2+y^2)$$
then write how radius r depend on height z. It's $$r=r_0(1-\frac{z}{h})$$
Now you are integrating: $$2\pi \int_{0}^{h}\int_{0}^{r_0(1-\frac{z}{h})} \sqrt{r^2+z^2} ~ r dr dz$$, first over r and then over z.
2 pi comes from integrating over phi since problem has axial symmetry so volume element is 2 pi r dr dz.

Hope I was clear enough