# Double integral of mass of circular cone

 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 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).
 P: 51 you can write it also as double integral $$\rho \propto \sqrt{r^2 +z^2}$$, and then write the volume element as $$dV=2\pi r dr dz$$
 P: 162 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?
 P: 51 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