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?
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[itex]\sqrt{x^2+y^2+z^2}[/itex]. Yes, this is a triple integral (not the double integral your title implies).
you can write it also as double integral [tex] \rho \propto \sqrt{r^2 +z^2} [/tex], and then write the volume element as [tex]dV=2\pi r dr dz [/tex]
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 [tex]\int^{2pi}_{0}\int^{pi/4}_{0}\int^{}_{} p dp dphi dtheta[/tex] is this correct so far?
how do i go about looking for the limit of p? also it is suppose to say : p dp dphi dtheta in integral
you should do it this way. [tex]\rho \propto \sqrt {r^2+z^2}~~~~(r^2 = x^2+y^2) [/tex] then write how radius r depend on height z. It's [tex] r=r_0(1-\frac{z}{h}) [/tex] Now you are integrating: [tex] 2\pi \int_{0}^{h}\int_{0}^{r_0(1-\frac{z}{h})} \sqrt{r^2+z^2} ~ r dr dz [/tex], 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