Register to reply

Double integral of mass of circular cone

by braindead101
Tags: circular, cone, double, integral, mass
Share this thread:
braindead101
#1
Sep27-07, 01:07 AM
P: 162
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?
Phys.Org News Partner Science news on Phys.org
Fungus deadly to AIDS patients found to grow on trees
Canola genome sequence reveals evolutionary 'love triangle'
Scientists uncover clues to role of magnetism in iron-based superconductors
HallsofIvy
#2
Sep27-07, 06:47 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,500
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).
fikus
#3
Sep27-07, 06:56 AM
P: 51
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]

fikus
#4
Sep27-07, 06:58 AM
P: 51
Double integral of mass of circular cone

you would have problems with limits calculating this integral in x,y,z system.
braindead101
#5
Sep27-07, 08:47 AM
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

[tex]\int^{2pi}_{0}\int^{pi/4}_{0}\int^{}_{} p dp dphi dtheta[/tex]

is this correct so far?
braindead101
#6
Sep27-07, 08:47 AM
P: 162
how do i go about looking for the limit of p?

also it is suppose to say : p dp dphi dtheta in integral
fikus
#7
Sep27-07, 10:13 AM
P: 51
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


Register to reply

Related Discussions
Flux Integral Help through abnormal cone Calculus & Beyond Homework 1
Centre of mass of a solid cone Calculus & Beyond Homework 6
Center of mass of a cone Advanced Physics Homework 3
Center of mass of a cone Introductory Physics Homework 4
Circular Cone Pendulum General Physics 0