double integral of mass of circular cone


by braindead101
Tags: circular, cone, double, integral, mass
braindead101
braindead101 is offline
#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?
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HallsofIvy
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#2
Sep27-07, 06:47 AM
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PF Gold
P: 38,879
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
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#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
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#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
braindead101 is offline
#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
braindead101 is offline
#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
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#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


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