- #1

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**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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- Thread starter braindead101
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- #1

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does this mean that density function = K sqrt (x^2 + y^2 + z^2) ?

is it a triple integral problem?

- #2

HallsofIvy

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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).

- #3

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[tex] \rho \propto \sqrt{r^2 +z^2} [/tex], and then write the volume element as [tex]dV=2\pi r dr dz [/tex]

- #4

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

- #5

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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?

- #6

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

- #7

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[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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