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Spherical coordinates

Calculate the z-component of the centre of mass for a northern hemisphere of radius

R

with constant density

\rho_0 > 0

using spherical coordinates

(r,\theta, \varphi )

defined by:

x(r,\theta, \varphi) = r\sin\theta\cos\varphi \;\;\;\;\;\;0 \leq r < \infty

y(r,\theta, \varphi) = r\sin\theta\sin\varphi \;\;\;\;\;\;\, 0 \leq \theta \leq \pi

z(r,\theta, \varphi) = r\cos\theta \;\;\;\;\;\;\;\;\;\;\;\;\;\; 0 \leq \varphi < 2\pi

For a solid with density

\rho (\bf{r})

occupying a region

\cal{R}

z_{cm} = \frac{1}{M} \iiint_{\cal{R}} z \rho ({\bf r})\;dV

where

M= \iiint_{\cal{R}} \rho ({\bf r})\;dV

I have the solution but I'm wondering why the limits of

\theta

[0,\pi /2]

for calculating M then it changes to

[0,\pi]

when calculating

z_{cm}

M = \iiint _{\cal{R}} \rho_0\;dV = \int_0^R dr \int_0^{\frac{\pi}{2}} d\theta \int_0^{2\pi} d\varphi \; r^2\sin\theta = \frac{2\pi}{3}R^3 \rho_0

Mz_{cm} = \iiint _{\cal{R}} z \rho_0\;dV = \int_0^R dr \int_0^{\pi} d\theta \int_0^{2\pi} d\varphi \; r\cos\theta r^2\sin\theta = M \frac{3}{8} R

(edited 13 years ago)

Reply 1

nuodai

I'm guessing that it's a mistake; it doesn't look like they've done anything other than multiply the integrand by

z

, so it shouldn't affect the limits in any way.

Reply 2

TheEd

Original post by nuodai

I'm guessing that it's a mistake; it doesn't look like they've done anything other than multiply the integrand by

z

, so it shouldn't affect the limits in any way.

So I'm guessing both integrals should have

\theta

limits

[0,\pi]

for a 'northern hemisphere'?

Reply 3

nuodai

Original post by TheEd

So I'm guessing both integrals should have

\theta

limits

[0,\pi]

for a 'northern hemisphere'?

No, they should both be

[0, \frac{\pi}{2}]

. If P is a point in

\mathbb{R}^3

then

\theta

tells you the angle that OP makes with the positive z-axis, so if it went up to

\pi

then you'd have a full sphere.

Reply 4

TheEd

Original post by nuodai

No, they should both be

[0, \frac{\pi}{2}]

. If P is a point in

\mathbb{R}^3

then

\theta

tells you the angle that OP makes with the positive z-axis, so if it went up to

\pi

then you'd have a full sphere.

Yeah makes sense now. Ta.

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Spherical coordinates

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