Question

Find the centroid of the hemisphere
x^2+y^2+z^2 <= R^2, z>=0

Can anyone help me? I am clueless.

Answer 1

All you have to find is z since x=0 and y=0 by symmwtry

Answer 2

Recall that the centroid of a solid \(\large G\) is given by the point \(\large (\overline{x},\overline{y},\overline{z})\), where the volume of the solid is given by \(\large\displaystyle V= \iiint\limits_G \,dV\) and
\[\large \begin{aligned}\overline{x} &= \frac{1}{V}\iiint\limits_G x\,dV\\ \overline{y} &= \frac{1}{V}\iiint\limits_G y\,dV\\ \overline{z} &= \frac{1}{V}\iiint\limits_G z\,dV.\end{aligned}\]To compute \(\large V\), we can evaluate the triple integral
\[V=\large \int_{-R}^{R}\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \int_0^{\sqrt{R^2-x^2-y^2}} \,dz\,dy\,dx\]which, in cylindrical coordinates is\[\large \int_0^{2\pi}\int_0^R\int_0^{\sqrt{R^2-r^2}}r\,dz\,dr\,d\theta = \frac{2}{3}\pi R^3\]Or just get this using geometry since the volume of the upper half of a sphere is half the volume of a sphere, i.e. \(\large V=\dfrac{1}{2}\left(\dfrac{4}{3}\pi R^3\right)=\dfrac{2}{3}\pi R^3\). Therefore, we see that the coordinates of the centrioid are
\[\large \begin{aligned}\overline{x} &= \frac{3}{2\pi R^3}\int_{-R}^{R}\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \int_0^{\sqrt{R^2-x^2-y^2}}x \,dz\,dy\,dx\\ &= \frac{3}{2\pi R^3}\int_0^{2\pi}\int_0^R\int_0^{\sqrt{R^2-r^2}}r^2\cos\theta\,dz\,dr\,d\theta\\ &\phantom{X} \\ \overline{y} &= \frac{3}{2\pi R^3}\int_{-R}^{R}\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \int_0^{\sqrt{R^2-x^2-y^2}}y \,dz\,dy\,dx\\ &= \frac{3}{2\pi R^3}\int_0^{2\pi}\int_0^R\int_0^{\sqrt{R^2-r^2}}r^2\sin\theta\,dz\,dr\,d\theta\\ &\phantom{X} \\ \overline{z} &= \frac{3}{2\pi R^3}\int_{-R}^{R}\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \int_0^{\sqrt{R^2-x^2-y^2}}z \,dz\,dy\,dx\\ &= \frac{3}{2\pi R^3}\int_0^{2\pi}\int_0^R\int_0^{\sqrt{R^2-r^2}}zr\,dz\,dr\,d\theta.\end{aligned}\]

Answer 3

@eliassaab Wow...can't believe I over looked that fact. XD

Answer 4

Yea, it turns out that the \(\large \overline{x}\) and \(\large \overline{y}\) integrals are zero.

Answer 5

It is better to do it using spherical coordinates see
http://mathworld.wolfram.com/Hemisphere.html

Answer 6

Good point. In spherical coordinates, the integrals become
\[\large \begin{aligned}\overline{x} &= \frac{3}{2\pi R^3}\int_0^{2\pi}\int_0^{\pi/2}\int_0^R \rho^3\sin^2\phi\cos\theta\,d\rho\,d\phi\,d\theta = 0 \\ \overline{y} &= \frac{3}{2\pi R^3}\int_0^{2\pi}\int_0^{\pi/2}\int_0^R \rho^3\sin^2\phi\sin\theta\,d\rho\,d\phi\,d\theta = 0 \\ \overline{z} &= \frac{3}{2\pi R^3}\int_0^{2\pi}\int_0^{\pi/2}\int_0^R \rho^3\sin\phi\cos\phi\,d\rho\,d\phi\,d\theta = \ldots\end{aligned}\]which appear nicer to evaluate.