Double Integral 267
Easy Applications

Find the centroid of a homogeneous hemisphere x²+y²+z²≤

Find the centroid of a homogeneous hemisphere $x^2+y^2+z^2\le a^2,\,z\ge0$.

Explanation of Method

Rotational symmetry about the $z$-axis gives $\bar x=\bar y=0$ immediately; the moment $M_{xy}$ is computed both in cylindrical and in spherical coordinates as a cross-check.

  1. Use symmetry to get $\bar x=\bar y=0$.
  2. Compute the volume $V$.
  3. Compute $M_{xy}$ in cylindrical coordinates.
  4. Verify $M_{xy}$ in spherical coordinates.

Step 1 · Symmetry and volume

By symmetry about the $z$-axis, $\bar x=\bar y=0$. The volume is $V=\dfrac23\pi a^3$.

Step 2 · Moment $M_{xy}$ in cylindrical coordinates

With $x=r\cos\theta,\,y=r\sin\theta,\,dV=r\,dr\,d\theta\,dz$:

$$ M_{xy} = \int_0^{2\pi}\int_0^a\int_0^{\sqrt{a^2-r^2}}zr\,dz\,dr\,d\theta = \pi\int_0^ar(a^2-r^2)\,dr = \pi\left(\frac{a^4}2-\frac{a^4}4\right) = \frac{\pi a^4}4. $$

Step 3 · Verification in spherical coordinates

With $z=\rho\cos\phi$, $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$, $0\le\phi\le\pi/2$:

$$ M_{xy} = 2\pi\int_0^{\pi/2}\sin\phi\cos\phi\,d\phi\cdot\int_0^a\rho^3\,d\rho = 2\pi\cdot\frac12\cdot\frac{a^4}4 = \frac{\pi a^4}4.\ \checkmark $$

Step 4 · Centroid coordinate $\bar z$

$$ \bar z = \frac{M_{xy}}V = \frac{\pi a^4/4}{2\pi a^3/3} = \frac{3a}8. $$


Additional Notes

Result

The centroid of the homogeneous hemisphere is:

$$ \bar x = 0, \qquad \bar y = 0, \qquad \bar z = \frac{3a}8. $$