By symmetry, the centroid ( x ˉ , y ˉ , z ˉ ) (\bar x,\bar y,\bar z) ( x ˉ , y ˉ , z ˉ ) x ˉ = y ˉ = z ˉ \bar x=\bar y=\bar z x ˉ = y ˉ = z ˉ
r ( θ , φ ) = a sin φ cos θ ı ^ + a sin φ sin θ ȷ ^ + a cos φ k ^ \mathbf{r}(\theta,\varphi) = a\sin\varphi\cos\theta\,\hat{\boldsymbol{\imath}}
+a\sin\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} +a\cos\varphi\,\hat{\mathbf{k}} r ( θ , φ ) = a sin φ cos θ ^ + a sin φ sin θ ^ + a cos φ k ^
Then
∂ r ∂ θ = − a sin φ sin θ ı ^ + a sin φ cos θ ȷ ^ ∂ r ∂ φ = a cos φ cos θ ı ^ + a cos φ sin θ ȷ ^ − a sin φ k ^ \begin{aligned}
\frac{\partial \mathbf{r}}{\partial \theta}
=-a\sin\varphi\sin\theta\,\hat{\boldsymbol{\imath}}+a\sin\varphi\cos\theta\,\hat{\boldsymbol{\jmath}}\qquad
\frac{\partial \mathbf{r}}{\partial \varphi}
=a\cos\varphi\cos\theta\,\hat{\boldsymbol{\imath}}+a\cos\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a\sin\varphi\,\hat{\mathbf{k}} \end{aligned} ∂ θ ∂ r = − a sin φ sin θ ^ + a sin φ cos θ ^ ∂ φ ∂ r = a cos φ cos θ ^ + a cos φ sin θ ^ − a sin φ k ^
so that
∂ r ∂ θ × ∂ r ∂ φ = det [ ı ^ ȷ ^ k ^ − a sin φ sin θ a sin φ cos θ 0 a cos φ cos θ a cos φ sin θ − a sin φ ] = − a 2 sin 2 φ cos θ ı ^ − a 2 sin 2 φ sin θ ȷ ^ − a 2 sin φ cos φ k ^ ⟹ d S = ∣ ∂ r ∂ θ × ∂ r ∂ φ ∣ d θ d φ = a 2 sin φ d θ d φ \begin{aligned}
\frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial \varphi}
&=\det\left[\begin{matrix}
\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\
-a\sin\varphi\sin\theta & a\sin\varphi\cos\theta & 0 \\
a\cos\varphi\cos\theta & a\cos\varphi\sin\theta & -a\sin\varphi
\end{matrix}\right]
\\
&=-a^2\sin^2\varphi\cos\theta\,\hat{\boldsymbol{\imath}}-a^2\sin^2\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a^2\sin\varphi\cos\varphi\,\hat{\mathbf{k}}\\
\implies \mathrm{d}S&=\left|\frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial \varphi}\right|
\,\mathrm{d}\theta\,\mathrm{d}\varphi
=a^2\sin\varphi\,\mathrm{d}\theta\,\mathrm{d}\varphi
\end{aligned} ∂ θ ∂ r × ∂ φ ∂ r ⟹ d S = det ^ − a sin φ sin θ a cos φ cos θ ^ a sin φ cos θ a cos φ sin θ k ^ 0 − a sin φ = − a 2 sin 2 φ cos θ ^ − a 2 sin 2 φ sin θ ^ − a 2 sin φ cos φ k ^ = ∂ θ ∂ r × ∂ φ ∂ r d θ d φ = a 2 sin φ d θ d φ
As the surface area of the part of the sphere in the first octant is
1 8 4 π a 2 = π a 2 2 \frac{1}{8}4\pi a^2=\frac{\pi a^2}{2} 8 1 4 π a 2 = 2 π a 2
x ˉ = y ˉ = z ˉ = ∬ S z d S ∬ S d S = 1 π a 2 / 2 ∫ 0 π / 2 d θ ∫ 0 π / 2 d φ ( a 2 sin φ ) ( a cos φ ⏞ z ) = 2 a π π 2 ∫ 0 π / 2 d φ sin φ cos φ = a [ 1 2 sin 2 φ ] 0 π / 2 = a 2 \begin{aligned}
\bar x=\bar y=\bar z
&=\frac{\iint _S z\,\mathrm{d}S}{\iint _S\,\mathrm{d}S}
=\frac{1}{\pi a^2/2}\int_0^{\pi/2}\mathrm{d}\theta\int_0^{\pi/2}\mathrm{d}\varphi \
(a^2\sin\varphi)(\overbrace{a\cos\varphi}^{z}) \\
&=\frac{2a}{\pi}\frac{\pi}{2}\int_0^{\pi/2}\mathrm{d}\varphi \ \sin\varphi\cos\varphi
=a\bigg[\frac{1}{2}\sin^2\varphi\bigg]_0^{\pi/2}
=\frac{a}{2}
\end{aligned} x ˉ = y ˉ = z ˉ = ∬ S d S ∬ S z d S = π a 2 /2 1 ∫ 0 π /2 d θ ∫ 0 π /2 d φ ( a 2 sin φ ) ( a cos φ z ) = π 2 a 2 π ∫ 0 π /2 d φ sin φ cos φ = a [ 2 1 sin 2 φ ] 0 π /2 = 2 a
