(a) Let's use spherical coordinates. As S S S a a a
r ( θ , φ ) = a sin φ cos θ ı ^ + a sin φ sin θ ȷ ^ + a cos φ k ^ ∂ 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 ^ n ^ d S = ± ∂ r ∂ θ × ∂ r ∂ φ d θ d 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 ] d θ d 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 θ d φ = ∓ a 2 sin φ ( sin φ cos θ ı ^ + sin φ sin θ ȷ ^ + cos φ k ^ ) d θ d φ \begin{aligned}
\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}} \\
\frac{\partial \mathbf{r}}{\partial \theta}
&=-a\sin\varphi\sin\theta\,\hat{\boldsymbol{\imath}} +a\sin\varphi\cos\theta\,\hat{\boldsymbol{\jmath}} \\
\frac{\partial \mathbf{r}}{\partial \varphi}
&= \phantom{-}a\cos\varphi\cos\theta\,\hat{\boldsymbol{\imath}} +a\cos\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a\sin\varphi\,\hat{\mathbf{k}} \\
\hat{\mathbf{n}}\,\mathrm{d}S
&=\pm\frac{\partial \mathbf{r}}{\partial \theta}
\times \frac{\partial \mathbf{r}}{\partial \varphi}
\ \mathrm{d}\theta\, d\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] \mathrm{d}\theta\, d\varphi\\
&=\pm\Big(-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}}\Big)\ \mathrm{d}\theta\, d\varphi \\
&=\mp a^2\sin\varphi \big(\sin\varphi\cos\theta\,\hat{\boldsymbol{\imath}} +\sin\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} +\cos\varphi\,\hat{\mathbf{k}}\Big)\ \mathrm{d}\theta\, d\varphi
\end{aligned} r ( θ , φ ) ∂ θ ∂ r ∂ φ ∂ r n ^ d S = a sin φ cos θ ^ + a sin φ sin θ ^ + a cos φ k ^ = − a sin φ sin θ ^ + a sin φ cos θ ^ = − a cos φ cos θ ^ + a cos φ sin θ ^ − a sin φ k ^ = ± ∂ θ ∂ r × ∂ φ ∂ r d θ d φ = 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 d θ d φ = ± ( − a 2 sin 2 φ cos θ ^ − a 2 sin 2 φ sin θ ^ − a 2 sin φ cos φ k ^ ) d θ d φ = ∓ a 2 sin φ ( sin φ cos θ ^ + sin φ sin θ ^ + cos φ k ^ ) d θ d φ
For the outward normal, we want the + + +
n ^ d S = a 2 sin φ ( sin φ cos θ ı ^ + sin φ sin θ ȷ ^ + cos φ k ^ ) d θ d φ F ⋅ n ^ d S = z ( θ , φ ) k ^ ⋅ n ^ d S = a 3 sin φ cos 2 φ d θ d φ \begin{aligned}
\hat{\mathbf{n}}\,\mathrm{d}S
&= a^2\sin\varphi \big(\sin\varphi\cos\theta\,\hat{\boldsymbol{\imath}} +\sin\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} +\cos\varphi\,\hat{\mathbf{k}}\Big)\ \mathrm{d}\theta\, d\varphi \\
\mathbf{F}\cdot \hat{\mathbf{n}}\,\mathrm{d}S &= z(\theta,\varphi)\,\hat{\mathbf{k}}\cdot \hat{\mathbf{n}}\,\mathrm{d}S
= a^3 \sin\varphi\cos^2\varphi\ \mathrm{d}\theta\,d\varphi
\end{aligned} n ^ d S F ⋅ n ^ d S = a 2 sin φ ( sin φ cos θ ^ + sin φ sin θ ^ + cos φ k ^ ) d θ d φ = z ( θ , φ ) k ^ ⋅ n ^ d S = a 3 sin φ cos 2 φ d θ d φ
and
∬ S F ⋅ n ^ d S = a 3 ∫ 0 π d φ ∫ 0 2 π d θ sin φ cos 2 φ = 2 π a 3 ∫ 0 π d φ sin φ cos 2 φ = 2 π a 3 [ − 1 3 cos 3 φ ] 0 π = 4 3 π a 3 \begin{aligned}
\iint _S\mathbf{F}\cdot \hat{\mathbf{n}}\,\mathrm{d}S
&= a^3\int_0^\pi d\varphi\int_0^{2\pi}\mathrm{d}\theta\ \sin\varphi\,\cos^2\varphi\\
&=2\pi a^3 \int_0^\pi d\varphi \ \sin\varphi\,\cos^2\varphi
=2\pi a^3\left[-\frac{1}{3}\cos^3\varphi\right]_0^\pi \\
&=\frac{4}{3}\pi a^3
\end{aligned} ∬ S F ⋅ n ^ d S = a 3 ∫ 0 π d φ ∫ 0 2 π d θ sin φ cos 2 φ = 2 π a 3 ∫ 0 π d φ sin φ cos 2 φ = 2 π a 3 [ − 3 1 cos 3 φ ] 0 π = 3 4 π a 3
(b) Call the solid x 2 + y 2 + z 2 ≤ a 2 x^2+y^2+z^2\le a^2 x 2 + y 2 + z 2 ≤ a 2 V V V
∇ ⋅ F = ∂ ∂ x ( 0 ) + ∂ ∂ y ( 0 ) + ∂ ∂ z ( z ) = 1 { \boldsymbol{\nabla} }\cdot\mathbf{F} =\frac{\partial }{\partial x}(0)
+\frac{\partial }{\partial y}(0)
+\frac{\partial }{\partial z}(z)
=1 ∇ ⋅ F = ∂ x ∂ ( 0 ) + ∂ y ∂ ( 0 ) + ∂ z ∂ ( z ) = 1
the divergence theorem gives
∬ S F ⋅ n ^ d S = ∭ V ∇ ⋅ F d V = ∭ V d V = Volume ( V ) = 4 3 π a 3 \begin{aligned}
\iint _{\mathcal{S}}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S
&=\iiint _V{ \boldsymbol{\nabla} }\cdot\mathbf{F}\ \mathrm{d}V
=\iiint _V \ \mathrm{d}V
=\text{Volume}(V)
=\frac{4}{3}\pi a^3
\end{aligned} ∬ S F ⋅ n ^ d S = ∭ V ∇ ⋅ F d V = ∭ V d V = Volume ( V ) = 3 4 π a 3