Interesting dilemma, answer not matching with stewart, My work is Included

Question :

Compute flux through the upper hemisphere of $x^2+y^2+z^2 = 1$ .

Where $$\textbf{F} = \left( z^2x\right)\textbf{ i }+\left[\dfrac{1}{3}y^3+ \tan z\right]\textbf{ j } + \left(x^2z+y^2 \right)\textbf{ k }$$

ANSWER GIVEN AT BACK OF STEWART : $\dfrac{13\pi}{20}$

MY WORK

$\textbf{Divergence theorem }$

$$\textbf{Flux} = \int\int_S \textbf{F}\cdot d\textbf{S} = \int\int\int_E \text{div }\textbf{F} \hspace{2mm} dV$$

Where $S$ is a closed surface.

And $E$ is the region inside that surface.

In this problem, instead of computing the surface is not closed

But we want to use divergence theorem, because divergence of the given vector field is cute.

We will over come this problem by attaching a disk at the bottom of the hemisphere, we call this closed this closed surface $S_2$ and the disk as $S_1$

We can use divergence theorem for $S_2$

We will then find flux through $S_1$.

Then the flux through $S$ = Flux through $S_2$ $-S_1$

\begin{align} & \text{div }\textbf{F} = \dfrac{\partial \left( z^2x\right)}{\partial x}+\dfrac{\partial }{\partial y}\left[\dfrac{1}{3}y^3+ \tan z\right] + \dfrac{\partial \left(x^2z+y^2 \right)}{\partial z} \\ & \text{div }\textbf{F} = z^2+y^2 +x^2 \end{align}

$$\textbf{Compute Flux through S_2 using divergence theorem }$$

$$\int\int\int_E \text{div }\textbf{F} \hspace{2mm} dV= \int\int\int_E x^2+y^2+z^2 \hspace{2mm} dV$$



We can define $E$ in spherical as follows :

\begin{align} & \left(\rho, \theta, \phi \right)\in E \hspace{1mm} | \hspace{2mm} 0< \rho < 1, \hspace{2mm} 0< \phi < \dfrac{\pi}{2}, \hspace{2mm} 0< \theta < 2\pi \\ & \text{Therefore,} \quad \int_0^{\pi/2}\int_0^{2\pi}\int_0^{1} \rho^2 \quad (\rho) d\rho d\theta d \phi \\ & =\int_0^{\pi/2}\int_0^{2\pi}\int_0^{1} \rho^3 \hspace{2mm} d\rho d\theta d\phi \\ & =\int_0^{\pi/2}\int_0^{2\pi}\left[ \dfrac{\rho^4}{4} \right]_0^{1} \hspace{2mm} d\theta d\phi \\ & =\dfrac{1}{4}\int_0^{\pi/2}\int_0^{2\pi} d\theta d\phi = \dfrac{1}{4}\times \dfrac{\pi}{2}\times 2\pi = \dfrac{\pi^2}{4} \end{align}

$$\textbf{Compute Flux through S_1 }$$



Note that $$\int\int_{S_1} \textbf{F}\cdot d\textbf{S} = \int\int_{S_1} \textbf{F}\cdot \textbf{n}\hspace{1mm}dS$$

Note that $S_1$ is part of the plane $z=0$

$$= \int\int_{D} \textbf{F}\cdot \textbf{n}\sqrt{\left( \dfrac{\partial z}{\partial x}\right)^2+\left( \dfrac{\partial z}{\partial y}\right)^2+1}\hspace{1mm}dA$$

Where $\textbf{n} = \textbf{k}$ [because $\textbf{n}$ is normal unit vector to $S_1$ ]

And $D$ is the region inside the circle $x^2+y^2=1$ [ Because $D$ is projection of $S_1$ on $xy$ plane ]

\begin{align} & = \int\int_{D} \textbf{F}\cdot \textbf{k}\sqrt{\left( 0\right)^2+\left( 0\right)^2+1}\hspace{1mm}dA \\ & = \int\int_{D} x^2z+y^2 \hspace{1mm}dA \end{align}

Substitute $z=0$, since $S_1$ is part of the plane $z=0$

$$= \int\int_{D} y^2 \hspace{1mm}dA$$

In polar coordinates the Integral becomes

\begin{align} & = \int_0^{2\pi}\int_{0}^1 r^2\sin^2\theta \hspace{1mm}(r)drd\theta \\ & =\left( \int_0^{2\pi}\sin^2\theta \hspace{1mm}d\theta \right) \left(\int_{0}^1 r^3 \hspace{1mm}dr \right) \\ & =\dfrac{1}{2}\left[ \theta-\dfrac{\sin2\theta}{2}\right]_0^{2\pi} \left[ \dfrac{r^4}{4} \right]_{0}^1 \\ & =\dfrac{1}{2}\times 2\pi \times \dfrac{1}{4} = \dfrac{\pi}{4} \\ \end{align}

Therefore, answer should be $$\dfrac{\pi^2-\pi}{4}$$

• TeX tip : Try to align your equations using begin{align} & I Equation \\ & II equation (and so on) \end{align} instead of using  again and again. Aligning the equations makes the work look better! :) Good Luck. – Kushashwa Ravi Shrimali Jun 29 '14 at 6:59
• Your volume element for spherical coordinates is not correct. It should be $dV = \rho^2 \sin(\phi) d\rho d\phi d\theta$. – Spencer Jun 29 '14 at 7:01
• thank you u made my day, u can post this comment as an answer, i will upvote it – Holy cow Jun 29 '14 at 7:08
• Did you mean to define the divergence of $\mathbf{F}$ as the vector field $\text{div }\textbf{F} = \left( z^2x\right)\textbf{ i }+\left[\dfrac{1}{3}y^3+ \tan z\right]\textbf{ j } + \left(x^2z+y^2 \right)\textbf{ k }$??? If $\mathbf{F}$ is supposed to be a vector field, then its divergence should be a scalar field. – David H Jun 29 '14 at 9:17
• It's $-{\bf k}$ in the $S_{1}$ flux. – Felix Marin Jun 29 '14 at 17:03

Your volume element for spherical coordinates is not correct. It should be $$dV=\rho^2 \sin(\varphi) \, d \rho \, d \varphi \, d \theta.$$