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Calculate the flux of the field $F(x,y,z) = (yz, xz, xy)$ through the sphere: $$ x,y,z > 0, \space x^2 + y^2 + z^2 = a^2 $$ With outer normal.

Solution says: The normal is $N = \frac{1}{a}(x,y,z)$, hence $\langle F,N \rangle = \frac{3}{a}xyz$, and using spherical coordinates, we get: $$ flux_F(M) = \int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} \frac{3}{a}(a\cos(\varphi)\sin(\theta)\times a\sin(\varphi)\sin(\theta)\times a\cos(\theta))\times {\color{Red} \sin(\theta)} d\theta d\varphi = ... = \frac{3a^2}{8} $$

My question is - shouldn't the part marked in red be $a^2\sin(\theta)$?

Because $r=(a\cos(\varphi)\sin(\theta), a\sin(\varphi)\sin(\theta), a\cos(\theta))$ is a mapping $r:\mathbb{R}^2 \rightarrow \mathbb{R}^3$, the formula for surface integrals is: $$ \int_{M}f = \int_{\Omega}f\circ r \sqrt{det(D_r^T D_r)} $$ and $\sqrt{det(D_r^T D_r)} = a^2\sin(\theta)$.


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Yes, it should: you mean the Jacobian of the transformation to spherical coordinates, right? –  DonAntonio Sep 6 '12 at 15:48
@DonAntonio Yes, this is what I mean. –  Hila Sep 6 '12 at 18:47
You mean the half sphere, with $z > 0$? Otherwise, the flux is zero since $\nabla \bullet F = 0$. –  Eric Angle Sep 7 '12 at 3:23
@BabakSorouh I am not sure - this is what the solution says. It is entirely possible that the solution is wrong. –  Hila Sep 8 '12 at 9:38

1 Answer 1

up vote 2 down vote accepted

You certainly know that if $z=f(x,y)$ be an equation of a surface $\mathcal{S}$ then: $$\hat{N}= \pm\frac{-\frac{\partial f}{\partial x}\mathbf{i}-\frac{\partial f}{\partial y}\mathbf{j}+\mathbf{k}}{\sqrt{1+\big(\frac{\partial f}{\partial x}\big)^2+\big(\frac{\partial f}{\partial y}\big)^2}}$$ $$= \frac{1}{a}\big(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\big),a>0$$ and this is what you noted. Now if we put $dS$ a small area element at point $P$ on the surface, we have: $$dS=\sqrt{1+\big(\frac{\partial f}{\partial x}\big)^2+\big(\frac{\partial f}{\partial y}\big)^2}dxdy$$ and $d\mathbf{S}$, the vector area element on $\mathcal{S}$, will be: $$d\mathbf{S}=\hat{N} dS=\big(-\frac{\partial f}{\partial x}\mathbf{i}-\frac{\partial f}{\partial y}\mathbf{j}+\mathbf{k}\big)dxdy$$ Here we have $$\iint_{\mathcal{S}}F(x,y,z)\bullet\mathbf{S}=\iint_{D} F(x,y,z)\cdot d\mathbf{S}\\\ = \iint_{D}(yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k})\cdot (\frac{x}{z}\mathbf{i}+\frac{y}{z}\mathbf{j}+\mathbf{k})=\iint_{D}3xy$$ wherein $D:x^2+y^2\leq a^2$. I used the rectangular coordinate instead.

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$\quad \checkmark \;\;\ddot\smile \;\; +1\;\;$ –  amWhy Mar 16 '13 at 0:50

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