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I want to compute the flux of the vector field $F(x,y,z)=(y,x,x^3)$ across the hemisphere $S\ldots x^2+y^2+z^2=1.$

My thoughts:

Since the orientation isn't mentioned, I took it to mean the unit normal has a non-negative $z$ coordinate. I noticed $\operatorname{div F}=0,$ so I thought the divergence theorem would be convenient here. Let $V=\{(x,y,z)\in\Bbb R^2\mid 0\le x^2+y^2+z^2\le 1,z\ge 0\}$ and let $P_S=\{(x,y,0)\in\Bbb R^3\mid x^2+y^2\le1\}.$ By the divergence theorem, $$\int_{P_S}F\cdot NdA+\int_SF\cdot\vec NdA=\int_{\partial V}F\cdot\vec NdA=\int_V\operatorname{div}Fdxdydz=0\implies \int_SF\cdot\vec NdA=-\int_{P_S}F\cdot\vec NdA,$$ where $\vec N$ denotes the outer unit normal.

We can write $P_S=\Psi([0,2\pi]\times[0,1]),\Psi(\varphi,r)=(r\cos\varphi,r\sin\varphi,0)$ so that $$\partial_\varphi\Psi\times\partial_r=\begin{vmatrix}\vec i&\vec j&\vec k\\-r\sin\varphi&r\cos\varphi&0\\ \cos\varphi&\sin\varphi&0\end{vmatrix}=(0,0,-r).$$

On the other hand, $F\circ\psi\mid_{P_S}=(r\sin\varphi,r\cos\varphi,r^3\cos^3\varphi),$ so the integral boils down to $$\begin{aligned}\int_SF\cdot\vec NdA&=-\int_0^{2\pi}\int_0^1-r^4\cos^3\varphi drd\varphi\\&=\int_0^{2\pi}(1-\sin^2\varphi)\cos\varphi d\varphi\int_0^1 r^4dr\\&=\int_0^0(1-t^2)dt\cdot\frac15\\&=0\end{aligned}$$

Is my answer correct?

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1 Answer 1

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Everything is correct. That said, there's some imprecision in this part: $$ \int_0^{2\pi}\cos^3\varphi d\varphi = \int_0^0(1-t^2)dt. $$ While that statement is true, as both sides are $0$, the implication that you get this by the simple substitution $\varphi = \cos^{-1}t$ is not. That only works on the interval $(0,\pi)$. On the interval $(\pi,2\pi)$, the substitution would be $\phi = \pi - \cos^{-1}t$. Splitting the integral and applying those substitions would lead you to the RHS in the end. That said, you could also just argue the vanishing of the integrand by symmetry, as $\cos^3\varphi$ is odd on both of those intervals.

Also, while using cross products to find the normal vector to $P_S$ does work, you could also note that it's obviously $\hat{z}$.

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  • $\begingroup$ Did you maybe mean $\varphi=\pi+\arccos t$? $\endgroup$
    – PinkyWay
    May 23, 2022 at 15:25

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