Calculate flux of the given vector field

Question

I need to calculate the flux of the vector field

$ F(x,y,z) = (xy^2, yz^2 + xze^{sin(z^2)}, zx^2+e^{x^2}) $

Through the surface

S = {$(x, y, z) | x^2+y^2+z^2 = 9, x \ge 0$}

When the $x$ component of the normal vector is non-negative.

I tried to calculate it in the regular way, but it's involve integrals that I couldn't solve.

Is the a theory that can help (like stokes or the divergence)?

Answer 1

We can definitely use the divergence theorem in this problem. Note that the divergence of the field is given by

$$\mathrm{div} \vec{F} = \partial_x F_x + \partial_y F_y + \partial_z F_z = y^2 + z^2 + x^2$$

Now we can calculate the flux through the surface as a volume integral:

$${\int \int}_S \vec{F} \cdot d\vec{S} = {\int \int \int}_V \mathrm{div} \vec{F} \, d{V}$$

Where $V$ is the volume contained by surface $S$. To be more clear, $V$ is a half ball of radius $R = 3$. If we take spherical co-ordinates this integral becomes much easier to calculate:

$$\int \int \int_V \mathrm{div} \vec{F} \, d{V} = {\int \int \int}_V r^2 \, d{V} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\pi} \int_{0}^{R} \, r^4 dr \sin(\theta) d\theta d\phi = \frac{1}{5}R^5 (2) (\pi)= \frac{486 \pi}{5}$$

Therefore, the flux is $\frac{486 \pi}{5}$. Hope this helps!

Answer 2

Stokes isn't really going to help because $\vec F=\vec\nabla\times\vec A+\vec\nabla\Phi$ where $$\vec A=\langle-ye^{x^2}-\frac12x^2yz,-\frac12xy^2z,-\frac12x^2ze^{\sin(z^2)}-\frac12xyz^2\rangle$$ and $$\Phi=\frac{x^2y^2+x^2z^2+y^2z^2}4$$ So even though $$\int\int_S\vec A\cdot d^2\vec S=0$$ Is easy because the $x$-component of $d\vec r$ is zero along the boundary and the $y$- and $z$-components of $\vec A$ are zero there as well, we would still have to integrate $\vec\nabla\Phi$ over the surface.

The direct approach works: on the surface let $\vec r=\langle x,y,z\rangle=\langle R\sin\theta\cos\phi,R\sin\theta\sin\phi,R\cos\theta\rangle$. Then $$d\vec r=\langle R\cos\theta\cos\phi,R\cos\theta\sin\phi,-R\sin\theta\rangle\,d\theta+\langle-R\sin\theta\sin\phi,R\sin\theta\cos\phi,0\rangle\,d\phi$$ So $$\begin{align}d^2\vec S&=\langle R\cos\theta\cos\phi,R\cos\theta\sin\phi,-R\sin\theta\rangle\,d\theta\times\langle-R\sin\theta\sin\phi,R\sin\theta\cos\phi,0\rangle\,d\phi\\ &=R^2\sin\theta\langle\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta\rangle\,d\theta\,d\phi\end{align}$$ And $$\begin{align}\vec F&=\langle R^3\sin^3\theta\cos\phi\sin^2\phi,R^3\sin\theta\cos^2\theta\sin\phi+R^2\sin\theta\cos\theta\cos\phi e^{\sin(R^2\cos^2\theta)},\\ &\quad R^3\sin^2\theta\cos\theta\cos^2\phi+e^{R^2\sin^2\theta\cos^2\phi}\rangle\end{align}$$ So to evaluate $\int\int_S\vec F\cdot d^2\vec S$ we consider $5$ terms: $$\begin{align}I_1&=\int_0^{\pi}\int_{-\pi/2}^{\pi/2}R^5\sin^5\theta\cos^2\phi\sin^2\phi\,d\phi\,d\theta\\ &=R^5\int_0^{\pi}\sin^5\theta\int_{-\pi/2}^{\pi/2}\frac14\sin^22\phi\,d\phi\,d\theta=\frac{\pi R^5}8\int_0^{\pi}(\cos^4\theta-2\cos^2\theta+1)\sin\theta\,d\theta\\ &=\frac{\pi R^5}8\left[-\frac15\cos^5\theta+\frac23\cos^3\theta-\cos\theta\right]_0^{\pi}=\frac{2\pi R^5}{15}\end{align}$$ $$\begin{align}I_2&=\int_0^{\pi}\int_{-\pi/2}^{\pi/2}R^5\sin^3\theta\cos^2\theta\sin^2\phi\,d\phi\,d\theta=\frac{\pi R^5}2\int_0^{\pi}(\cos^2\theta-\cos^4\theta)\sin\theta\,d\theta\\ &=\frac{\pi R^5}2\left[-\frac13\cos^3\theta+\frac15\cos^5\theta\right]_0^{\pi}=\frac{2\pi R^5}{15}\end{align}$$ $$\begin{align}I_3&=\int_0^{\pi}\int_{=\pi/2}^{\pi/2}R^4\sin^3\theta\cos\theta\sin\phi\cos\phi e^{\sin(R^2\cos^2\theta)}d\phi\,d\theta\\ &=R^4\int_0^{\pi}\sin^3\theta\cos\theta e^{\sin(R^2\cos^2\theta)}\left[-\cos^2\phi\right]_{-\pi/2}^{\pi/2}d\theta=0\end{align}$$ $$\begin{align}I_4&=\int_0^{\pi}\int_{-\pi/2}^{\pi/2}R^5\sin^3\theta\cos^2\theta\cos^2\phi\,d\phi\,d\theta\\ &=\frac{\pi R^5}2\int_0^{\pi}(\cos^2\theta-\cos^4\theta)\sin\theta\,d\theta=\frac{\pi R^5}2\left[-\frac13\cos^3\theta+\frac15\cos^5\theta\right]_0^{\pi}=\frac{2\pi R^5}{15}\end{align}$$ $$\begin{align}I_5&=\int_{-\pi/2}^{\pi/2}\int_0^{\pi}R^2\sin\theta\cos\theta e^{R^2\sin^2\theta\cos^2\phi}d\theta\,d\phi\\ &=\int_{-\pi/2}^{\pi/2}\frac12\sec^2\phi\left[e^{R^2\sin^2\theta\cos^2\phi}\right]_0^{\pi}d\phi=0\end{align}$$ So $$\int\int_S\vec F\cdot d^2\vec S=I_1+I_2+I_3+I_4+I_5=\frac{2\pi R^5}5=\frac{486\pi}5$$ Because $R=3$.

Alternatively we could use the divergence theorem. Here we have to consider $2$ surfaces. On the flat surface, $\vec r=\langle x,y,z\rangle=\langle0,r\sin\theta,r\cos\theta\rangle$ so $$d\vec r=\langle0,\sin\theta\cos\theta\rangle\,dr+\langle0,r\cos\theta,-r\sin\theta\rangle\,d\theta$$ $$\begin{align}d^2\vec S&=\langle0,\sin\theta\cos\theta\rangle\,dr\times\langle0,r\cos\theta,-r\sin\theta\rangle\,d\theta\\ &=\langle-r,0,0\rangle\,dr\,d\theta\end{align}$$ And since the $x$-conponent of $\vec F$ is zero on this surface $$I_6=\int_0^{2\pi}\int_0^R\vec F\cdot d^2\vec S=0$$ So by the divergence theorem $$\begin{align}I&=I+I_6=I_7=\int_0^{\pi}\int_{-\pi/2}^{\pi/2}\int_0^R\vec\nabla\cdot\vec Fd^3V\\ &=\int_0^{\pi}\int_{-\pi/2}^{\pi/2}\int_0^Rr^2\cdot r^2dr\,d\phi\sin\theta\,d\theta=(2)(\pi)\left(\frac{R^5}5\right)=\frac{486\pi}5\end{align}$$ Once again.

EDIT: On second examination, Stokes' theorem can work out if we change gauge to $$\begin{align}\vec A&=\langle-ye^{x^2}+\frac1{20}yz(-13x^2-3y^2+z^2),\frac1{20}xz(x^2-13y^2-3z^3),\\ &\quad-\frac12x^2ze^{\sin(z^2)}+\frac1{20}xy(-3x^2+y^2-13z^2)\rangle\end{align}$$ and $$\vec\nabla\Phi=\frac{x^2+y^2+z^2}{5}\langle x,y,z\rangle$$ Because still $A_y=A_z=0$ on the boundary we have $$\int\int_S\vec\nabla\times\vec A\cdot d^2\vec S=0$$ and $$\int\int_S \vec\nabla\Phi\cdot d^2\vec S=\int\int_S\frac15r^3\hat r\cdot\hat rd^2S=\frac15R^3S=\frac15R^3\cdot2\pi R^2=\frac{486\pi}5$$ So Stokes' theorem does the job without any real integration but it's ugly trying to find that vector potential $\vec A$.