I have a small confusion with the applicability of the fundamental theorem of line integral. The theorem states that if $F=P\vec{i}+Q\vec{j}\;$ is a gradient field ($\nabla g$) of some function $g$ ,then line integrals over any curve from $P$ to $Q$ under the influence of $\vec{F}$ is just $g(Q)- g(P)$.
Am I allowed to use this to calculate flux . In calculating flux in a planar vector field, we are evaluating line integral . So I think we can use this theorem to calculate flux.
But prof. Denis in his lecture says that we can't use this theorem to calculate flux. Refer the following lecture: 45:00 - 46:30 min
https://www.youtube.com/watch?v=PnPIqh7Frlw&list=PL4C4C8A7D06566F38&index=24
Transcript:
So, important: this is only for work. There is no statement like that for flux, ok, so don't try to apply this in a problem about flux.
More precisely , suppose I have a force field $\vec{F}=y\vec{i}-x\vec{j}$. I want to calculate the flux across the curve $C$ which is a straight line from $(0,0)$ to $(1,1)$.
My approach:
$\int _{\scriptsize C}\vec{F}\cdot \vec{n}\;dS$=$\int_{\scriptsize C} xdx+ydy=\int_{\scriptsize C} d(\frac{1}{2}x^2+\frac{1}{2}y^2)=[\frac{1}{2}x^2+\frac{1}{2}y^2]_{(0,0)}^{(1,1)}=1 $
I think I can use the gradient theorem here because the proof of gradient theorem is independent of work and flux.
So My questions:
Whether my approach is correct or not?
If correct, What does Prof. Denis try to say in that clip.( see the frames from 45:00-46:30 in the above video.) More precisely: why doesn't it contradict with my approach?
How to calculate the Flux of the Vector field $$\vec{F}=\bigg(2\pi x+\dfrac{2x^2y^2}{\pi}\bigg)\vec{i}+\bigg(2\pi xy-\cfrac{4y}{\pi}\bigg)\vec{j}$$ along the outward normal, across the ellipse $x^2+16y^2=4.$
I think the question is long but anyhow I am forced to ask these three questions. Thanks in advance.
In your example there is no closed curve so the direction of the normal to the curve must be explicitly specified. It looks like you point it down and to the right, so along the curve let $\vec r=\langle x,y,z\rangle=\langle x,x,0\rangle$, so $d\vec r=\langle1,1,0\rangle dx=\hat Tds$. To point down and to the right we need $d\vec r\times\hat k=\langle1,-1,0\rangle dx=\hat Nds$. Then along the curve, $\vec F=\langle y,-x,0\rangle=\langle x,-x,0\rangle$ so we have $$\begin{align}\int_C\vec F\cdot\hat Nds&=\int_0^1\langle x,-x,0\rangle\cdot\langle1,-1,0\rangle ds=\int_0^12x\,dx=\left.x^2\right|_0^1=1\\ &=\int_C\vec F\cdot\left(d\vec r\times\hat k\right)=\int_C\left(\hat k\times\vec F\right)\cdot d\vec r=\int_C\langle x,y,0\rangle\cdot d\vec r\\ &=\int_C\left(\vec{\nabla}\left(\frac12x^2+\frac12y^2\right)\right)\cdot d\vec r=\left.\left(\frac12x^2+\frac12y^2\right)\right|_{(0,0)}^{(1,1)}=1\end{align}$$ So you seem to be kinda right. I'm not sure what the professor is saying in the video; his accent is too difficult to understand.
In your third question, the ellipse is $$\frac{x^2}{2^2}+\frac{y^2}{\left(\frac12\right)^2}=1$$ If we parameterize it as $\vec r=\langle x,y,z\rangle=\langle 2\cos\theta,\frac12\sin\theta,0\rangle$ then $d\vec r=\langle-2\sin\theta,\frac12\cos\theta,0\rangle d\theta$ and outward normal will be $\hat Nds=d\vec r\times\hat k=\langle\frac12\cos\theta,2\sin\theta,0\rangle d\theta$, so $$\int_C\vec F\cdot\hat Nds=\int_C\vec F\cdot\left(d\vec r\times\hat k\right)=\int_C\left(\hat k\times\vec F\right)\cdot d\vec r$$ Now this time, $$\left(\hat k\times\vec F\right)=\langle\frac{4y}{\pi}-2\pi xy,2\pi x+\frac{2x^2y^2}{\pi},0\rangle$$ But $$\vec{\nabla}\times\left(\hat k\times\vec F\right)=\left(2\pi+\frac{4xy^2}{\pi}-\frac4{\pi}+2\pi x\right)\hat k$$ is not the zero vector, so $\hat k\times\vec F$ is not the gradient of a scalar function. So we can either go back to our parameterization of the curve where $$\vec F(x,y,z)=\vec F\left(2\cos\theta,\frac12\sin\theta,0\right)=\langle4\pi\cos\theta+\frac{2\cos^2\theta\sin^2\theta}{\pi},2\pi\cos\theta\sin\theta-\frac{2\sin\theta}{\pi},0\rangle$$ Then we get $$\begin{align}\oint_C\vec F\cdot\hat Nds&=\oint_C\left(\hat k\times\vec F\right)\cdot d\vec r\\ &=\int_0^{2\pi}\langle-2\pi\cos\theta\sin\theta+\frac{2\sin\theta}{\pi},4\pi\cos\theta+\frac{2\cos^2\theta\sin^2\theta}{\pi},0\rangle\cdot\langle-2\sin\theta,\frac12\cos\theta,0\rangle d\theta\\ &=2\pi\left(4\pi(0)-\frac4{\pi}\left(\frac12\right)+2\pi\left(\frac12\right)+\frac1{\pi}(0)\right)=2\pi^2-4\end{align}$$ Alternatively we could use the divergence theorem: $$\vec{\nabla}\cdot\vec F=2\pi+\frac{4xy^2}{\pi}+2\pi x-\frac4{\pi}$$ So $$\begin{align}\oint_C \vec F\cdot\hat Nds&=\int\int_A\left(\vec{\nabla}\cdot\vec F\right)d^2A=\int\int_A\left(2\pi+\frac{4xy^2}{\pi}+2\pi x-\frac4{\pi}\right)d^2A\\ &=\int\int_A\left(2\pi-\frac4{\pi}\right)d^2A=\left(2\pi-\frac4{\pi}\right)\int\int_Ad^2A\\ &=\left(2\pi-\frac4{\pi}\right)A=\left(2\pi-\frac4{\pi}\right)\pi(2)\left(\frac12\right)=2\pi^2-4\end{align}$$ In the above we used identities like $$\begin{align}\int\int_Axy^2d^2A&=\int_{-1/2}^{1/2}\int_{-\sqrt{4-16y^2}}^{\sqrt{4-16y^2}}xy^2\,dx\,dy\\ &=\int_{-1/2}^{1/2}\int_{\sqrt{4-16y^2}}^{-\sqrt{4-16y^2}}(-u)y^2\,(-du)\,dy\\ &=-\int_{-1/2}^{1/2}\int_{-\sqrt{4-16y^2}}^{\sqrt{4-16y^2}}uy^2\,du\,dy\\ &=0\end{align}$$ Because anything that equals minus itself is zero except in a field of characteristic $2$. We used symmetry to get rid of this term.