Suppose there is a vector field A. Now the line integral over some curve c equals double integration of curl A over surface S enclosed by C. Now if we put the divergence theorem in this then it becomes triple integral of divergence of curl Adv over the volume enclosed by the surface. Now divergence of curl of vector A becomes zero. Now so the volume integral should be zero. So the surface and line integral is also zero. So we can do this for any line integral (where the vector is defined in the surface). So does it imply all line integrals are zero. Where am I doing the mistake? Sorry for mathjax.
Are all line integrals zero by divergence theorem
vector analysisVector Fields
Related Solutions
You are mixing up two different things; the surface integral is not a generalization of the line integral.* This is easiest to see in two dimensions, where everything is an integral along curves and yet you will still find a difference.
For simplicity, let's consider a constant wind field blowing to the right, $\mathbf f(x,y)=(1,0)$. Also consider two curves, $A$ a horizontal line segment from $(0,0)$ to $(1,0)$, and $B$ a vertical line segment $B$ from $(0,0)$ to $(0,1)$. In 2D, given a vector field and a curve there are two different kinds of integral you can consider.
Interpret the curve as a wire on which a bead is threaded. If you move the bead from one end to the other, how much does the wind help or hinder the motion of the bead? This is the usual line integral $\int \mathbf f\cdot\mathrm d\mathbf r$. It is large for curve $A$ and zero for curve $B$.
Interpret the curve as a butterfly net being held stationary while the wind blows through it. How much air passes through it per unit time? This is the flux integral $\int \mathbf f\cdot\mathbf n\,\mathrm d\ell$, where $\mathbf n$ is the unit vector perpendicular to the curve tangent. It is zero for curve $A$ and large for curve $B$.
These two notions generalize to higher dimensions in different ways. The line integral remains an integral over a $1$-dimensional object, i.e. a curve. The flux integral becomes an integral over a over an $(n-1)$-dimensional object, i.e. a surface.
*In an abstract sense one could argue that they are both specializations of the same thing, but that will take us too far into the theory of differential forms.
I think some confusion can arise when using the divergence theorem to treat, say, a point particle. Consider the force field
\begin{equation*} \mathbf{F} = \frac{\hat{\mathbf{r}}}{r^{2}}. \end{equation*}
Just calculating the divergence, as you say, does yield zero
\begin{equation*} \nabla \cdot \mathbf{F} = \frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\cdot\frac{1}{r^{2}}\right) = 0. \end{equation*}
But if we pick a sphere centered at the origin, we have the surface integral
\begin{equation*} \iint{\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S} = 4\pi \end{equation*}
so the divergence theorem, which says that
\begin{equation*} \iint{\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S} = \iiint{\nabla\cdot\mathbf{F}\,\mathrm{d}V} \end{equation*}
seems to fail.
The problem is that $\nabla\cdot\mathbf{F}$ is zero everywhere except at the origin. To correctly define the divergence of $\mathbf{F}$, we need to write
\begin{equation*} \nabla\cdot\left(\frac{\hat{\mathbf{r}}}{r^{2}}\right) = 4\pi\delta^{3}(\mathbf{r}). \end{equation*}
This way, we obtain
\begin{equation*} \iiint{\nabla\cdot\mathbf{F}\,\mathrm{d}V} = \iiint{4\pi\delta^{3}(\mathbf{r})\,\mathrm{d}V} = 4\pi, \end{equation*}
and the divergence theorem holds after all.
(Actually, the divergence theorem together with the fact that the surface integral is $4\pi$ implies the form of the divergence of $\mathbf{F}$, just to keep the logic correct.)
Best Answer
In order to apply the divergence theorem, your surface $S$ needs to enclose a solid. This, in particular, implies that $S$ itself has no boundary. But using Stokes's theorem relates the integral over a surface to an integral over its boundary. Thus, there is no surface $S$ for which both theorems apply.