Consider the following statements :

Stokes' theorem is valid irrespective of 

1. Shape of closed curve C

2. Type of vector A

3. Type of coordinate system

4. Whether the surface is closed or open

Which of the above statements are correct?

Concept:

Stokes’s theorem:

: It states that the circulation of a vector field \(\vec A\) around a (closed) path L is equal to the surface integral of the curl of \(\vec A\) over the open surface, S bounded by L (fig).

 provided \(\vec A\) and

\(\nabla \times \vec A\) are continuous on S.

 \(\mathop \oint \limits_L \vec A \cdot d\vec l = \mathop \smallint \limits_S \left( {\nabla \times \vec A} \right) \cdot d\vec S\) 

Fig: Determining the sense of  \(d \vec l\)and \(d \vec s\) involved in Stokes’s theorem

The curl of \(\vec A\) is an axial (or rotational) vector whose magnitude is the maximum circulation of \(\vec A\) per unit area as the area tends to zero and whose direction is the normal direction of the area is oriented to make the circulation maximum.

That is the curl of the vector defined as:

 \(curl\;of\;\vec A = \nabla \times \vec A\) 

\( = {\left( {\mathop {\lim }\limits_{{\rm{\Delta }}S \to 0} \frac{{\mathop \oint \nolimits_L \vec A \cdot d\vec l}}{{{\rm{\Delta }}S}}} \right)_{max}}\widehat {{a_n}}\).

With the stokes theorem, we can convert closed line integral into surface integral.

Stokes’s theorem does not apply to the closed surface.