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C F · d r = a b ( P x ( t ) + Q y ( t ) + R z ( t ) ) d t = a b [ P x ( t ) + Q y ( t ) + R ( z x d x d t + z y d y d t ) ] d t = a b [ ( P + R z x ) x ( t ) + ( Q + R z y ) y ( t ) ] d t = C ( P + R z x ) d x + ( Q + R z y ) d y = D [ x ( Q + R z y ) y ( P + R z x ) ] d A = D ( Q x + Q z z x + R x z y + R z z x z y + R 2 z x y ) ( P y + P z z y + R z z y z x + R 2 z y x ) d A .

By Clairaut’s theorem, 2 z x y = 2 z y x . Therefore, four of the terms disappear from this double integral, and we are left with

D [ ( R y Q z ) z x ( P z R x ) z y + ( Q x P y ) ] d A ,

which equals S curl F · d S .

We have shown that Stokes’ theorem is true in the case of a function with a domain that is a simply connected region of finite area. We can quickly confirm this theorem for another important case: when vector field F is conservative. If F is conservative, the curl of F is zero, so S curl F · d S = 0. Since the boundary of S is a closed curve, C F · d r is also zero.

Verifying stokes’ theorem for a specific case

Verify that Stokes’ theorem is true for vector field F ( x , y ) = z , x , 0 and surface S , where S is the hemisphere, oriented outward, with parameterization

r ( ϕ , θ ) = sin ϕ cos θ , sin ϕ sin θ , cos ϕ , 0 θ π , 0 ϕ π as shown in the following figure.

A diagram in three dimensions of a hemisphere in a vector field. The arrows of the vector field follow the shape of the hemisphere, which is located in quadrants 2 and 3 of the (x, y) plane and stretches up and down into the z-plane. The center of the hemisphere is at the origin. The normal N is drawn stretching up and away from the hemisphere.
Verifying Stokes’ theorem for a hemisphere in a vector field.

Let C be the boundary of S. Note that C is a circle of radius 1, centered at the origin, sitting in plane y = 0 . This circle has parameterization cos t , 0 , sin t , 0 t 2 π . By [link] ,

C F · d r = 0 2 π sin t , cos t , 0 · sin t , 0 , cos t d t = 0 2 π sin 2 t d t = π .

By [link] ,

S curl F · d S = D curl F ( r ( ϕ , θ ) ) · ( t ϕ × t θ ) d A = D 0 , −1 , 1 · cos θ sin 2 ϕ , sin θ sin 2 ϕ , sin ϕ cos ϕ d A = 0 π 0 π ( sin ϕ cos ϕ sin θ sin 2 ϕ ) d ϕ d θ = π 2 0 π sin θ d θ = π .

Therefore, we have verified Stokes’ theorem for this example.

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Verify that Stokes’ theorem is true for vector field F ( x , y , z ) = y , x , z and surface S , where S is the upwardly oriented portion of the graph of f ( x , y ) = x 2 y over a triangle in the xy -plane with vertices ( 0 , 0 ) , ( 2 , 0 ) , and ( 0 , 2 ) .

Both integrals give 136 45 .

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Applying stokes’ theorem

Stokes’ theorem translates between the flux integral of surface S to a line integral around the boundary of S . Therefore, the theorem allows us to compute surface integrals or line integrals that would ordinarily be quite difficult by translating the line integral into a surface integral or vice versa. We now study some examples of each kind of translation.

Calculating a surface integral

Calculate surface integral S curl F · d S , where S is the surface, oriented outward, in [link] and F = z , 2 x y , x + y .

A diagram of a complicated surface S in a three dimensional vector field. The surface is a cylindrical tube that twists about in the three-dimensional space arbitrarily. The upper end of the tube is an open circle leading to inside the tube. It is centered on the z-axis at a height of z=1 and has a radius of 1. The bottom end of the tube is closed with a hemispherical cap on the end. The vector arrows are best described by their components. The x component is positive everywhere and becomes larger as z increases. The y component is positive in the first and third octants and negative in the other two. The z component is zero when y=x and becomes more positive with more positive x and y values and more negative in the other direction.
A complicated surface in a vector field.

Note that to calculate S curl F · d S without using Stokes’ theorem, we would need to use [link] . Use of this equation requires a parameterization of S . Surface S is complicated enough that it would be extremely difficult to find a parameterization. Therefore, the methods we have learned in previous sections are not useful for this problem. Instead, we use Stokes’ theorem, noting that the boundary C of the surface is merely a single circle with radius 1.

The curl of F is 1 , 1 , 2 y . By Stokes’ theorem,

S curl F · d S = C F · d r ,

where C has parameterization cos t , sin t , 1 , 0 t < 2 π . By [link] ,

S curl F · d S = C F · d r = 0 2 π 1 , 2 sin t cos t , cos t + sin t · sin t , cos t , 0 d t = 0 2 π ( sin t + 2 sin t cos 2 t ) d t = [ cos t 2 cos 3 t 3 ] 0 2 π = cos ( 2 π ) 2 cos 3 ( 2 π ) 3 ( cos ( 0 ) 2 cos 3 ( 0 ) 3 ) = 0.
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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