In vector calculus, Stokes' theorem relates the flux of the curl of a vector field through surface to the circulation of along the boundary of It is a generalization of Green's theorem, which only takes into account the component of the curl of Mathematically, the theorem can be written as below, where refers to the boundary of the surface.
The true power of Stokes' theorem is that as long as the boundary of the surface remains consistent, the resulting surface integral is the same for any surface we choose. Intuitively, this is analogous to blowing a bubble through a bubble wand, where the bubble represents the surface and the wand represents the boundary. Because the wand remains the same, the surface integral will be the same no matter what the shape of the bubble is.
EditDeriving the Surface Element
- Consider an arbitrary vector function . Below, we let
- Calculate differentials. For is being held constant, and vice versa. We use the notation
- Take the cross product of the two differentials. Surface integrals are a generalization of line integrals. A surface element therefore contains information about both its area and orientation. Thus, the goal is to compute a cross product.
- The formula above is the surface element for general surfaces defined by It is important to note that the nature of surfaces (more accurately, the cross product) still allows one ambiguity - the way the normal vector is pointing. The result that we have derived applies to outward normals, as recognized by the positive component, and for most applications, this will always be the case.
- Find the surface integral of over the surface . The surface below has a boundary of an ellipse, not a circle. If we choose to do the surface integral, then we will need to use Jacobian change of variables in order to properly convert into polar coordinates. Therefore, we will choose to parameterize the boundary directly.
- Parameterize the boundary. As always, verify that the chosen parameters work before proceeding.
- Calculate differentials.
- Substitute these parameters into the vector field, and take the resulting dot product . Since our boundary is on the xy-plane, so cross out all terms that contain Additionally, we are performing a closed loop integral, so our interval is
- Cancel out terms. The second term is 0 if we perform a u-substitution.
- Evaluate using any means possible. It is useful to memorize
- To check that this answer is correct, simply do the surface integral. The process will be longer, since you have to take the curl of a vector field and do Jacobians when you convert to the area integral.
- Verify Stokes' theorem. Use the surface above the xy-plane with the given vector field below.
- The goal of verification is to evaluate both integrals and check that their answers are the same. First, we will parameterize the boundary and compute the line integral. Then, we'll evaluate the surface integral. With enough practice using Stokes' theorem, you will be able to rewrite a problem into something that is easier to solve.
- Parameterize the boundary. When we set we find that the boundary is a circle of radius on the xy-plane. Therefore, the following parameters are appropriate. These are the components of
- Calculate differentials.
- Calculate the dot product . The vector field contains terms with in them, but since on the xy-plane, neglect those terms.
- Set the boundaries and simplify the integrand. Stokes' theorem tells us that is being integrated on the interval It is useful to recognize that which allows us to annihilate that term. Even though it is being multiplied by that does not affect being odd over the interval because is even.
- Evaluate using any means possible. Here, we recognize that which, while they can be found using trig identities, are worth memorizing regardless.
- Find the surface element . We recall the formula converting the surface integral into an easier-to-manage area integral as In this case, refers to the surface
- Find the curl of and compute the resulting dot product . During the dot product, we find that we have three variables, yet we are integrating over just two dimensions. Simply substitute to solve this.
- Cancel out terms. The function is symmetric over both the and axes. Therefore, any terms with an odd function of either variable will cancel out. In this problem, notice that is an even function. Therefore, we don't even need to do the multiplication for the term, because is odd, so the entire term cancels out. This step greatly simplifies the integral to be evaluated.
- Simplify and convert to polar coordinates. Our problem has now been reduced to an area integral on the xy-plane, for we have taken advantage of Stokes' theorem and recognized that this "surface" - the disk on the plane - will yield the same result as our elliptic paraboloid.
- Evaluate using any means possible.
- Our answer agrees with our answer obtained in step 6, so Stokes' theorem has been verified.