Use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field 𝐅 acr

Use the surface integral in Stokes' Theorem to calculate the...

Key Concepts

Stokes' Theorem

Stokes' Theorem is a fundamental result in vector calculus that relates a surface integral of the curl of a vector field over an oriented surface to a line integral of the vector field along the boundary of that surface. This theorem provides an elegant way to convert problems in surface integrals into often simpler line integral problems and plays a crucial role in understanding circulation and rotational properties of fields.

Curl of a Vector Field

The curl of a vector field quantifies the rotation or circulation at a point in the field. It is obtained by taking a specific combination of partial derivatives of the field components and gives a new vector field. In the context of Stokes' Theorem, the curl operation is central because it links the behavior of a vector field within a surface to the behavior along its boundary.

Surface Integral

A surface integral extends the concept of integration to functions defined over a surface, allowing measurements like flux across that surface. When applied to vector fields, the surface integral computes the total amount of field passing through the surface. In problems involving Stokes' Theorem, the surface integral is taken over the curl of a vector field, combined with the surface's orientation.

Surface Parametrization

Surface parametrization is the method of representing a surface using coordinate parameters, converting a complex surface into a more tractable form for analysis and integration. By expressing the surface in terms of parameters, one can calculate necessary quantities like the normal vector and the differential area element, which are essential when performing surface integrals.

Surface Orientation

Surface orientation refers to the assignment of a direction to a surface by choosing a unit normal vector at each point. This orientation is crucial in determining the sign and value of surface integrals, especially when using theorems like Stokes'. The 'outward' condition or another specified direction needs to be maintained consistently along the entire surface to ensure accurate calculation of flux.

Transcript

00:01 Okay, so we have that stokes theorem tells us that flux of the curl of f taken over the surface is equal to the integral of f dotted with the parameterization of the boundary.

00:53 So what we need to do, so c is the boundary curve, so we need a parameterization of the boundary.

01:11 So f is the surface described by our theta as being our cos theta i -hat plus our sin theta j -hat plus 5 minus our k -hat.

01:44 And the thing to notice about this shape is that it's actually a hemisphere.

01:51 You can do a little sketch of it.

01:59 This is 5 right here, that's the top of it, and extends down like that, and goes around in a circle.

02:26 And so it looks something like this, where this is also 5 right here.

02:33 So that's 5, and this is 5.

02:39 So the boundary curve of this shape is when r is equal to 5.

02:46 After all, we're interested where r goes from 0 to 5, and theta goes from 0 to 2 pi.

03:00 You can see when r is 5, you get this portion of the hemisphere.

03:15 So a parameterization of the curve c is given by 5 cos theta i -hat plus 5 sin theta j -hat plus 0 k -hat.

03:43 I'm going to do a little abusive notation because i want whatever i'm doing to match this equation right here.

03:50 So let's call r theta.

03:57 Let's let that be s when r is 5, and that's going to be equal to our curve on the boundary.

04:13 Now what i want to do is note that the integral around c of the vector field f dot dr is equal to...

04:36 The curve runs from 0 to 2 pi, since r is this guy right here, of the vector field evaluated at r of theta.

05:08 And that's going to be dotted with r prime of theta d theta, because everything will be in terms of theta after we do that.

05:22 And recall that dr can be expressed as dr d theta d theta.

05:36 Sorry about all the noise right there.

05:44 Continuing, let's plug in what we have.

05:49 So from 0 to 2 pi, we want f to be in terms of theta.

05:58 And therefore, let me first say that on the boundary, the x -coordinate is going to be equal to 5 cos theta.

06:13 The y -coordinate is going to be equal to 5 sin theta.

06:19 And the z -coordinate is always going to be 0.

06:23 So f in terms of r of theta is going to be 5 cos theta minus 5 sin theta i -hat plus 5 sin theta minus 0 j -hat.

06:58 I'm not going to write the minus 0.

07:01 Plus negative 0 minus x.

07:06 So that's just negative x, which is just negative 5 cos theta k -hat.

07:16 And all of that is going to be dotted with the expression we found for r prime of theta.

07:33 Which, where did i write that down? the answer is nowhere, because i didn't write it down yet.

07:40 Let's do that now.

07:42 Well, let's scroll up a bit.

07:51 The vector r expressed in terms of theta is 5 cos i -hat plus 5 sin j -hat plus 0 k -hat.

07:59 Then we just take the derivative with respect to theta for each of those.

08:05 So we get negative 5 sin i -hat plus 5 cos j -hat plus 0.

08:23 I don't know why i wrote theta right there.

08:29 Alright, get out of here.

08:33 Alrighty, where was i? 5 cos j -hat plus 0 k -hat.

08:40 I'm taking the dot product of these two vectors.

08:44 And then we're going to integrate from 0 to 2 pi.

08:50 So what will that be? let's scroll down a bit.

08:55 This is going to equal, from 0 to 2 pi, 5 cos theta minus 5 sin theta times negative 5 sin...

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