# Use Stokes' theorem on intersection of two surfaces

In summary: The surface element is given by$$d\vec S = (\partial \vec x/\partial t)\times(\partial \vec x/\partial u) dt\,du$$What line integral are you supposed to calculate?The line that's created by the intersection of surface A and B.
Homework Statement
A = (yz + 2z, xy -x + z, xy + 5y)
Surface A: x^2 + z^2 = 4
Surface B: x + y = 2

The intersection of A and B creates a curve. Use stokes theorem to calculate the line integral along this curve.
Relevant Equations
Stokes Theorem
I parameterize surface A as:
$$A = (2cos t, 0, 2sin t), t: 0 \rightarrow 2pi$$

Then I get y from surface B:
$$y = 2 - x = 2 - 2cos t$$

$$r(t) = (2cost t, 2 - 2cos t, 2sin t)$$

Now I'm asked to integral over the surface, not solve the line integral.
So I create a new function to cover the surface, call it g.
$$g(u, t) = u * r(t), u: 0 \rightarrow 1$$

$$\oint A dr = \iint rot A dS$$

$$\iint rot A dS = \iint rot A * \hat n * |J| du dr$$

$$J = d(x, y, z)/d(u, r)$$
I can't calculate the jacobian |J| because it's not a square matrix.
Idk what to do, this is where I get stuck.

What line integral are you supposed to calculate?

The line that's created by the intersection of surface A and B.

There are several issues with your approach.
I parameterize surface A as:
$$A = (2cos t, 0, 2sin t), t: 0 \rightarrow 2pi$$
This is not the parametrization of a surface, it only has a single parameter.

Then I get y from surface B:
$$y = 2 - x = 2 - 2cos t$$
You have now introduced y, which could be used as the second parameter for A.
$$r(t) = (2cost t, 2 - 2cos t, 2sin t)$$

Now I'm asked to integral over the surface, not solve the line integral.
Not over the surface. You are asked to apply Stokes’ theorem. There are many surfaces with your given curve as its boundary. You need to pick one such surface. Preferably one that makes the integration easy.

So I create a new function to cover the surface, call it g.
$$g(u, t) = u * r(t), u: 0 \rightarrow 1$$
You have now implicitly chosen a surface. The surface formed by straight lines from the origin to your curve. This is not guaranteed to give you a nice integral. You should wait with choosing the surface until you know what the curl looks like.

$$\oint A dr = \iint rot A dS$$

$$\iint rot A dS = \iint rot A * \hat n * |J| du dr$$
You should use a better notation. Don’t use * for any type of multiplication. To make things worse you have here used it for two different types of multiplication and you are not using ##\cdot## for the inner product as should be required.

Furthermore, it is not clear what you mean by the Jacobian appearing here. You seem to mix the surface integral with the volume integral. The surface element is given by
$$d\vec S = (\partial \vec x/\partial t)\times(\partial \vec x/\partial u) dt\, du$$

$$J = d(x, y, z)/d(u, r)$$
I can't calculate the jacobian |J| because it's not a square matrix.
Idk what to do, this is where I get stuck.
Because you are mixing up the Jacobian appearing in a coordinate transformation of a volume integral with the parametrization of the surface element.

## 1. What is Stokes' theorem?

Stokes' theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a surface to the line integral of the same vector field along the boundary of the surface.

## 2. How is Stokes' theorem used on the intersection of two surfaces?

To use Stokes' theorem on the intersection of two surfaces, you first need to find the boundary curve of the intersection. Then, you can use the theorem to convert the surface integral over the intersection to a line integral along the boundary curve.

## 3. What is the significance of using Stokes' theorem on the intersection of two surfaces?

Using Stokes' theorem on the intersection of two surfaces allows us to evaluate complicated surface integrals by converting them into simpler line integrals. This can save time and effort in solving problems in vector calculus.

## 4. Are there any limitations to using Stokes' theorem on the intersection of two surfaces?

Stokes' theorem can only be applied to surfaces that are smooth and have a well-defined boundary. If the surfaces are not smooth or the boundary is not well-defined, the theorem may not hold.

## 5. Can Stokes' theorem be used in higher dimensions?

Yes, Stokes' theorem can be generalized to higher dimensions. In three dimensions, it is known as the generalized Stokes' theorem or the divergence theorem, and in four dimensions, it is known as the generalized Stokes' theorem or the generalized Stokes' theorem.

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