Vector Calculus II: Flux Integrals

[Tylerdhamlin]

Homework Statement

Calculate the flux of the vector field through the surface: F = i + 2j through a square of side length 2, lying in the plane x + y + z = 1, oriented away from the origin.

Homework Equations

The Attempt at a Solution

So, I need to take the integral over the surface of F dotted with dA. I've solved for dA already. I'm trying to think of the limits of integrations, but I can't figure them out. Often these relatively simply things seem to slip my mind.

 

[mighty2000]

To solve this problem, we can use the formula for flux through a surface, which is the integral of the dot product of the vector field and the surface normal vector, dA. In this case, since the surface is a square, we can use the parametrization method to find the surface normal vector.

First, let's parametrize the surface using two variables, u and v, with the following equations:

x = u
y = v
z = 1 - u - v

Next, we can find the partial derivatives of x, y, and z with respect to u and v:

∂x/∂u = 1
∂x/∂v = 0
∂y/∂u = 0
∂y/∂v = 1
∂z/∂u = -1
∂z/∂v = -1

Using these partial derivatives, we can find the cross product of the two tangent vectors, which will give us the surface normal vector:

n = (∂x/∂u, ∂y/∂u, ∂z/∂u) x (∂x/∂v, ∂y/∂v, ∂z/∂v)
= (1, 0, -1) x (0, 1, -1)
= (-1, -1, -1)

Now, we can set up the integral:

∫∫F⋅dA = ∫∫(i + 2j)⋅(-1, -1, -1)dA
= ∫∫-i - 2j - k dA

Since the surface is a square with side length 2, the limits of integration for u and v will be from -1 to 1. Therefore, the integral becomes:

∫∫-i - 2j - k dA = ∫-1^1 ∫-1^1 -i - 2j - k du dv

Evaluating this integral, we get a flux of -8 through the surface. I hope this helps! Let me know if you have any further questions or if you need clarification on any of the steps.