Question

Find the flux through through the boundary of the rectangle 0≤ x ≤ 2, 0 ≤ y ≤ 4 for fluid flowing along the vector field ⟨x^4 + 5,ycos(6x)⟩

Answer 1

The flux through through the boundary of the rectangle is -sin(6x)/6.

According to the statement

we have given that the limits and the equation, and we have to find the flux of the given equation.

For this purpose, we know that the

Flux is a the surface integral of the perpendicular component of a vector field over a surface.

And the given equation is :

x^4 + 5,ycos(6x)

The given limits are:

0≤ x ≤ 2, 0 ≤ y ≤ 4

So, Integrate with the given imits,

$\int\limits^2_0 \int\limits^4_0 {x^4 + 5,ycos(6x)} \, dx$

When integrate these terms with dx and dy one by one then the equation become:

For the x:

$\int\limits^2_0 {x^4 + 5,ycos(6x)} \, dx \\\bracket\limits^2_0 [{\frac{x^5}{5},{-ysin(6x)}}/{6]$

And then integrate w.r.t the y.

Then the flux through the boundary become:

-sin(6x)/6.

So, The flux through through the boundary of the rectangle is -sin(6x)/6.

Learn more about Flux here

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