It Often Goes the Other Way

Calculus Level 3

There is an open surface consisting of a half unit-sphere. It's circular base is parallel to the x y xy plane, and has its center at the origin.

There is a 3D vector field F ⃗ \vec{F} with the following components:

F x = x 3 y 4 z 5 F y = x y s i n ( z ) F z = x ( 1 − e z ) \large{F_x = x^3 y^4 z^5 \\ F_y = x y \, sin(z) \\ F_z = x \, (1 - e^z)}

Determine the surface integral of the curl of F ⃗ \vec{F} over the open surface.

∫ ∫ S ( ∇ × F ⃗ ) ⋅ d S ⃗ \large{\int \int_S (\nabla \times \vec{F}) \cdot \vec{dS}}

Note: The constant e e is Euler's number


The answer is 0.0.

Steven Chase by Steven Chase , 37, USA

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1 solution

Steven Chase
Dec 14, 2017

By Stokes theorem, the surface integral of the curl is equal to the line integral of the vector field over the boundary curve. The boundary curve here is the unit-circle in the x y xy plane.

∫ C F ⃗ ⋅ d l ⃗ = ∫ ∫ S ( ∇ × F ⃗ ) ⋅ d S ⃗ \large{\int_C \vec{F} \cdot \vec{dl} = \int \int_S (\nabla \times \vec{F}) \cdot \vec{dS}}

Because z = 0 z = 0 over the entire boundary curve, so is the vector field at all points along the curve. Since the line integral is zero, the surface integral must be also.

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