Distinguish surface area from flux through a surface
For a graph z = g(x,y), the surface element is dS = √(1 + gₓ² + gᵧ²) dx dy. A scalar surface integral adds a quantity such as mass per unit surface area over the sloping surface. A flux integral instead uses the component of a vector field normal to the surface; its sign depends on the chosen orientation. Do not replace dS by dx dy unless the surface is horizontal or the correct projection factor is already included.
Worked practice example
A thin panel follows z = 2x + y over 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2. Find its surface area. The coordinates are in metres.
| Step | Working |
|---|---|
| Differentiate the surface | gₓ = 2 and gᵧ = 1. |
| Build dS | dS = √(1 + 2² + 1²) dx dy = √6 dx dy. |
| Integrate over the base | A = ∫ from 0 to 1 ∫ from 0 to 2 √6 dy dx = 2√6 m². |
| Optional mass calculation | At a uniform surface density of 3 kg/m², mass = 3A = 6√6 kg. |
Check your method
The actual sloping area 2√6 is larger than its projected area 2. For an upward flux calculation the vector area element is (−gₓ, −gᵧ, 1) dx dy. Reversing orientation reverses flux, but cannot make the scalar area negative.
Try it yourself
Find the area of z = x over the unit square. Then find the upward flux of the constant field F = (0,0,5) through that surface.
Show the answer and method
Area = √2 square units. The upward vector area is (−1,0,1) dx dy, so F dot this vector = 5; integrating over a base area of 1 gives flux 5.
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.