The patch X(u,v)=(2u,3v,0) maps a unit parameter square to a rectangle of area six. Its tangent vectors span a parallelogram with area ||X_u×X_v||=6. On a curved surface the same tangent parallelogram gives the local area scale.
The area element is dA=√(EG−F²)du dv. It uses the square root of the metric determinant, not the determinant itself. This quantity is positive on a regular patch and is independent of orientation.
Predict
Predict whether reversing the order of coordinates changes unsigned area. The cross product reverses, but its magnitude does not. Oriented surface integrals retain the sign separately through the chosen normal.
Predict whether integrating over a parametrization that wraps twice gives the area of the image only once. It counts multiplicity. An area formula needs a one-to-one patch or an explicit accounting of repeated coverage.
Experiment
Compare three area densities.
Patch
Density
(2u,3v,0)
6
(u,v,f(u,v))
√(1+f_u²+f_v²)
Radius-R sphere in polar angles
R²sin θ, 0<θ<π
For the graph f=u+2v, the density is √6 everywhere. A unit parameter square therefore has area √6 even though its projection has area one. Explain the difference using the tilted tangent parallelogram.
Understand
Under a coordinate change with Jacobian J, the metric becomes JᵀgJ. Its determinant is det(J)²det g. Taking the positive square root yields the absolute Jacobian factor |det J|√det g. Combined with ordinary change of variables, this proves that integrating a scalar function against dA gives the same result in overlapping valid charts.
For a graph, E=1+f_u², F=f_u f_v and G=1+f_v². The determinant simplifies to 1+f_u²+f_v². This derives the graph density without relying on a perspective drawing.
For a sphere, integrate R²sin θ over 0<θ<π and an angular interval of length 2π. The θ integral is two and the angular integral is 2π, giving total area 4πR². The poles and seam are not regular points of that one parametrization; they form area-zero sets, and one may justify the calculation by limiting patches or an atlas.
Apply
Repair “An integral over any parameter rectangle equals the embedded surface area.” If a cylinder parameter runs through two full turns, almost every surface point is counted twice. Restrict to one turn up to its negligible seam, or report a multiplicity-weighted area.
On a surface requiring several charts, use a finite or locally finite decomposition with a partition of unity to avoid double counting overlapping regions. Existence of suitable partitions is an analysis/manifold theorem; the local change-of-variables calculation explains why the result does not depend on the chosen decomposition.
For optional depth, unsigned area is available even on nonorientable surfaces. Integrating a globally oriented flux form requires more structure. The first form determines area while a global normal may not exist.
Continue
Verify the chart domain, density and coverage count before trusting a numerical area integral. Continue to the second fundamental form to measure the normal component of acceleration that the first form alone does not display.
Apply · explain · revise
Practice and reasoning
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A planar patch X(u,v)=(2u,3v,0), 0≤u,v≤1, has what area?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
The area density in a regular chart is
Show the area density transforms by the absolute value of the coordinate Jacobian.
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Integrate the standard radius-R sphere area element R²sin θ dθ dφ to obtain 4πR², and address the coordinate poles.
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Why does integrating a cylinder parametrization over two full angular turns give twice its surface area?
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Derive the area density of z=f(x,y) and explain when projecting onto the xy-plane preserves area.
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Reflect and continue
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