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Polar change of variables on a compact annular sector gives the Jacobian factor rr and its area

Example

On K=[1,2]×[π/6,π/3]K=[1,2]\times[\pi/6,\pi/3], the polar map P(r,θ)=(rcosθ,rsinθ)P(r,\theta)=(r\cos\theta,r\sin\theta) is injective with Jacobian factor rr. Its annular-sector image has area π/4\pi/4.

Facts & Assumptions

Given: The polar map and compact parameter rectangle KK.

[L1]

Sine and cosine have their standard derivatives and satisfy sin2+cos2=1\sin^2+\cos^2=1 (The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine).

[L2]

Cosine is strictly decreasing on [0,π][0,\pi] (Signs, monotonicity intervals, and ranges of sine and cosine).

[L3]

Compact-Jordan change of variables uses the absolute Jacobian determinant (Change of variables for an injective C1C^1 map on a compact Jordan set).

Verification

technique · computation
1.1

Differentiation and [L1] give DP=(cosθrsinθsinθrcosθ),detDP=r.DP=\begin{pmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{pmatrix},\qquad \det DP=r.

L1
2.1

Equality of two images first gives equality of radii by [L1], then equality of cosines; [L2] gives equality of angles. The same recovery works on the open neighborhood (1/2,5/2)×(π/12,5π/12)(1/2,5/2)\times(\pi/12,5\pi/12), where detDP\det DP never vanishes. Thus the compact theorem's neighborhood hypotheses hold.

L1L2step 1.1
3.1

Applying [L3] to the constant-one function and integrating rr gives area(P(K))=π/6π/312rdrdθ=32π6=π4.\operatorname{area}(P(K))=\int_{\pi/6}^{\pi/3}\int_1^2r\,dr\,d\theta=\frac32\cdot\frac\pi6=\frac{\pi}{4}.

L3step 2.1

Depends on

Used by

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