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A closed disc of radius $r \geq 0$ has Jordan content $π r^{2}$

Statement

A closed disc of radius $r \geq 0$ has Jordan content $π r^{2}$ .

Facts & Assumptions

Given: A real radius $r \geq 0$ and the closed disc $D_{r} : = {(x, y) \in R^{2} : x^{2} + y^{2} \leq r^{2}}$ .

[L1]

For every $r > 0$ , the Riemann area of the closed disc of radius $r$ is $π r^{2}$ (A disc of radius r has Riemann area pi r squared; in particular the unit disc has area pi).

[L2]

Every degenerate rectangle has Jordan content $0$ (Jordan inner and outer content and Jordan measurable bounded sets in $R^{m}$ ).

[L3]

A region between continuous graphs is compact and Jordan measurable, and its graph area equals its Jordan content (Riemann area between continuous graphs equals Jordan content).

Proof

technique · cases

1.1assume-case zeroL2

In the case $r = 0$ , the disc is the singleton ${(0, 0)} = [(0, 0), (0, 0)]$ , a degenerate rectangle, so [L2] gives $cont (D_{0}) = 0 = π 0^{2}$ .

1.2assume-case posL1L3

In the case $r > 0$ , the disc is the region on $[- r, r]$ between the continuous graphs $y = - \sqrt{r^{2} - x^{2}}$ and $y = \sqrt{r^{2} - x^{2}}$ ; [L3] identifies its Jordan content with its Riemann graph area, which [L1] evaluates as $π r^{2}$ .

2.1step 1.1step 1.2cases-exhaustive∎

The cases $r = 0$ and $r > 0$ exhaust $r \geq 0$ , and each gives $cont (D_{r}) = π r^{2}$ .

Depends on

Used by

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Sources

W. F. Trench, Introduction to Real Analysis, §§7.2–7.3 (standard reference, not scraped)

A closed disc of radius r≥0r≥0 has Jordan content πr2πr2