2.2: Plane Geometry - The Area of an Ellipse
- Page ID
- 58562
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The second application of easy cases is from plane geometry: the area of an ellipse. This ellipse has semimajor axis \(a\) and semiminor axis \(b\). For its area \(A\) consider the following candidates:
(a) \(ab^{2}\) (b) \(a^{2}\) + \(b^{2}\) (c) \(a^{3}/b\) (d) 2ab (e) \(\pi\)ab.
What are the merits or drawbacks of each candidate?
The candidate \(A = ab^{2}\) has dimensions of \(L^{3}\), whereas an area must have dimensions of \(L^2\). Thus ab must be wrong. The candidate \(A = a^2 + b^2\) has correct dimensions (as do the remaining candidates), so the next tests are the easy cases of the radii \(a\) and \(b\). For a, the low extreme \(a = 0\) produces an infinitesimally thin ellipse with zero area. However, when \(a = 0\) the candidate \(A = a^{2} + b^{2}\) reduces to \(A = b^{2}\) rather than to 0; so \(a^2 + b^2\) fails the \(a = 0\) test.
The candidate \(A = a^3/b\) correctly predicts zero area when \(a = 0\). Because \(a = 0\) was a useful easy case, and the axis labels a and b are almost interchangeable, its symmetric counterpart \(b = 0\) should also be a useful easy case. It too produces an infinitesimally thin ellipse with zero area; alas, the candidate \(a^{3}/b\) predicts an infinite area, so it fails the \(b = 0\) test. Two candidates remain.
The candidate \(A = 2ab\) shows promise. When \(a = 0 \text{ or } b = 0\), the actual and predicted areas are zero, so \(A = 2ab\) passes both easy-cases tests. Further testing requires the third easy case: \(a = b\). Then the ellipse 2 becomes a circle with radius a and area \(\pi a\). The candidate \(2ab\), however, reduces to \(A=2a^2\), so it fails the \(a = b\) test.
The candidate \(A = \pi ab\) passes all three tests: \(a = 0, b = 0, \text{ and } a = b\). With each passing test, our confidence in the candidate increases; and \(pi ab\) is indeed the correct area (Problem 2.4).
Problem 2.4 Area by calculus
Use integration to show that \(A = \pi ab\).
Problem 2.5 Inventing a passing candidate
Can you invent a second candidate for the area that has correct dimensions and passes the \(a = 0, b = 0, \text{ and } a = b\) tests?
Problem 2.6 Generalization
Guess the volume of an ellipsoid with principal radii \(a, b, \text{ and } c\).