- #1
nonequilibrium
- 1,412
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Spot the error:
The circumference of an ellipse with semi-major axis = 2 and semi-minor axis = 1 (i.e. \left( \frac{x}{2} \right)^2 + y^2 = 1) is 4 \pi. Proof:
\textrm{circumference} = \iint_{\mathbb R^2}{ \delta \left( \left( \frac{x}{2} \right)^2 + y^2 - 1 \right) } \mathrm d x \mathrm d y = 2 \iint_{\mathbb R^2}{ \delta \left( \left( \frac{x}{2} \right)^2 + y^2 - 1 \right) } \mathrm d \frac{x}{2} \mathrm d y = 2 \iint_{\mathbb R^2}{ \delta \left( x^2 + y^2 - 1 \right) } \mathrm d x \mathrm d y = 2 \times 2 \pi = 4 \pi
(I couldn't post this in the Math forum since they would just yell at me for using the dirac delta, and it's also not homework as I know the answer.)
The circumference of an ellipse with semi-major axis = 2 and semi-minor axis = 1 (i.e. \left( \frac{x}{2} \right)^2 + y^2 = 1) is 4 \pi. Proof:
\textrm{circumference} = \iint_{\mathbb R^2}{ \delta \left( \left( \frac{x}{2} \right)^2 + y^2 - 1 \right) } \mathrm d x \mathrm d y = 2 \iint_{\mathbb R^2}{ \delta \left( \left( \frac{x}{2} \right)^2 + y^2 - 1 \right) } \mathrm d \frac{x}{2} \mathrm d y = 2 \iint_{\mathbb R^2}{ \delta \left( x^2 + y^2 - 1 \right) } \mathrm d x \mathrm d y = 2 \times 2 \pi = 4 \pi
(I couldn't post this in the Math forum since they would just yell at me for using the dirac delta, and it's also not homework as I know the answer.)
