easy calculation of the area of an ellipse

Consider the unit circle $\\left\\{\\right(x,y)\\in\\mathbb{R}^{2}:x^{2}+y^{2}\\leq 1\\}$ . It’s a well known fact that the area of this set is $\\pi$ .

Now consider the following linear transformation $(x,y)\\to(u,v)=(ax,by)$ .

The determinant of the transformation is $a b$ and the transformed circle is:

$\\left\\{\\right(u,v)\\in\\mathbb{R}^{2}:\\left(\\frac{u}{a}\\right)^{2}+\\left(\\frac{v%}{b}\\right)^{2}\\leq 1\\}$ an ellipse of axis $(a,b)$ .

Now since the Jacobian of the transformation is constant, the change of variables in integral theorem (http://planetmath.org/ChangeOfVariablesInIntegralOnMathbbRn) allows us to say the area of the transformed set is $a b$ times the area of the original set.

Thus, the area of an ellipse is $\\pi ab$ .

单词	EasyCalculationOfTheAreaOfAnEllipse
释义	easy calculation of the area of an ellipse Consider the unit circle $\\left\\{\\right(x,y)\\in\\mathbb{R}^{2}:x^{2}+y^{2}\\leq 1\\}$ . It’s a well known fact that the area of this set is $\\pi$ . Now consider the following linear transformation $(x,y)\\to(u,v)=(ax,by)$ . The determinant of the transformation is $a b$ and the transformed circle is: $\\left\\{\\right(u,v)\\in\\mathbb{R}^{2}:\\left(\\frac{u}{a}\\right)^{2}+\\left(\\frac{v%}{b}\\right)^{2}\\leq 1\\}$ an ellipse of axis $(a,b)$ . Now since the Jacobian of the transformation is constant, the change of variables in integral theorem (http://planetmath.org/ChangeOfVariablesInIntegralOnMathbbRn) allows us to say the area of the transformed set is $a b$ times the area of the original set. Thus, the area of an ellipse is $\\pi ab$ .
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