Home page for accesible maths 3 Chapter 3 contents 3.9 The ellipse E 3.11 Perimeter of the ellipse E

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3.10 Area of the ellipse E

Proposition.

The area that is bounded by the ellipse $E$ is $\pi ab$ .

Proof. The area $A$ inside the region that is bounded by the ellipse $E$ is

A=4\int_{0}^{a}y(x)\,dx=4b\int_{0}^{a}\sqrt{1-{{x^{2}}\over{a^{2}}}}\,dx.

Set $x=a\sin t$ . Then the range of values $0<x<a$ can be given by $0<t<\frac{\pi}{2}$ . Also, $\frac{dx}{dt}=a\cos t$ and $\sqrt{1-\frac{x^{2}}{a^{2}}}=\sqrt{1-\sin^{2}t}=\cos t$ by our assumption on $t$ . Thus

A=\,{4ab\int_{0}^{\frac{\pi}{2}}\cos^{2}t\,dt}\,{=2ab\int_{0}^{\frac{\pi}{2}}(% 1+\cos 2t)\,dt}\,{=2ab\left[t+\frac{1}{2}\sin 2t\right]_{0}^{\frac{\pi}{2}}}

which equals $\pi ab$ .