Cum se calculeaza aria unei elipse?
Cum se calculeaza aria unei elipse?
Salut!Ma numesc Stefan,sunt elev in clasa a-12-a si recent am ajuns la mate la aplicabilitatea integralelor definite.O problema propusa de prof a fost sa calculam aria unei elipse si nu am nici o idee de start.Sper ca ma poate ajuta cineva.
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Virgil Nicula
- Euler
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- Joined: Fri Sep 28, 2007 11:23 pm
\( \frac {x^2}{a^2}+\frac {y^2}{b^2}=1\ \Longrightarrow\ |y|=\frac ba\cdot\sqrt{a^2-x^2}\ \Longrightarrow\ S=4\cdot\int_0^a\frac ba\cdot\sqrt {a^2-x^2}\ \mathrm{dx}=4ab\cdot\int_0^1\sqrt {1-x^2}\ \mathrm{dx}=4ab\cdot \frac {\pi}{4}=\pi ab \) deoarece
\( \int_0^1\sqrt {1-x^2}\ \mathrm{dx}=\frac {\pi}{4} \) este aria unui sector de cerc cu raza \( 1 \) . In concluzie, aria \( S \) a elipsei cu semiaxe \( a \) , \( b \) este \( \underline{\overline{\left\|\ S=\pi ab\ \right\|}} \) .
Observatie. \( x=\phi (t)=\sin t\ \Longrightarrow\ \int_0^1\sqrt {1-x^2}\ \mathrm{dx}=\int_0^{\frac {\pi}{2}}\cos t\cdot\sqrt{1-\sin^2t}\ \mathrm{dx}=\int_0^{\frac {\pi}{2}}\cos^2t\ \mathrm{dx}=\frac 12\cdot \int_0^{\frac {\pi}{2}}(1+\cos 2t)\ \mathrm{dx}=\frac {\pi}{4} \) .
\( \int_0^1\sqrt {1-x^2}\ \mathrm{dx}=\frac {\pi}{4} \) este aria unui sector de cerc cu raza \( 1 \) . In concluzie, aria \( S \) a elipsei cu semiaxe \( a \) , \( b \) este \( \underline{\overline{\left\|\ S=\pi ab\ \right\|}} \) .
Observatie. \( x=\phi (t)=\sin t\ \Longrightarrow\ \int_0^1\sqrt {1-x^2}\ \mathrm{dx}=\int_0^{\frac {\pi}{2}}\cos t\cdot\sqrt{1-\sin^2t}\ \mathrm{dx}=\int_0^{\frac {\pi}{2}}\cos^2t\ \mathrm{dx}=\frac 12\cdot \int_0^{\frac {\pi}{2}}(1+\cos 2t)\ \mathrm{dx}=\frac {\pi}{4} \) .