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Math Help - Area bounded by the ellipse

  1. #1
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    Area bounded by the ellipse

    Show that the area bounded by the ellipse
    <br />
\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1<br />
    is given by A = \pi a b

    So I take the area in the first quadrant (with boundaries from 0 to a) and multiply it by 4

    <br />
A = 4b\int_0^a \sqrt {1 - \frac {x^2}{a^2}}dx<br />
    <br />
A = 4b(\frac {x}{2a} \sqrt {1 - \frac {x^2}{a^2}} + \frac {1}{2} sin^{-1} \frac {x}{a})<br />
    from 0 to a

    However, as the final result I get A = \pi b and I am missing that "a".

    Any ideas would be greatly appreciated!
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  2. #2
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    Hello, hasanbalkan!

    Show that the area bounded by the ellipse:  \frac{x^2}{a^2} + \frac{y^2}{b^2} \:= \:1
    . . is given by: . A \:=\: \pi a b
    I would solve for y like this . . .

    Multiply by a^2b^2\!:\;\;b^2x^2 + a^2y^2 \:=\:a^2b^2\quad\Rightarrow\quad a^2y^2 \:=\:a^2b^2-b^2x^2

    . . a^2y^2 \:=\:b^2(a^2-x^2) \quad\Rightarrow\quad y^2 \:=\:\frac{b^2}{a^2}(a^2-x^2) \quad\Rightarrow\quad y \:=\:\pm\frac{b}{a}\sqrt{a^2-x^2}


    Then: . A \;=\;4\cdot\frac{b}{a}\int^a_0\left(a^2-x^2\right)^{\frac{1}{2}}\,dx

    Let: x \:=\:a\sin\theta\quad\Rightarrow\quad dx \:=\:a\cos\theta\,d\theta

    Substitute: . A \;=\;\frac{4b}{a}\int a\cos\theta(a\cos\theta\,d\theta) \;=\;4ab\int\cos^2\theta\,d\theta

    . . = \;2ab\int(1 + \cos2\theta)\,d\theta \;=\;2ab\left(\theta + \frac{1}{2}\sin2\theta\right) \;=\;2ab(\theta + \sin\theta\cos\theta)

    Back-substitute:. . A \;=\;2ab\left(\sin^{-1}\frac{x}{a} + \frac{x\sqrt{a^2-x^2}}{a^2}\right)\,\bigg]^a_0

    Evaluate: . A \;=\;2ab\left[\sin^{-1}(1) + 0\right] - 2ab\left[\sin^{-1}(0) + 0\right] \;=\;2ab\left(\frac{\pi}{2}\right) \;=\;\boxed{\pi ab}

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  3. #3
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