Use the parametric equations of an ellipse, x=acosθ, y=bsinθ, 0≤θ≤2π , to find the area that it encloses.

Question

Use the parametric equations of an ellipse, x=acosθ, y=bsinθ, 0≤θ≤2π , to find the area that it encloses.

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Jezebel 1 year 2021-07-23T11:38:15+00:00 1 Answers 500 views 0

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    2021-07-23T11:39:50+00:00

    Answer:

    Area of ellipse=[tex]\pi ab[/tex]

    Step-by-step explanation:

    We are given that

    [tex]x=acos\theta[/tex]

    [tex]y=bsin\theta[/tex]

    [tex]0\leq\theta\leq 2\pi[/tex]

    We have to find the area enclose by it.

    [tex]x/a=cos\theta, y/b=sin\theta[/tex]

    [tex]sin^2\theta+cos^2\theta=x^2/a^2+y^2/b^2[/tex]

    Using the formula

    [tex]sin^2x+cos^2x=1[/tex]

    [tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]

    This is the equation of ellipse.

    Area of ellipse

    =[tex]4\int_{0}^{a}\frac{b}{a}\sqrt{a^2-x^2}dx[/tex]

    When x=0,[tex]\theta=\pi/2[/tex]

    When x=a, [tex]\theta=0[/tex]

    Using the formula

    Area of ellipse

    =[tex]\frac{4b}{a}\int_{\pi/2}^{0}\sqrt{a^2-a^2cos^2\theta}(-asin\theta)d\theta[/tex]

    Area of ellipse=[tex]-4ba\int_{\pi/2}^{0}\sqrt{1-cos^2\theta}(sin\theta)d\theta[/tex]

    Area of ellipse=[tex]-4ba\int_{\pi/2}^{0} sin^2\theta d\theta[/tex]

    Area of ellipse=[tex]-2ba\int_{\pi/2}^{0}(2sin^2\theta)d\theta[/tex]

    Area of ellipse=[tex]-2ba\int_{\pi/2}^{0}(1-cos2\theta)d\theta[/tex]

    Using the formula

    [tex]1-cos2\theta=2sin^2\theta[/tex]

    Area of ellipse=[tex]-2ba[\theta-1/2sin(2\theta)]^{0}_{\pi/2}[/tex]

    Area of ellipse[tex]=-2ba(-\pi/2-0)[/tex]

    Area of ellipse=[tex]\pi ab[/tex]

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