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    for part d, narb..

    You integrated between π/4 and π/2 for your values of theta.

    Surely for the area within the ellipse between P and the point where the ellipse touches the x-axis i.e: (x,y) (5,0).. you need to integrate between theta=0 and theta=π/4 .. right? Because when y=0, theta=0


    so first if you integrate under the ellipse between @=0 and @=π/4

    = INT [y.(dx/[email protected])[email protected]]

    = INT [([email protected])([email protected])[email protected]]

    = INT [-20.([email protected])^[email protected]]

    = -10. INT [1 - [email protected]@]

    = -10.[@ - 1/2 (sin [email protected])]

    = -10.{[0 - 1/2 (sin 0)] - [π/4 - 1/2 (sin(π/2)]}

    = (5π/2 - 5) units^2

    ________________

    and then find the area of the triangle PQR, where Q ((5/2)root2 , 0)..

    = [(5root2) - (5/2)root2]. root2

    = 10 - 5 = 5 units^2

    _______

    So shaded area..

    = 5 - (5π/2 - 5) = (10 - 5π/2) units^2

    anyone agree?! (ppppplllllllease?!)
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    I seem to be getting a different answer to everyone else....I integrated the curve between x=5√2 and x=5 (i.e. from the point P to where the curve crosses the x-axis). So in terms of 'θ' the integration is from pi/4 to 0
    So, in the end my answer was: 0.5 (20 - 5pi)
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    (Original post by Sahir)
    for part d, narb..

    You integrated between π/4 and π/2 for your values of theta.

    Surely for the area within the ellipse between P and the point where the ellipse touches the x-axis i.e: (x,y) (5,0).. you need to integrate between theta=0 and theta=π/4 .. right? Because when y=0, theta=0


    so first if you integrate under the ellipse between @=0 and @=π/4

    = INT [y.(dx/[email protected])[email protected]]

    = INT [([email protected])([email protected])[email protected]]

    = INT [-20.([email protected])^[email protected]]

    = -10. INT [1 - [email protected]@]

    = -10.[@ - 1/2 (sin [email protected])]

    = -10.{[0 - 1/2 (sin 0)] - [π/4 - 1/2 (sin(π/2)]}

    = (5π/2 - 5) units^2

    ________________

    and then find the area of the triangle PQR, where Q ((5/2)root2 , 0)..

    = [(5root2) - (5/2)root2]. root2

    = 10 - 5 = 5 units^2

    _______

    So shaded area..

    = 5 - (5π/2 - 5) = (10 - 5π/2) units^2

    anyone agree?! (ppppplllllllease?!)
    hehe, practically simultaneous posting....and yeah, I agree with your answer
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    (Original post by mockel)
    hehe, practically simultaneous posting....and yeah, I agree with your answer
    yay, good stuff that was well tough!
 
 
 
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