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Math Help - Area problem by using the Trigonometric Substitution

  1. #1
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    Area problem by using the Trigonometric Substitution

    Let R be the smaller of the two regions enclosed by the elipse 169x^2 +100y^2=16900 and the line x= 5 squar root of (2).

    Find the area of the region R.


    thanks!~
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by shannon1111 View Post
    Let R be the smaller of the two regions enclosed by the elipse 169x^2 +100y^2=16900 and the line x= 5 squar root of (2).

    Find the area of the region R.


    thanks!~
    \frac{x^2}{10^2} + \frac{y^2}{13^2} = 1

    x = 5\sqrt{2}

    The ellipse is centered at the origin, stretched in the y-direction, and intersects with the vertical line at:

    y^2 = \frac{169}{2}
    y = \pm \frac{13 \sqrt{2}}{2}

    Solving for x from the ellipse's upper half, in terms of y

    x^2 = 100 (1 - \frac{y^2}{169})
    x = \frac{10}{13} * \sqrt{169-y^2}

    The equation above represents the curve to the right. The curve to the left is simply:
    x= 5\sqrt{2}

    Setup the integral now as
    INTEGRAL (X_right - X_left) dy where y ranges from 0 to \frac{13 \sqrt{2}}{2}

    After you find the integral remember to multiply it by two to account for the other half of the area below the horizontal axis.

    I hope this helps.
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