Calculus

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find the region enclosed by the given curves:

4x+y^2=9, x=2y

  • Calculus -

    I assume you want the area of the region

    you will need their intersection
    use substitution ....
    4(2y) + y^2 = 9
    y^2 + 8y - 9 = 0
    (y-1)(y+9) = 0
    y = 1 or y -9

    if y = 1 , x = 2 --> (2,1)
    if y = -9, x = -18 --> (-18, -9)

    use horizontal slices
    x = -(1/4)y^2 + 9/2
    the value of x for the region
    = (-1//4)y^2 + 9/2 - 2y

    area = ∫(-1/4 y^2 + 9/2 - 2y ) dy from -9 to 1

    = [(-1/20)y^5 + (9/2)y - y^2 ] from -9 to 1

    = (-1/20) + 9/2 - 1 - ( (-1/20)(-59049) + 81/2 - 18)
    = ....

    you do the button - pushing.
    (you might also want to check my arithmetic, my weakness)

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