Calculus: can someone check this answer?

Should I have put pi(3.14) next to my answer and is it correct?

Attachment image

1 Answer

Relevance
  • Ash
    Lv 7
    2 months ago
    Favorite Answer

    Lets consider the area is made up of thin slices of width dy

    The length will be the the difference between x = (y-3)²/3 and the line x = 9 - y

    To find y-limits find the intersection 

    x = (y-3)²/3 ...............(1)

    x = 9 - y ...................(2)

    from (1) and (2)

    (y-3)²/3 = 9-y

    y² - 6y + 9 = 27 - 3y

    y² - 3y -18 = 0

    (y-6)(y+3) = 0

    y = 6 and y = -3

    The lower limit is -3 , while upper limit is +6

    Area = ₋₃∫⁶ [(9 - y ) - (y-3)²/3]  dy

    A = ₋₃∫⁶ [(9 - y ) - (y² - 6y + 9)/3] dy

    A = ₋₃∫⁶ [27 - 3y - y² + 6y - 9)/3] dy

    A = ⅓ ₋₃∫⁶ (18  + 3y - y²) dy

    A = ⅓ [18y + (3/2)y² - ⅓y³] ₋₃|⁶

    A = ⅓[18(6) +(3/2)(6²) - ⅓(6³) - (18(-3) + (3/2)(-3)² - ⅓(-3)³) ]

    A = ⅓[108 + 54 - 72 + 54 - (27/2) - 9 ]

    A = ⅓[135 - (27/2) ]

    A = 81/2

    Your answer is correct !!!

Still have questions? Get your answers by asking now.