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# Calculus: can someone check this answer?

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

### 1 Answer

- AshLv 72 months agoFavorite 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 !!!