I need help using intergals to find volumes. If you have the equations y=9-x^2, y=0,x=0. Then you rotate the region bounded by those equations around x=3 and use intergals to get the volume of that region. I'm gettting 42.4113 for the volume but maple keeps telling me something else. Can somone tell me how to get the right answer.
Originally Posted by flash9286
I've attached a diagram to show what I've calculated.
The area which is painted green in my diagram is a circle with radius R minus a smaller circle with radius r:
From y = 9 - x² you get: x = √(9 - y)
R = 3, (constant)
r = 3 - (√(9-y))
The outer circle has the area A_(oc) = π * R² that means A_(oc) = 9π
The inner circle has the area A_(ic) = π * (3 - (√(9-y))² that means A_(ic) = π*(y - 6(√(9-y) +18)
The green area is the difference between the 2 circles:
A = 9π - π*(y - 6(√(9-y) +18) = π*(y + 6(√(9-y) -9)
The volume you are looking for is
V = ∫(π*(y + 6(√(9-y) -9))dy from(0) to(9). Use substitution method. You'llget:
V = (-4π(9-y)^(3/2)+π/2 y² - 9πy) from(0) to(9).
V = 135/2 π
V ≈ 212.06
Your result is 13.5 π. So I guess you made a typo somewhere in your calculations.
could you help me with one more? If you have the same region as in the first problem, expect that you rotate it around y=9