The region between the circle x^(2/3) + y^(2/3) = a^(2/3) and the x-axis in the 1st quad. is revolved around the x-axis. Show that the surface area of the solid generated is 6/5*pi*a^3.
I would like to know how to properly set up this problem. I want to make sure I am able to read a question and set up the integral correctly. I have been having a hard time doing that sometimes. :( So many formulas are just different enough and I seem to get them mixed up. Thank you so much for your help!!
I would like to know how to properly set up this problem. I want to make sure I am able to read a question and set up the integral correctly. I have been having a hard time doing that sometimes. :( So many formulas are just different enough and I seem to get them mixed up. Thank you so much for your help!!
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Since the region is in the 1st, quad, the limits will be 0 to a.
We can re-write y in terms of x:
y = [a^(2/3) - x^(2/3)]^(3/2)
Note that since this is rotated about the x-axis, the surface area is equal to:
2π * ∫y√(1 + (dy/dx)²) dx
In this problem:
dy/dx = 3/2*√[a^(2/3) - x^(2/3)] * (-2/3)*x^(-1/3) = -√[a^(2/3) - x^(2/3)]/x^(1/3)
Surface Area = A = 2π * ∫[a^(2/3) - x^(2/3)]^(3/2) * √(1 + (a^(2/3) - x^(2/3))/x^(2/3)) dx from 0 to a
= 2π*a^(1/3) * ∫[a^(2/3) - x^(2/3)]^(3/2)/x^(1/3) dx from 0 to a
u = a^(2/3) - x^(2/3)
-3 du/2 = dx/(x^(1/3))
= -3π * a^(1/3) * ∫u^(3/2) eval. from a^(2/3) to 0
= -3π *a^(1/3) * (2/5) *(-a^(5/3)) = 6π/5*a^2
We can re-write y in terms of x:
y = [a^(2/3) - x^(2/3)]^(3/2)
Note that since this is rotated about the x-axis, the surface area is equal to:
2π * ∫y√(1 + (dy/dx)²) dx
In this problem:
dy/dx = 3/2*√[a^(2/3) - x^(2/3)] * (-2/3)*x^(-1/3) = -√[a^(2/3) - x^(2/3)]/x^(1/3)
Surface Area = A = 2π * ∫[a^(2/3) - x^(2/3)]^(3/2) * √(1 + (a^(2/3) - x^(2/3))/x^(2/3)) dx from 0 to a
= 2π*a^(1/3) * ∫[a^(2/3) - x^(2/3)]^(3/2)/x^(1/3) dx from 0 to a
u = a^(2/3) - x^(2/3)
-3 du/2 = dx/(x^(1/3))
= -3π * a^(1/3) * ∫u^(3/2) eval. from a^(2/3) to 0
= -3π *a^(1/3) * (2/5) *(-a^(5/3)) = 6π/5*a^2