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March 27, 2015

March 27, 2015

Posted by **Leona** on Sunday, February 27, 2011 at 6:38pm.

y=x^3, x=y^3; rotated about x-axis

I first made y=x^1/3, then x^3=x^1/3

Then I tried to do an integration from 0 to 1 with x^1/3-x^3, but I can't get one of the possible answers, which are:

A. 16/35π

B. 16/7π

C. 18/35

D. 7/2

E. 16π

Any advice?

- Calculus -
**Helper**, Sunday, February 27, 2011 at 7:52pm| = integrate symbol

y = x^3

y^3 = x, y = x^(1/3)

Using the formula

pi | (f(x))^2 - (g(x))^2 dx

pi | (x^3)^2 - x^(1/3)^2 dx

pi | (x^6) - x^(2/3) dx

pi ( 1/7 x^7 - 3/5 x^(5/3))

When I evaluated from 0 to 1,

I got,

- 16/35 pi

- Calculus -
**Helper**, Sunday, February 27, 2011 at 8:03pmCorrection for above,

| = integrate symbol

y = x^3

y^3 = x, y = x^(1/3)

Using the formula

pi | (f(x))^2 - (g(x))^2 dx

pi | x^(1/3)^2 - (x^3)^2 dx

pi | x^(2/3) - (x^6) dx

pi ( 3/5 x^(5/3) - 1/7 x^7 )

When I evaluated from 0 to 1,

I got,

16/35 pi

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