Asked by Leona
Volume problem:
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?
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?
Answers
Answered by
Helper
| = 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
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
Answered by
Helper
Correction 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
| = 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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