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?
| = 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
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
To find the volume of the solid generated by rotating the curve y = x^3, x = y^3 about the x-axis, you can use the method of cylindrical shells.
First, let's find the points of intersection between the two curves.
Given: y = x^3 and x = y^3
To find the intersection points, we set the two equations equal to each other:
x^3 = x
Now, let's solve for x:
x^3 - x = 0
Factoring out x:
x(x^2 - 1) = 0
This equation is satisfied when x = 0, x = 1, or x = -1.
Now, let's find the upper and lower limits of integration. Since we want to rotate about the x-axis, the upper and lower limits will correspond to the x-values of the intersection points.
The lower limit is x = -1, and the upper limit is x = 1.
Next, let's set up the integral for the volume of the solid:
V = ∫[a,b] 2πx * (y - g(x)) dx
Where a and b are the lower and upper limits of integration, y represents the curve "y = x^3" and g(x) represents the curve "x = y^3"
Since we want to integrate with respect to x, we need to express y and g(x) as functions of x.
Rearranging the equation x = y^3, we get:
y = x^(1/3)
Now, let's rewrite the integral:
V = ∫[-1,1] 2πx * (x^(1/3) - x) dx
Simplifying further:
V = 2π ∫[-1,1] (x^(4/3) - x^2) dx
Now, integrate with respect to x:
V = 2π * [ (3/7)x^(7/3) - (1/3)x^3 ] | [-1,1]
Evaluating the integral:
V = 2π * [ (3/7)(1)^(7/3) - (1/3)(1)^3 ] - [ (3/7)(-1)^(7/3) - (1/3)(-1)^3 ]
V = 2π * [ 3/7 - 1/3 - (-3/7 - 1/3) ]
V = 2π * [ 3/7 - 1/3 + 3/7 + 1/3 ]
V = 2π * [ 6/7 + 7/7 ]
V = 2π * 13/7
V = 26π/7
Therefore, the correct answer is not listed among the given options. The volume of the solid generated by rotating the curve y = x^3, x = y^3 about the x-axis is 26π/7.