The region R is bounded by the x-axis, x = 1, x = 3, and y = 1/x^3. C. Find the volume of the solid generated when R is revolved about the x-axis.
We've done a couple of these for you. You gonna show what you got, or just keep moochin'?
Here is what I have so far:
Radius = 1/(x^3)
Area of Cross Section = pi(1/(x^3))^2
Simplified: pi(1/(x^6))
Volume = (definite integral from 1 to 3) pi(1/(x^6))
= pi( -1 / 5(3)^5) - pi(-1 / 5(1)^5)
= pi (-1 / 1215) - pi (-1 / 5)
= pi(242 / 1215) = 0.625732858
Sincerely,
Mooch
Very good-- perfect!
To find the volume of the solid generated when region R is revolved about the x-axis, we can use the method of cylindrical shells.
The formula for the volume of a solid generated by revolving a region R about the x-axis using cylindrical shells is:
V = 2π ∫ [a, b] x * f(x) * dx
In this case, the region R is bounded by the x-axis, x = 1, x = 3, and y = 1/x^3.
To solve this problem, we need to find the limits of integration, the function f(x), and then integrate using the formula mentioned above.
1. Limits of integration:
The region R is bounded by x = 1 and x = 3. Therefore, our limits of integration are a = 1 and b = 3.
2. Function f(x):
The solid is generated by revolving the region R about the x-axis, so the radius of each cylindrical shell is the distance from the x-axis to the function y = 1/x^3.
To find the function f(x), we need to express y in terms of x. From y = 1/x^3, we can rewrite it as x^3 = 1/y, and taking the reciprocal of both sides, we get y = 1/x^3.
Therefore, the function f(x) for this problem is f(x) = 1/x^3.
3. Integrate:
Now that we have our limits of integration (a = 1, b = 3) and our function f(x) = 1/x^3, we can substitute these values into the formula:
V = 2π ∫ [1, 3] x * (1/x^3) * dx
Simplifying the expression inside the integral, we have:
V = 2π ∫ [1, 3] (1/x^2) * dx
To integrate this expression, we can rewrite it as:
V = 2π ∫ [1, 3] x^-2 * dx
Integrating x^-2 gives us:
V = 2π * [-x^-1] evaluated from 1 to 3
V = 2π * [(1/3) - (1/1)]
V = 2π * [(1/3) - 1]
V = 2π * [-2/3]
V = -4π/3
Therefore, the volume of the solid generated when R is revolved about the x-axis is -4π/3, or approximately -4.18879 cubic units.