find the volume of the solid formed by revolving the region bounded by the graphs of y=x^3,y=1, and x=2 about the x-axis using the disk method. 9 express the answer in terms of pie)
i got the answer to be 3/5 pie but im not sure if is right. i used the formula V=(pie)(r)^2W
V=πr²W applies to the volume of a uniform solid cylinder of length W and radius r.
The volume in question is very much like a hollow truncated cone.
Since y=x^3 and y=1 intersect at (1,1), the volume of the solid starts at x=1.
The elemental disk has a volume of
2π(x^3-1)dx
So integrating from x=1 to 2 gives you the volume of revolution.
I get 11π/2.
Check my thinking.
To find the volume of the solid formed by revolving the region bounded by the graphs of y = x^3, y = 1, and x = 2 about the x-axis using the disk method, you are on the right track.
First, we need to determine the limits of integration. The region is bounded by y = x^3 and y = 1, so we need to find the x-values where these two curves intersect.
Setting x^3 = 1, we get x = 1. Therefore, the limits of integration are from x = 0 to x = 1.
Next, let's consider an infinitesimally small disk created by rotating an infinitesimally small segment of the region about the x-axis. The radius of each disk is the distance between the x-axis and the curve y = x^3 at a given x-value. In this case, the radius is x^3.
To calculate the volume of each disk, we use the formula V = πr^2Δx, where r is the radius and Δx is the thickness of the disk (infinitesimally small width). In this case, Δx = dx, which represents the infinitesimally small change in x.
So, V = π(x^3)^2 dx.
To find the total volume, we integrate the expression from x = 0 to x = 1:
V = ∫[0 to 1] π(x^3)^2 dx
Simplifying the expression:
V = π∫[0 to 1] x^6 dx
Integrating:
V = π[(1/7)x^7] from 0 to 1
Evaluating the integral:
V = π(1/7)
So, the volume of the solid formed is (1/7)π, which is the correct answer.
Therefore, your answer of 3/5π is incorrect. The correct volume is (1/7)π.