Consider the solid obtained by rotating the region bounded by the given curves about the line x = 6.
y = sqrt(x), y = x
Find the volume V of this solid.
What are the bounds on x?
To find the volume of the solid obtained by rotating the region between the curves about the line x = 6, we can use the method of cylindrical shells.
The first step is to sketch the region and the line of rotation.
The given curves are y = √x and y = x. Let's find the points of intersection of these curves:
√x = x
Squaring both sides:
x = x^2
Rearranging:
x^2 - x = 0
Factoring:
x(x - 1) = 0
x = 0 or x = 1
So, the curves intersect at x = 0 and x = 1.
Now, let's sketch the region and the line of rotation:
The region is bounded by the two curves and extends from x = 0 to x = 1. The line of rotation is x = 6, which is a vertical line.
Next, we need to determine the height of a typical cylindrical shell. Consider a horizontal strip of width Δx within the region. As we rotate this strip about the line of rotation, it generates a cylindrical shell. The height of this shell is the difference between the y-values of the two curves at a given x-value.
For a typical x-value within the region, the height of the cylindrical shell is y = x - √x.
Now that we have the height, we need to determine the radius of the cylindrical shell. The radius is the distance between the line of rotation (x = 6) and the x-value of the strip (x). So, the radius is r = 6 - x.
The volume of a cylindrical shell is given by the formula:
Vshell = 2πrhΔx
Now, we can express the volume of the solid as the sum of all the individual cylindrical shells:
V = ∫(from 0 to 1) of 2π(6 - x)(x - √x)dx
To integrate this expression, we need to use the techniques of calculus. The integral will give us the total volume of the solid formed by rotating the region between the curves about the line x = 6.
Evaluating this integral will give us the volume V of the solid.