the volume of the solid generated by revolving infinite region bounded by x-axis, x=k, and y=1/x+2 in the first quadrant about the x-axis to generate a solid. Find the volume of the solid.
To find the volume of the solid generated by revolving the given region about the x-axis, we can use the method of cylindrical shells.
First, let's visualize the region in the first quadrant. We have the x-axis, the line x = k (where k is a constant), and the curve y = 1/(x+2).
The region between the x-axis and the curve y = 1/(x+2) corresponds to the values of x from k to infinity. However, since we are revolving the region about the x-axis, we need to find the point of intersection between the curve and the x-axis. Setting y = 0, we can find this point:
0 = 1/(x+2)
0 = 1
x + 2 = 0
x = -2
Therefore, the region of interest is from x = k to x = -2.
Now, let's consider a vertical strip within this region. The width of the strip is Δx (where Δx approaches zero), and its height is given by the difference between the y-coordinates of the two curves at that x-value. In this case, the height of the strip is y = 1/(x+2).
To obtain the volume of each cylindrical shell, we multiply the height by the circumference of the shell and by the width Δx. The circumference of each shell is given by 2πx since we're revolving the region about the x-axis.
The volume of each cylindrical shell is then given by:
dV = 2πx(1/(x+2))Δx
To find the total volume of the solid, we integrate this expression over the interval [k, -2]:
V = ∫[k, -2] 2πx(1/(x+2)) dx
Now, we can solve this integral to find the volume of the solid generated.