Final answer:
The volume of the solid of revolution formed by revolving the region bounded by y = 10/x², y = 0, x = 1, and the y-axis, is calculated using the disk method with an improper integral from 1 to infinity of f(x) = 10/x².
Step-by-step explanation:
The volume of a solid of revolution can be calculated using the method of disks or washers. Since we’re revolving the region bounded by y = 10/x², y = 0, x = 1, and the y-axis, we will use the disk method. The formula for the volume V of a solid of revolution generated by revolving a function y=f(x) about the x-axis from a to b is given by:V = π ∫_a^b f(x)^2 dxIn our case, a = 1, b = infinity (since the area extends indefinitely above the y-axis), and f(x) = 10/x², the volume can be calculated as:V = π ∫_1^∞ (10/x²)^2 dxThis improper integral represents the volume of the solid of revolution.
Let's divide the region into infinitely many thin cylindrical shells with radius x and height dx. The volume of each shell is given by dV = 2πx(10/x²)dx = 20π/x dx.The total volume of the solid is obtained by integrating the volumes of all the shells, which gives: V = ∫(from 1 to ∞) 20π/x dx. This integral converges and can be evaluated as V = 20π ln(x) |(from 1 to ∞) = ∞.