Question

Let R be the region in the first quadrant bounded by the x- and y-axes, the horizontal line y=1, and the graph of y=In x, as shown in the figure above. What is the volume of the solid generated when region R is revolved about the Y axis?

Answer 1

Step-by-step explanation:

To find the volume of the solid generated when region R is revolved about the y-axis, we can use the method of cylindrical shells.

First, let's determine the limits of integration. The region R is bounded by the x-axis, y-axis, the line y=1, and the graph of y=ln(x). To find the limits, we set up the integral with respect to y.

Since the graph of y=ln(x) intersects the line y=1 at x=e, the limits of integration for y will be from 0 to 1.

Now, let's consider an infinitesimally thin vertical strip within region R. The height of this strip is dy, and its radius is the distance from the y-axis to the graph of y=ln(x), which is x=e^y.

The circumference of the cylindrical shell at height y is given by 2π(e^y), and the height (or thickness) is dy. Therefore, the volume of this shell is 2π(e^y)dy.

To find the total volume, we integrate this expression with respect to y over the limits of integration (0 to 1):

V = ∫[0,1] 2π(e^y)dy

Evaluating this integral gives:

V = 2π ∫[0,1] e^y dy

= 2π [e^y] [0,1]

= 2π (e^1 - e^0)

= 2π (e - 1)

Hence, the volume of the solid generated when region R is revolved about the y-axis is 2π (e - 1) cubic units.