Question

Let R be the region enclosed by the graph of y=ln x, the line x=3, and the x-axis. (a) Fine the area of region R. (b) Find the volume of the solid generated by revolving region R about the x-axis. (c) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated by revolving region R about the line x=3. ...?

Answer 1

The area of region R can be found by integrating the function y=ln(x) from x=0 to x=3. The volume of the solid generated by revolving region R about the x-axis can be determined using the method of cylindrical shells. To set up an integral expression for the volume of the solid generated by revolving region R about the line x=3, consider the offset caused by the rotation.

Step-by-step explanation:

To find the area of region R, we need to integrate the function y=ln(x) from x=0 to x=3. The region is bounded by the graph of y=ln(x), the line x=3, and the x-axis. The integral expression for the area is given by:

A = ∫ₕₒ ₒ ln(x) dx, where the limits of integration are from 0 to 3.

To find the volume of the solid generated by revolving region R about the x-axis, we can use the method of cylindrical shells. We integrate the expression 2πx(ln(x)) dx from x=0 to x=3. This gives us the integral expression:

V = 2π∫ₕₒ ₒ x(ln(x)) dx, where the limits of integration are from 0 to 3.

To set up an integral expression for the volume of the solid generated by revolving region R about the line x=3, we need to consider the offset caused by the rotation. The radius of rotation is 3-x. So, the integral expression becomes:

V = 2π∫ₑₕ ₒ [(3-x)(ln(x))] dx, where the limits of integration depend on the intersection points of the curve and the line x=3.

Answer 2

a.) Area of region R =

`\int\limits^3_1 {ln(x)} \, dx =x(ln(x)-1) \left \} {{y=3} \atop {x=1}} \right. \\=3(ln(3)-1)-(ln(1)-1) = 0.2958 + 1=1.2958\,square\,units`