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.

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.

anonymous · Answer

The area under the curve is just the integral from wherever y = lnx crosses the x axis to 3

In this case, lnx crosses the x axis at x =1, so you'll be integrating lnx from 1 to 3