Well, Idon't! Because you are missing a left boundary. Did you intend x= 0 as a boundary?

The first of those is easy because now you can assume that x= 0 is a boundary. Imagine a horizontal line, representing a thin disk as it is rotated around the y-axis. Its boundary will be a circle having radius from x= 0 to x= ln(y)/10 (solving for x). Such a circle has area and, taking the disk to have thickness dy, the disk will have volume [tex]\frac{\pi}{100}(ln(y))^2dy[tex] and the whole figure will have volume [tex]\frac{\pi}{100}\int_1^3 (ln(y))^2dy[/itex]But what if I am told to find the volume if:

revolved bout the y-axis or the x-axis

how would I find these volumes?

For rotating around the x-axis, again, you will need to give some left boundary, perhaps the y-axis, x= 0. Now, the simplest thing to do is to look at separate parts: The upper boundary, y= 3, is a horizontal line and rotating around the x-axis gives a cylinder with radius 3 and height ln(10)/3 (where crosses y= 3. That cylinder has volume . The lower boundary is and if we rotate the area below that around the x-axis, each thin disk will have radius so area and volume so the total volume is . That is NOT the volume you want- the volume you want is the volume of the cylinder minus that integral. (This is exactly the same thing as using the washer method, doing that subtraction before the integral.)