This is a calculus question involving solids of revolution. It might be an idea to talk to your teacher and ask for more help and tuition because - and I don't mean to be rude - the question is very clear and I fear that the method for solving the question might be difficult to understand if deciphering the question is giving you problems.
But here goes:
If a shape is BOUNDED by several functions (here we have four), then you want to sketch all of these functions on a graph and see if the lines they produce form a closed shape. Now, I can't draw here so bear with me.
The graph of y = 1/x has two parts to it. In the positive regions of the graph (x > 0, y > 0), you have a curved L-shape. It doesn't touch either of the axes but comes very close to them (hence it doesn't intersect with y = 0 at all). In the negative region of the graph (x < 0, y < 0), you have the same thing but it's been rotated by 180º. It's an upside-down and back-to-front curved L. Again, this doesn't touch the axes.
Here you go, it looks like this: http://www.revisioncentre.co.uk/gcse/maths/1overx.gif
Now, on this same graph, draw the lines x = 3 and x = 1. These are simply two vertical lines that pass through x = 3 and x = 1.
Our fourth function is y = 0. Well, this is just the x-axis, so it's already drawn on.
If you've done this, can you see that there is a distinct shape (or boundary) that's been created ? It's like a rectangle (between x = 1 and x = 3) but it has a curved piece missing off the top (the bit that's bounded by y = 1/x).
This is the shape we now need to rotate around the line y = -1 and the question is asking us to calculate its volume.
Because we're rotating it around a line, this 3D shape we're making is called a solid (or volume) of revolution. For any function y = f(x) rotated around the x-axis 360º, the volume of the shape it creates is given by:
∫ πy² dx
and the limits of this integral are the values on the x-axis where the shape starts and ends. You actually guessed that part of the question correctly - x = 3 and x = 1 do turn out to be left/right limits, but be careful because that won't always be the case. Always make a sketch first to see what you're dealing with!
Now, although it asks us to rotate it around the line y = -1, I'm going to start by imagining I am rotating it around the x-axis (ie. the line y = 0). That way, I can use quite easily the formula I gave you above. The function f(x) we're dealing with is y = 1/x because this is the function that's bounding the top of the shape. So, the volume of this shape is:
∫ π(1/x)² dx = π ∫ 1/x² dx = -π(1/x)
Now plug in the values x = 3 and x = 1 and take the difference:
-π * (1/3 - 1/1) = -π * (-2/3) = 2π/3
We're actually done and that's our answer. Although I rotated the shape around the line y = 0 instead of the line y = -1, the only difference between the two solids created by these two different rotations is that the second one (around the line y = -1) has a big hole in the middle. However, this hole hasn't taken anything away from the volume of the solid we just considered, so both volumes would end up being the same. Hence the answer is:
2π/3
Please rate me best answer :o) And do consider asking your teacher for help. It'll make everything so much clearer and less scary, and you'll be able to do these types of questions in exams no problem.