Hi,
To find limits for multivariable integrals, you will almost always have to draw a sketch of the curves that form the boundaries of the region you are integrating over. This is because it will often not be enough to merely know at what points the curves intersect, but you will also have to decide which coordinate you would like to integrate over first, and in what order the boundary curves should be used as limits for the innermost integrals.
So the first step is either using a calculator, computer, or your own knowledge of curve sketching from earlier calculus courses to sketch the given curves.
The first boundary curve in your problem is y = 0. The points we want to consider for our region are all points for which y ≥ 0, so we have to recall that points with y-coordinates greater than a linear function all lie above the graph of that line. So we put little hatch marks on the upper side of the boundary curve y = 0 to remember which side of the boundary curve we are interested in. We do not shade in all of these points because we have more boundaries to consider.
The next boundary curve is y = x, and we want to make little hatch marks on the upper side of the line y = x to represent which side of the boundary contains the points that solve the inequality y ≥ x.
The third boundary curve is 1 = x² + y², which we recall is the equation of a unit circle centered at the origin. We must also recall that x² + y² = r² represents, in general, a circle with radius r centered at the origin. Therefore, the set of points that solve the inequality x² + y² ≥ 1 (note that this is the same inequality as 1 ≤ x² + y²) must belong to all circles centered at the origin with radii greater than 1 unit. So this is basically all points outside the circle of radius 1. So once more we make little hatch marks on the outside edge of the circle of radius 1 to signify which points we are interested in.
The fourth boundary curve is the set of points that solve the inequality x² + y² ≤ 4, which, as we may recall from the previous paragraph's discussion, is the set of all points inside the circle of radius 2 that is centered at the origin.
Once we have drawn all of the boundary curves and ascertained which side of each curve we want to pay attention to, we can see that there is only one region of points that solve all 4 inequalities simultaneously (find the closed region where all your hatch marks are facing inwards). This is the region of integration, which is called R in your problem.
Once you have sketched this region, you can see that it will be easier to describe this region in polar coordinates, since it is a region between two circles, and between two angular rays (see the attached graph if you are not sure why that is). Translate the curves into polar coordinates to see why: the line y = 0 is simply θ = 0, the line y = x is also simple: θ = π/4.
Circles are especially simple: they are the constant curves of the radial coordinate. The circle x² + y² = 1 is the set of points that solve the equation r = 1 (the set of points that are 1 unit away from the origin), while the circle x² + y² = 4 is the set of points that solve the equation r = 2 (the set of points that are 2 units away from the origin).
In addition, since the function you want to integrate over this region, ln(x² + y²), contains the expression x² + y², which as you might recall from the definition of polar coordinates is r², this integral will be greatly simplified in polar coordinates.
To find the limits for each polar coordinate, r and θ, first decide which coordinate will be your "outermost" integral. This is important because this outermost integral must always be an ordinary definite integral, where the two limits are numbers, not functions.
In polar coordinates, the usual choice for the variable being integrated over on the outermost integral is the angular coordinate θ. Looking at our sketch, we can see that our region lies between θ = π/4 and θ = π. So the limits of your outermost integral, which will be dθ, are θ = π/4 and θ = π.
To get the limits for the next innermost integral, which is your radial coordinate integral, fix a single arbitrary value of θ, and see what two r-coordinates you must integrate over for each value of θ.
In this particular case, it is straightforward: for each angular ray corresponding to a fixed θ value, the function must be integrated over the part of the ray between r = 1 and r = 2. These are then your limits for the innermost dr integral.
Remember that when you use polar coordinates, the area element is not just dr dθ, but r dr dθ.
So your integral will be:
\[\int_{\frac{\pi}{4}}^{\pi} \, \int_1^2\, \ln \left( r^2 \right) r \, dr \, dθ\]
