Question

Double integral formula when converting into polar coordinate

Answer 1

To this point we’ve seen quite a few double integrals. However, in every case we’ve seen to this point the region DD could be easily described in terms of simple functions in Cartesian coordinates. In this section we want to look at some regions that are much easier to describe in terms of polar coordinates. For instance, we might have a region that is a disk, ring, or a portion of a disk or ring. In these cases, using Cartesian coordinates could be somewhat cumbersome. For instance, let’s suppose we wanted to do the following integral,

∬Df(x,y)dA,D is the disk of radius 2∬Df(x,y)dA,D is the disk of radius 2

To this we would have to determine a set of inequalities for xx and yy that describe this region. These would be,

−2≤x≤2−√4−x2≤y≤√4−x2−2≤x≤2−4−x2≤y≤4−x2

With these limits the integral would become,

∬Df(x,y)dA=∫2−2∫√4−x2−√4−x2f(x,y)dydx∬Df(x,y)dA=∫−22∫−4−x24−x2f(x,y)dydx

Due to the limits on the inner integral this is liable to be an unpleasant integral to compute.

However, a disk of radius 2 can be defined in polar coordinates by the following inequalities,

0≤θ≤2π0≤r≤20≤θ≤2π0≤r≤2

These are very simple limits and, in fact, are constant limits of integration which almost always makes integrals somewhat easier.

So, if we could convert our double integral formula into one involving polar coordinates we would be in pretty good shape. The problem is that we can’t just convert the dxdx and the dydy into a drdr and a dθdθ. In computing double integrals to this point we have been using the fact that dA=dxdydA=dxdy and this really does require Cartesian coordinates to use. Once we’ve moved into polar coordinates dA≠drdθdA≠drdθand so we’re going to need to determine just what dAdA is under polar coordinates.

So, let’s step back a little bit and start off with a general region in terms of polar coordinates and see what we can do with that. Here is a sketch of some region using polar coordinates.



So, our general region will be defined by inequalities,

α≤θ≤βh1(θ)≤r≤h2(θ)α≤θ≤βh1(θ)≤r≤h2(θ)

Now, to find dAdA let’s redo the figure above as follows,



As shown, we’ll break up the region into a mesh of radial lines and arcs. Now, if we pull one of the pieces of the mesh out as shown we have something that is almost, but not quite a rectangle. The area of this piece is ΔAΔA. The two sides of this piece both have length Δr=ro−riΔr=ro−ri where rorois the radius of the outer arc and riri is the radius of the inner arc. Basic geometry then tells us that the length of the inner edge is riΔθriΔθ while the length of the out edge is roΔθroΔθ where ΔθΔθ is the angle between the two radial lines that form the sides of this piece.

Now, let’s assume that we’ve taken the mesh so small that we can assume that ri≈ro=rri≈ro=r and with this assumption we can also assume that our piece is close enough to a rectangle that we can also then assume that,

ΔA≈rΔθΔrΔA≈rΔθΔr

Also, if we assume that the mesh is small enough then we can also assume that,

dA≈ΔAdθ≈Δθdr≈ΔrdA≈ΔAdθ≈Δθdr≈Δr

With these assumptions we then get dA≈rdrdθdA≈rdrdθ.

In order to arrive at this we had to make the assumption that the mesh was very small. This is not an unreasonable assumption.Recall that the definition of a double integral is in terms of two limits and as limits go to infinity the mesh size of the region will get smaller and smaller. In fact, as the mesh size gets smaller and smaller the formula above becomes more and more accurate and so we can say that,

dA=rdrdθdA=rdrdθ

We’ll see another way of deriving this once we reach the Change of Variables section later in this chapter. This second way will not involve any assumptions either and so it maybe a little better way of deriving this.

Before moving on it is again important to note that dA≠drdθdA≠drdθ. The actual formula for dAdA has an rr in it. It will be easy to forget this rr on occasion, but as you’ll see without it some integrals will not be possible to do.

Now, if we’re going to be converting an integral in Cartesian coordinates into an integral in polar coordinates we are going to have to make sure that we’ve also converted all the xx’s and yy’s into polar coordinates as well. To do this we’ll need to remember the following conversion formulas,

x=rcosθy=rsinθr2=x2+y2x=rcos⁡θy=rsin⁡θr2=x2+y2

We are now ready to write down a formula for the double integral in terms of polar coordinates.

∬Df(x,y)dA=∫βα∫h2(θ)h1(θ)f(rcosθ,rsinθ)rdrdθ∬Df(x,y)dA=∫αβ∫h1(θ)h2(θ)f(rcos⁡θ,rsin⁡θ)rdrdθ

It is important to not forget the added rr and don’t forget to convert the Cartesian coordinates in the function over to polar coordinates.

Let’s look at a couple of examples of these kinds of integrals.