Double Integration in Polar Coordinates

Polar Double Integration Formula

Many of the double integrals that we have encountered so far have involved circles or at least expressions with x2 + y2.  When we see these expressions a bell should ring and we should shout, "Can't we use polar coordinates."  The answer is, "Yes" but only with care.  Recall that when we changed variables in single variable integration such as u  =  2x, we needed to work out the stretching factor du  =  2dx.  The idea is similar with two variable integration.  When we change to polar coordinates, there will also be a stretching factor.  This is evident since the area of the "polar rectangle" is not just as one may expect .  The picture is shown below.

           

Even if Dr and Dq are very small the area is not the product (Dr)(Dq).  This comes from the definition of radians.  An arc that extends Dq radians a distance r out from the origin has length rDq.  If both Dr and Dq are very small then the polar rectangle has area

        Area  =  r Dr Dq 

This leads us to the following theorem

Theorem:  Double Integration in Polar Coordinates

Let f(x,y) be a continuous function defined over a region R bounded in polar coordinates by 

               r1(q)  <  r  <  r2(q)                    q1  <  q  <  q2

Then 

              

Notice the extra "r" in the theorem

 

Using Polar Coordinates

 

Example

Find the volume to the part of the paraboloid 

        z  =  9 - x2 - y2

that lies inside the cylinder

        x2 + y2  =  4

 

Solution  

The surfaces are shown below.

           

This is definitely a case for polar coordinates.  The region R is the part of the xy-plane that is inside the cylinder.  In polar coordinates, the cylinder has equation

        r2  =  4

Taking square roots and recalling that r is positive gives

        r  =  2

The inside of the cylinder is thus the polar rectangle 

        0  <  r  <  2        0  <  q  <  2p

The equation of the parabola becomes 

        z  =  9 - r2 

We find the integral 

       

This integral is a matter of routine.  It evaluates to 28p.

 

Example

Find the volume of the part of the sphere of radius 3 that is left after drilling a cylindrical hole of radius 2 through the center.

 

Solution

The picture is shown below

       

The region this time is the annulus (washer) between the circles r  =  2 and r  =  3 as shown below.

           

The sphere has equation

        x2 + y2 + z2  =  9

In polar coordinates this reduces to 

        r2 + z2  =  9

Solving for z by subtracting r2 and taking a square root we get top and bottom surfaces of 

       

We get the double integral 

       

This integral can be solved by letting

        u  =  9 - r2           du  =  -2rdr

After substituting we get

       

 

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