Section 15.4
Double Integrals in Polar Coordinates
We have learned to deal with integrating multivariate functions over several different types of
regions in the xy plane; in particular, we think of the boundaries of the region R in terms of x and
y and integrate accordingly. However, there will be occasions in which it is far more convenient
to think of the region R in polar coordinates as opposed to Cartesian coordinates. For example,
consider what would happen if we attempted to integrate the function f (x, y) over the region R
below in Cartesian coordinates:
Since the bounds of the region are not given either in terms of functions of x or of functions of
y, it would be difficult to set up the integral in Cartesian coordinates; we would need to separate
the region into smaller pieces, then set up an integral for each piece.
However, setting up the integral in terms of polar coordinates would simplify the problem
significantly. In this section we will learn how to use polar coordinates to evaluate certain integrals.
Before we look at the details of the method, let’s recall a few facts about polar coordinates.
Any point in the plane can be described by the Cartesian coordinates (x, y), where x and y are
measured along the corresponding axes. However, this is not the only way to represent points in
the plane; we may think of them in terms of the polar coordinates r and θ.
To build up the polar coordinate system, we will fix a point O, the origin, and an initial ray
(which generally corresponds to the positive part of the x-axis). We describe a point P in the plane
using (1) its directed distance r from the origin, and (2) the directed angle θ from the initial ray
to the segment OP :
To graph the point (r, θ), simply go out r units along the initial ray, then rotate through the
angle θ. The point (1, 5π
6 ) is graphed below:
1
Section 15.4
If we wish to convert a point’s polar coordinates to Cartesian coordinates, or vice-versa, we
can use a basic trigonometry to help us out. Recall that, if (x, y) is the Cartesian coordinate of
a point with angle θ from the initial ray, and if x2 + y 2 = r2 , then sin θ = yr and cos θ = xr :
So if P has polar coordinates (r, θ), then we can rewrite the coordinates using the conversions
x = r cos θ and y = r sin θ.
Alternatively, if we have Cartesian coordinates (x, y), then we can determine r and θ using the
formulas x2 + y 2 = r2 and tan θ = xy .
If we decide that we need to evaluate the double integral of f (x, y) over the region R in polar
coordinates as opposed to rectangular coordinates, we will need to rewrite all of the variables and
the bounds for the region in terms of r and θ. To set up the integral, start by rewriting f (x, y) as
f (r, θ) by using the conversions given above.
Next, we will need to rewrite the bounds. Sketch the region, then think of rotating a line
segment starting at the origin through the region:
2
Section 15.4
The outer radius and inner radius of the segments gives us the bounds on r, and the angles θ1
and θ2 where the segments start and stop give us the bounds on θ. Finally, we can use another
version of Fubini’s theorem to evaluate the integral:
Theorem 2. If f (x, y) is a continuous function over the region R whose bounds are given in polar
coordinates by 0 ≤ g1 (θ) ≤ r ≤ g2 (θ) and α ≤ θ ≤ β, then the double integral of f (x, y) over the
region R is
∫ β ∫ g2 (θ)
rf (r cos θ, r sin θ)drdθ.
α
g1 (θ)
Notice that the integral includes an extra copy of r in the integrand; we will not go into the
details of why this is the case, but it is vital to remember the r in order to evaluate the integral
correctly.
Examples
∫
2∫
Evaluate
0
√
4−y 2
e−x
2 −y 2
dxdy.
0
Try as we might, we will not be able to integrate e−x −y with respect to x or y. Thus our only
option for integrating the function is to change it to polar coordinates (hopefully this will help!).
Let’s start be rewriting the function in terms of r and θ. Since x = r cos θ and y = r sin θ,
2
2
2
2
2
2
2
e−x −y can be rewritten as e−r cos θ−r sin θ = e−r .
Now we need to rewrite the bounds; the region over which we wish to integrate is the semicircle
graphed below:
2
3
2
Section 15.4
Below, line segments are drawn through the region:
The inner radius comes from the beginning of the line segments, while the outer radius comes
from the ends of the line segments. In this case, the segments start at the origin (thus have radius
1), and end on the semicircle; thus the outer radius is 2.
The starting and stopping angles for the line segments give us the lower and upper bounds for
π
θ; in this case, the segments start at −π
2 and end at 2 .
4
Section 15.4
Thus the integral may be rewritten as
∫ ∫ √
4−y 2
2
e
0
−x2 −y 2
∫
dxdy =
0
π
2
∫
−π
2
π
2
∫
−π
2
π
2
=
∫
2
re−r drdθ
2
0
(
1
2 2 )
− e−r 0 d θ
2
1
− (e−4 − e0 )dθ
−π
2
2
π
1
2
= − (e−4 − 1)θ −π
2
2
1 −4
π −π
= − (e − 1)( −
)
2
2
2
1
= − (e−4 − 1)π
2
π
π
= − 4
2 2e
=
Set up an integral that will yield the volume of the solid bounded above by z = 1 − x2 − y 2 and
below by z = 0.
Let’s start by getting an idea of what the solid looks like. It is clear that the highest point of
the solid is (0, 0, 1). When y = 0, the solid looks like the parabola z = 1 − x2 , and similarly for
x = 0, the solid looks like z = 1 − y 2 . Thus the solid looks like the region graphed below:
The region over which we wish to integrate is the intersection of the surface with the xy plane,
i.e. all of the points satisfying 1 − x2 − y 2 = 0. Thus we wish to integrate over the circle x2 + y 2 = 1.
However, in order to do this in Cartesian coordinates we will need to set up two separate integrals;
it will be easier to use polar coordinates here.
The function 1 − x2 − y 2 can be rewritten as 1 − r2 cos2 θ − r2 sin2 θ = 1 − r2 . To rewrite the
bounds, note that line segments through the region have inner radius 0 and outer radius 1, and go
from angle 0 to angle 2π:
5
Section 15.4
Thus the volume of the region is
∫
0
π
∫
1
r(1 − r2 )drdθ.
0
( )
Evaluate the double integral of arctan xy over the region 1 ≤ x2 + y 2 ≤ 4, −x ≤ y ≤ x.
Again, this seems like an example
polar
would
( r sin
) integrals
( sin
) simplify the problem
( y )where using
θ
θ
significantly. Thus we rewrite arctan x as arctan r cos θ = arctan cos θ = arctan(tan θ) = θ (for
− π2 < θ < π2 ).
Let us graph the region indicated by the bounds:
Again using line segments to determine the bounds on r and θ, we see that 1 ≤ r ≤ 2, and
− π4 ≤ θ ≤ π4 .
6
Section 15.4
So the integral we need to evaluate is
∫
π
4
− π4
∫
∫
2
rθdrdθ =
π
4
− π4
1
∫
=
π
4
− π4
∫
=
π
4
− π4
1 ( 2 2 )
(r θ) 1 dθ
2
1
(4θ − θ)dθ
2
3
θdθ
2
3 π
= θ2 −4 π
4
4
2
(
π2 )
3 π
−
=
4 4
4
= 0.
7