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Notice that the expression for d A is replaced by r d r d θ when working in polar coordinates. Another way to look at the polar double integral is to change the double integral in rectangular coordinates by substitution. When the function f is given in terms of x and y , using x = r cos θ , y = r sin θ , and d A = r d r d θ changes it to

R f ( x , y ) d A = R f ( r cos θ , r sin θ ) r d r d θ .

Note that all the properties listed in Double Integrals over Rectangular Regions for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well, so we can use them without hesitation.

Sketching a polar rectangular region

Sketch the polar rectangular region R = { ( r , θ ) | 1 r 3 , 0 θ π } .

As we can see from [link] , r = 1 and r = 3 are circles of radius 1 and 3 and 0 θ π covers the entire top half of the plane. Hence the region R looks like a semicircular band.

Half an annulus R is drawn with inner radius 1 and outer radius 3. That is, the inner semicircle is given by x squared + y squared = 1, whereas the outer semicircle is given by x squared + y squared = 9.
The polar region R lies between two semicircles.
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Now that we have sketched a polar rectangular region, let us demonstrate how to evaluate a double integral over this region by using polar coordinates.

Evaluating a double integral over a polar rectangular region

Evaluate the integral R 3 x d A over the region R = { ( r , θ ) | 1 r 2 , 0 θ π } .

First we sketch a figure similar to [link] but with outer radius 2 . From the figure we can see that we have

R 3 x d A = θ = 0 θ = π r = 1 r = 2 3 r cos θ r d r d θ Use an iterated integral with correct limits of integration. = θ = 0 θ = π cos θ [ r 3 | r = 1 r = 2 ] d θ Integrate first with respect to r . = θ = 0 θ = π 7 cos θ d θ = 7 sin θ | θ = 0 θ = π = 0 .
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Sketch the region R = { ( r , θ ) | 1 r 2 , π 2 θ π 2 } , and evaluate R x d A .

14 3

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Evaluating a double integral by converting from rectangular coordinates

Evaluate the integral R ( 1 x 2 y 2 ) d A where R is the unit circle on the x y -plane.

The region R is a unit circle, so we can describe it as R = { ( r , θ ) | 0 r 1 , 0 θ 2 π } .

Using the conversion x = r cos θ , y = r sin θ , and d A = r d r d θ , we have

R ( 1 x 2 y 2 ) d A = 0 2 π 0 1 ( 1 r 2 ) r d r d θ = 0 2 π 0 1 ( r r 3 ) d r d θ = 0 2 π [ r 2 2 r 4 4 ] 0 1 d θ = 0 2 π 1 4 d θ = π 2 .
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Evaluating a double integral by converting from rectangular coordinates

Evaluate the integral R ( x + y ) d A where R = { ( x , y ) | 1 x 2 + y 2 4 , x 0 } .

We can see that R is an annular region that can be converted to polar coordinates and described as R = { ( r , θ ) | 1 r 2 , π 2 θ 3 π 2 } (see the following graph).

Two semicircles are drawn in the second and third quadrants, with equations x squared + y squared = 1 and x squared + y squared = 2.
The annular region of integration R .

Hence, using the conversion x = r cos θ , y = r sin θ , and d A = r d r d θ , we have

R ( x + y ) d A = θ = π / 2 θ = 3 π / 2 r = 1 r = 2 ( r cos θ + r sin θ ) r d r d θ = ( r = 1 r = 2 r 2 d r ) ( π / 2 3 π / 2 ( cos θ + sin θ ) d θ ) = [ r 3 3 ] 1 2 [ sin θ cos θ ] | π / 2 3 π / 2 = 14 3 .
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Evaluate the integral R ( 4 x 2 y 2 ) d A where R is the circle of radius 2 on the x y -plane.

8 π

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General polar regions of integration

To evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider two types of regions, analogous to Type I and Type II as discussed for rectangular coordinates in Double Integrals over General Regions . It is more common to write polar equations as r = f ( θ ) than θ = f ( r ) , so we describe a general polar region as R = { ( r , θ ) | α θ β , h 1 ( θ ) r h 2 ( θ ) } (see the following figure).

A region D is shown in polar coordinates with edges given by theta = alpha, theta = beta, r = h2(theta), and r = h1(theta).
A general polar region between α < θ < β and h 1 ( θ ) < r < h 2 ( θ ) .

Double integrals over general polar regions

If f ( r , θ ) is continuous on a general polar region D as described above, then

D f ( r , θ ) r d r d θ = θ = α θ = β r = h 1 ( θ ) r = h 2 ( θ ) f ( r , θ ) r d r d θ
Practice Key Terms 1

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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