1. DOUBLE INTEGRALS IN POLAR COORDINATES
SECTION 12.3
DOUBLE INTEGRALS IN POLAR COORDINATES

2. DOUBLE INTEGRALS IN POLAR COORDINATES
Suppose that we want to evaluate a double integral , where R is one of the regions shown in Figure 1.
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3. DOUBLE INTEGRALS IN POLAR COORDINATES
In either case, the description of R in terms of rectangular coordinates is rather complicated but R is easily described by polar coordinates.
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4. DOUBLE INTEGRALS IN POLAR COORDINATES
Recall from Figure 2 that the polar coordinates (r, ) of a point are related to the rectangular coordinates (x, y) by the equations r2 = x2 + y2 x = r cos  y = r sin 
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5. POLAR RECTANGLE
The regions in the first figure are special cases of a polar rectangle R = {(r, ) | a ≤ r ≤ b, a ≤  ≤ b} shown in Figure 3.
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6. POLAR RECTANGLE
To compute the double integral where R is a polar rectangle, we divide:
The interval [a, b] into m subintervals [ri–1, ri] of equal width ∆r = ri – ri–1.
The interval [a , b] into n subintervals [j–1, i] of equal width ∆ =  i – i–1.
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7. POLAR RECTANGLES
Then, the circles r = ri and the rays  = i divide the polar rectangle R into the small polar rectangles shown in Figure 4.
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8. The “center” of the polar subrectangle
Rij = {(r, ) | ri–1 ≤ r ≤ ri, j–1 ≤  ≤ j}
has polar coordinates
ri* = ½ (ri–1 + ri)
j * = ½ (j–1 + j)
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9. POLAR SUBRECTANGLE
We compute the area of Rij using the fact that the area of a sector of a circle with radius r and central angle  is ½r2.
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10. POLAR SUBRECTANGLE
Subtracting the areas of two such sectors, each of which has central angle ∆j = j – j–1, we find that the area of Rij is:
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11. We have defined the double integral in terms of ordinary rectangles.
POLAR RECTANGLES
We have defined the double integral in terms of ordinary rectangles.
However, it can be shown that, for continuous functions f, we always obtain the same answer using polar rectangles.
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12. So, a typical Riemann sum is:
POLAR RECTANGLES
The rectangular coordinates of the center of Rij are (ri* cos j*, ri* sin j*).
So, a typical Riemann sum is:
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13. POLAR RECTANGLES
If we write g(r, ) = rf(r cos , r sin ), the Riemann sum in Equation 1 can be written as:
This is a Riemann sum for the double integral
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14. POLAR RECTANGLES
Thus, we have:
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15. CHANGE TO POLAR COORDINATES IN A DOUBLE INTEGRAL
If f is continuous on a polar rectangle R given by
0 ≤ a ≤ r ≤ b, a ≤  ≤ b
where 0 ≤ b – a ≤ 2p, then
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16. CHANGE TO POLAR COORDS.
Formula 2 says that we convert from rectangular to polar coordinates in a double integral by:
Writing x = r cos  and y = r sin 
Using the appropriate limits of integration for r and 
Replacing dA by r dr d
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17. CHANGE TO POLAR COORDS.
Be careful not to forget the additional factor r on the right side of Formula 2.
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18. A classical method for remembering the formula is shown in Figure 5.
CHANGE TO POLAR COORDS.
A classical method for remembering the formula is shown in Figure 5.
The “infinitesimal” polar rectangle can be thought of as an ordinary rectangle with dimensions r d and dr.
So, it has “area” dA = r dr d.
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19. Example 1
Evaluate where R is the region in the upper half-plane bounded by the circles x2 + y2 = 1 and x2 + y2 = 4.
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20. The region R can be described as:
Example 1 SOLUTION
The region R can be described as:
R = {(x, y) | y ≥ 0, 1 ≤ x2 + y2 ≤ 4}
It is the half-ring shown in Figure 1(b).
In polar coordinates, it is given by:
1 ≤ r ≤ 2, 0 ≤  ≤ p
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21. Example 1 SOLUTION
Hence, by Formula 2,
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22. Example 1 SOLUTION
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23. Find the volume of the solid bounded by:
Example 2
Find the volume of the solid bounded by:
The plane z = 0
The paraboloid z = 1 – x2 – y2
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24. If we put z = 0 in the equation of the paraboloid, we get x2 + y2 = 1.
Example 2 SOLUTION
If we put z = 0 in the equation of the paraboloid, we get x2 + y2 = 1.
This means that the plane intersects the paraboloid in the circle x2 + y2 = 1.
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25. Example 2 SOLUTION
So, the solid lies under the paraboloid and above the circular disk D given by x2 + y2 ≤ 1.
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26. In polar coordinates, D is given by 0 ≤ r ≤ 1, 0 ≤  ≤ 2p.
Example 2 SOLUTION
In polar coordinates, D is given by 0 ≤ r ≤ 1, 0 ≤  ≤ 2p.
As 1 – x2 – y2 = 1 – r2, the volume is:
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27. Had we used rectangular coordinates instead, we would have obtained:
Example 2 SOLUTION
Had we used rectangular coordinates instead, we would have obtained:
This is not easy to evaluate because it involves finding ∫(1 – x2)3/2 dx
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28. CHANGE TO POLAR COORDS.
What we have done so far can be extended to the more complicated type of region shown in Figure 7.
It’s similar to the type II rectangular regions considered in Section
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29. CHANGE TO POLAR COORDS.
In fact, by combining Formula 2 in this section with Formula 5 in Section 12.2, we obtain the following formula.
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30. If f is continuous on a polar region of the form
Formula 3
If f is continuous on a polar region of the form
D = {(r, ) | a ≤  ≤ b, h1( ) ≤ r ≤ h2()} then
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31. CHANGE TO POLAR COORDS.
In particular, taking f(x, y) = 1, h1() = 0, and h2() = h() in the formula, we see that the area of the region D bounded by  = a,  = b, and r = h() is:
This agrees with Formula 3 in Section 9.4
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32. Find the volume of the solid that lies:
Example 3
Find the volume of the solid that lies:
Under the paraboloid z = x2 + y2
Above the xy-plane
Inside the cylinder x2 + y2 = 2x
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33. Example 3 SOLUTION
The solid lies above the disk D whose boundary circle has equation x2 + y2 = 2x.
After completing the square, that is: (x – 1)2 + y2 = 1.
See Figure 8 and 9.
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34. In polar coordinates, we have: x2 + y2 = r2 and x = r cos 
Example 3 SOLUTION
In polar coordinates, we have:
x2 + y2 = r2 and x = r cos 
So, the boundary circle becomes:
r2 = 2r cos  or r = 2 cos 
Thus, the disk D is given by:
D = {(r, ) | –p/2 ≤  ≤ p/2 , 0 ≤ r ≤ 2 cos }
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35. Example 3 SOLUTION
So, by Formula 3, we have:
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36. Example 3 SOLUTION
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