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SECTION 12.3

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

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SECTION 12.3

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  1. 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. 12.3

  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. 12.3

  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  12.3

  5. POLAR RECTANGLE • The regions in the first figure are special cases of a polar rectangleR = {(r, ) | a ≤r ≤b, a ≤ ≤b} shown in Figure 3. 12.3

  6. CHANGE TO POLAR COORDINATES IN A DOUBLE INTEGRAL If f is continuous on a polar rectangle R given by 0 ≤ a ≤ r ≤ b, a≤≤bwhere 0 ≤ b–a≤ 2p, then 12.3

  7. 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 dr d 12.3

  8. CHANGE TO POLAR COORDS. • Be careful not to forget the additional factor r on the right side of Formula 2. 12.3

  9. 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. 12.3

  10. Example 1 • Evaluate where R is the region in the upper half-plane bounded by the circles x2 + y2 = 1 and x2 + y2 = 4. 12.3

  11. 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 12.3

  12. Example 1 SOLUTION • Hence, by Formula 2, 12.3

  13. Example 1 SOLUTION 12.3

  14. Example 2 • Find the volume of the solid bounded by: • The plane z = 0 • The paraboloid z = 1 –x2–y2 12.3

  15. 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. 12.3

  16. Example 2 SOLUTION • So, the solid lies under the paraboloid and above the circular disk D given by x2 + y2≤ 1. 12.3

  17. Example 2 SOLUTION • In polar coordinates, D is given by 0 ≤r ≤ 1, 0 ≤ ≤ 2p. • As 1 –x2–y2 = 1 –r2, the volume is: 12.3

  18. 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 12.3

  19. 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 12.2. 12.3

  20. Formula 3 If f is continuous on a polar region of the form D = {(r, ) | a≤ ≤ b, h1( ) ≤r ≤h2()}then 12.3

  21. 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 12.3

  22. 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 12.3

  23. 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. 12.3

  24. 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 } 12.3

  25. Example 3 SOLUTION • So, by Formula 3, we have: 12.3

  26. Example 3 SOLUTION 12.3

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