Use geometry or symmetry, or both, to evaluate the double integral. (6x + 2)dA D={(x, y) | 0 ≤ y ≤ sqrt(49 − x^2)}?

2 Answers

• 　

  0 ≤ y ≤ √(49−x²)

  This is the region inside semi-circle x² + y² = 49 above the x-axis

  ∫∫ (6x+2) dA = ∫ [−7 to 7] ∫ [0 to √(49−x²)] (6x+2) dy dx

  D

  　　　　　　　　= ∫ [−7 to 7] ∫ [0 to √(49−x²)] 6x dy dx

  　　　　　　　　　+ ∫ [−7 to 7] ∫ [0 to √(49−x²)] 2 dy dx

  By symmetry, first integral = 0

  　　　　　　　　= ∫ [−7 to 7] ∫ [0 to √(49−x²)] 2 dy dx

  　　　　　　　　= 2 ∫ [−7 to 7] ∫ [0 to √(49−x²)] dy dx

  Geometrically, since we are integrating over a semi-circle of radius 7, then

  ∫ [−7 to 7] ∫ [0 to √(49−x²)] dy dx = area of semi-circle of radius 7, and

  2 ∫ [−7 to 7] ∫ [0 to √(49−x²)] dy dx = area of circle of radius 7

  　　　　　　　　= π(7)²

  　　　　　　　　= 49π

  ∫∫ (6x+2) dA = 49π

  D

  Check:

  http://www.wolframalpha.com/input/?i=%E2%88%AB+%5B…

• Thank you!