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

　
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%E2%88%927+to+7%5D+%E2%88%AB+%5B0+to+%E2%88%9A(49%E2%88%92x%C2%B2)%5D+(6x%2B2)+dy+dx

Thank you!