2 Answers

0 ≤ y ≤ √(49−x²)
This is the region inside semicircle x² + y² = 49 above the xaxis
∫∫ (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 semicircle of radius 7, then
∫ [−7 to 7] ∫ [0 to √(49−x²)] dy dx = area of semicircle 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:

Thank you!