Mathematics · Integral Calculus
Let the circle be (x - 6)2 + y2 = 36 and the parabola be y2 = 12x. The curves intersect at two distinct points. The area of the region which lies inside the circle, outside the parabola, and in the first and fourth quadrants only, is equal to:
- 118π - 24
- 212π - 24
- 318π - 36✓
- 424π - 48
Correct answer
18π - 36
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