Application of Integrals · Question 27

If the area of the region \(x, y) : |4 - x2 | y x2, y 4, x 0 \ is ( 80 2 - ), , N, then + is equal to .

Answer

22

Solution

|4 - x2| x2: for 0 x 2 this needs 4 - x2 x2 x 2; for x 2 it always holds.

For x 2: y 4 and y x2 - 4 x2 - 4 4 x 2 2

Part 1 ( 2 x 2): 4 - x2 y x2

A1 = 22 (2x2 - 4) dx = [ 2x33 - 4x ] 22

= ( 163 - 8 ) - ( 4 23 - 4 2 ) = 8 23 - 83

Part 2 (2 x 2 2): x2 - 4 y 4

A2 = 22 2 (8 - x2) dx = [8x - x33 ]22 2

= (16 2 - 16 23 ) - (16 - 83 ) = 32 23 - 403

Area = A1 + A2 = 40 23 - 16 = 80 26 - 16

= 6, = 16

+ = 22

Answer: 22