The area of the region R=( (x ,y ) : xy 8 ,1 y x^2 ,x 0) is

Question

Accepted Answer

Correct answer: A. (a) The region R is defined by xy 8 ,1 y x^2, and x 0. First, find the intersection points of the curves. Intersection of y=1 and y=x^2 :x^2=1 ⟹x=1 (since x 0 ). Intersection of y=x^2 and xy=8 :x (x^2 )=8 ⟹x^3=8 ⟹x=2. At x=2 ,y=4. Intersection of xy=8 and y=1 :x (1 )=8 ⟹x=8. The region is bounded below by y=1 from x=1 to x=8. The upper boundary consists of two parts: From x=1 to x=2, the upper curve is y=x^2. From x=2 to x=8, the upper curve is y= 8x. The area A is given by: A= _1^2 0.05em(x^2-1) dx+ _2^8 0.05em( 8x-1) dx A=( x^33-x)_1^2+(8 log_e ⁡x-x)_2^8 A=( 83-2)-( 13-1)+(8 log_e ⁡8-8)-(8 log_e ⁡2-2) A= ( 23 )-(- 23)+8 (3 log_e ⁡2 )-8-8 log_e ⁡2+2 A= 43+24 log_e ⁡2-8 log_e ⁡2-6 A=16 log_e ⁡2+ 43-6 A=16 log_e ⁡2- 143 A= 23(24 log_e ⁡2-7)

$The area of the region R = ((x, y) : xy \leq 8, 1 \leq y \leq x^{2}, x \geq 0) is$

Solution

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