Find the area of the region enclosed between the graphs of x+y=10 and (x^1/2)+(y^1/2)=4.
2007-03-26 02:09:15 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x+y=10 and (x^1/2)+(y^1/2)=4.
intersection points squaring 2nd
x+y+2(xy)^1/2 = 16 >>>> 2(xy)^1/2 = 16 - 10 = 6
(xy)^1/2 = 3 or xy = 9 >>> y = 9/x put in (1)
x + 9/x = 10 >>> x^2 -10 x +9=0
x = 10+- (100 - 36)^1/2 / 2 = )10+-8) / 2
x1= 1 , x2 = 9 limits of integration
Area A = ∫ [f(x) - g(x)] dx where f g are y's
A = ∫ { [10 - x] -[ 4 - sqrt(x)]^2 } dx
A = ∫ {10 - x - [16 + x - 8 sqrt(x)]} dx
A = ∫ - 2 x - 6 + 8 sqrt(x)] dx integrate
A = [(16/3) (x)^3/2 - x^2 - 6 x ] limit x =1 to 9
A = [(16/3) * 27 - 81 - 54] - [(16/3) - 1 - 6]
A = [144 - 135] + [5/3]
A = [9] + [5/3] = 32 /3
2007-03-26 03:05:35 · answer #1 · answered by anil bakshi 7 · 0⤊ 0⤋
x+y=10 (x^1/2) + (y ^1/2) =4
x=10 - y (x +y)^1/2 = 4
(x + y)^1/2 - 4 = 0
2007-03-26 09:20:59 · answer #2 · answered by pokemon maniac 6 · 0⤊ 1⤋