Respuesta :
To answer this question, it is necessary to use polar coordinates, and integration procedures.
The solution is:
A = ( 50/6)×π
( x - 5 )² + y² = 25 (1)
Equation (1) is a circle equation with a center at ( 5 ; 0 ), we are going to look for the common points of the circles which at the same time will give the limits of integration
Transforming that equation into polar coordinates, we get:
x = r × cosθ y = r×sinθ ⇒ x = 5×cosθ and y = 5×sinθ
Plugging these values in equation (1)
( r×cosθ - 5 )² + r× sin²θ = 25
r²×cos²θ + 25 - 10×r×cosθ + r²×sin²θ = 25
r²×cos²θ + r²×sin²θ - 10×r×cosθ = 0
x² + y² - 10×r×cosθ = 0
r² - 10×r×cosθ = 0 ⇒ r - 10×cos θ = 0 ⇒ r = 10×cos θ
The equation
x² + y² = 25 is the equation of a circle with center at the origin and radius 5. Plugging that value of radius in equation (2)
r = 5
25 - 50×cosθ = 0
cosθ = 25/50 = 1/2 θ = 30° θ = π/3
By symmetry the other common point is at θ = - π/3
Then using double integration
A = ∫∫ r×dr×dθ
Integration limits, for
5 ≤ r ≤ 10×cosθ -π/3 ≤ θ ≤ π/3
∫rdr = (r²/2)|₅10cosθ = (100×cos²θ/2) - (25/2)
∫[(100×cos²θ)/2 - (25/2)]dθ
A = 50×∫cos²θ× dθ - ∫(25/2)]×dθ ⇒ A = 50×∫cos²θ× dθ - (25/2)×θ|₋π/3 y π/3
-[ (25/2)× (π/3) - (25/2)× (-π/3)] = - (25/6)×π + (25/6)×π = - (50/6)×π
Now ∫ cos²θ× dθ = (1/2)× ( θ + (1/2)×sin2θ then:
50 ∫ cos²θ× dθ = (50/2)× θ - (50/4) ×sin2θ | [ -π/3 y π/3 ]
25 ×(π/3) - 25 ×(-π/3) - [ (50/4)×[sin (2×π/3) - sin (-2×π/3)
[ (50/4)×[sin (2×π/3) - sin (-2×π/3) = 0
Then
A = (50/3)×π - (50/6)×π
A = ( 50/6)×π
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