Use a double integral to find the area of the region. The region inside the circle (x − 5)2 + y2 = 25 and outside the circle x2 + y2 = 25

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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