Find the area of the region that lies inside the first curve and outside the second curve.

Question

r = 14 cos 𝜃,    r = 7

Answer 1

you could convert to rectangular coordinates, if you're more comfortable with them

r= 7 is a circle with center at the origin and radius = 7

r=14cos(theta) is the same circle shifted horizontally to the right, with center at (7,0)

r^2 = 14rcos(theta)

r^2 = x^2+ y^2

rcos(theta) = x

x^2 + y^2 = 14x

x^2 -14x + 49 + y^2 = 49

(x-7)^2 + y^2 = 7^2

find the overlap area and subtract that from 49pi=area of either circle

it helps to graph the circles

you could do that with integration or geometry

area of the overlap = 4 times the integral of sqr(49-x^2) evaluated from 0 to 3.5

=4[(xsqr(49-x^2) +49arcsin(x/7)]/2 from x=0 to 3.5

= about 22.13pi

area of the circle minus the overlap

= about 49pi-22.13pi = 26.87 pi

geometrically, calculating the area of a sector with angle .275pi = 13.465pi

49pi- 13.465 = 35.5 pi as an upper bound

area < 35.5pi

subtract areas of two chord segments

Answer 2

One way to do this is to draw the curves and see that the areal region can be described as the differences in the "wedge-shapes" of r₁ and the portion of the wedge shape in r₂ for the range of theta that is of interest.

You can solve for the intersection points: 14cosθ = 7 or cosθ = 1/2 θ = ±π/3

The wedge areas are (1/2)r²dθ

The integral from π/3 to 5π/3 of (1/2)((14cosθ)² - 7²)dθ

I get 49/6 * (4π - 3sqrt(3)) using Wolfram Alpha. (It is integrable explicitly using cos²θ = 1/2 (1+cos2θ) sub.)