Question

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles x2 y2=64 and x2−8x y2=0.

Answer 1

You mention circles, so I assume the missing symbols are plus signs.

The first circle has equation x² + y² = 64 = 8², so it's centered at the origin and has radius 8. In polar coordinates, we take x = r cos(θ) and y = r sin(θ). Then this circle's equation is

(r cos(θ))² + (r sin(θ))² = 64

r² (cos²(θ) + sin²(θ)) = 8²

r = 8

Complete the square in the second circle's equation to determine its center and radius:

x² - 8x + y² = x² - 8x + 16 + y² = 16   ⇒   (x - 4)² + y² = 4²

so the second circle is centered at (4, 0) with radius 4. In polar coordinates, this equation becomes

(r cos(θ) - 4)² + (r sin(θ))² = 16

r² (cos²(θ) + sin²(θ)) - 8r cos(θ) + 16 = 16

r² - 8r cos(θ) = 0

r - 8 cos(θ) = 0

r = 8 cos(θ)

Determine where these two circles intersect:

8 = 8 cos(θ)

cos(θ) = 1

θ = arccos(1) + 2nπ

θ = 2nπ

where n is any integer. So both circles intersect once at (8, 0), which means they are tangent to one another, and since the second circle's radius is smaller, it must lie completely within the larger one.

The region between the two circles is the set

R = {(r, θ) : 0 ≤ θ ≤ π/2 and 8 cos(θ) ≤ r ≤ 8}

The area of R is given by the double integral,

$\displaystyle \iint_R dA = \int_0^{\frac\pi2} \int_{8\cos(\theta)}^8 r \, dr \, d\theta$

or the single integral

$\displaystyle \iint_R dA = \frac12 \int_0^{\frac\pi2} \left(64 - 64\cos^2(\theta)\right) \, d\theta$

Recall the half-angle identity for cosine:

cos²(x/2) = (1 + cos(x))/2

Then the area is

$\displaystyle \iint_R dA = 16 \int_0^{\frac\pi2} \left(1 - \cos(2\theta)\right) \, d\theta$

$\displaystyle \iint_R dA = 16 \left(\theta - \frac12 \sin(2\theta)\right) \bigg|_0^{\frac\pi2}$

$\displaystyle \iint_R dA = \boxed{8\pi}$