Double integrals in polar coordinates?

Question

find Int(Int) x dA over region D, where D is the region in the first quadrant that lies between the circles x^2 + y^2 = 4 and x^2 + y^2 = 2x

Answer 1

∬x dA

The first thing we need to do is to find the bounds of integration. The first circle is centered at the origin and has radius 2, and the second circle is centered at (1, 0) and has radius 1, since:

x² + y² = 2x
x² - 2x + 1 + y² = 1
(x-1)² + y² = 1

Thus the second circle is contained completely inside the first, and is tangent to it at (2, 0). So the region in the first quadrant that is between the two circles will be the part above the top half of the second circle and below the top half of the first circle (we take the top halves only since we are only interested in the region in the first quadrant). This will be evaluated from x=0 to x=2 (again, we do not consider x<0 since we are only interested in the portion in the first quadrant). So the top half of the first circle is:

x² + y² = 4
y² = 4-x²
y = √(4-x²) (the positive root is taken, since we only want the top half of the circle)

And the top half of the second circle is:

(x-1)² + y² = 1
y² = 1-(x-1)²
y = √(1-(x-1)²)

So now the integral, with limits, is:

[0, 2]∫[√(1-(x-1)²), √(4-x²)]∫x dy dx

Now we just integrate directly:

[0, 2]∫xy |[√(1-(x-1)²), √(4-x²)] dx
[0, 2]∫x√(4-x²) - x√(1-(x-1)²) dx
[0, 2]∫x√(4-x²) dx - [0, 2]∫x√(1-(x-1)²) dx

For the first integral, let u=4-x², du=-2x dx, x=0 ⇒ u=4, x=2 ⇒ u=0. Then we have:

-1/2 [4, 0]∫√u du - [0, 2]∫x√(1-(x-1)²) dx
-1/3 u^(3/2) | [4, 0] - [0, 2]∫x√(1-(x-1)²) dx
8/3 - [0, 2]∫x√(1-(x-1)²) dx

Now, break apart the second integral thus:

8/3 - [0, 2]∫(x-1)√(1-(x-1)²) dx - [0, 2]∫√(1-(x-1)²) dx

For the second integral, let u=1-(x-1)², du=-2(x-1), x=0⇒u=0, x=2⇒u=0, so we have:

8/3 + 1/2 [0, 0]∫√u du - [0, 2]∫√(1-(x-1)²) dx
8/3 - [0, 2]∫√(1-(x-1)²) dx

That the previous integral would cancel makes sense, since (x-1)√(1-(x-1)²) is antisymmetric about the line x=1. Now, for the final integral, make the substitution θ = arcsin (x-1), so x-1 = sin θ. Then dx = cos θ dθ, x=0 ⇒ θ=-π/2, x=2⇒θ=π/2. So we have:

8/3 - [-π/2, π/2]∫√(1-sin² θ) cos θ dθ
8/3 - [-π/2, π/2]∫√(cos² θ) cos θ dθ
8/3 - [-π/2, π/2]∫cos² θ dθ

The inference √(cos² θ) = cos θ follows because cos θ is nonnegative on [-π/2, π/2]. Now, we use the identity cos² θ - = (cos (2θ) + 1)/2 to obtain:

8/3 - 1/2 [-π/2, π/2]∫cos (2θ) + 1 dθ
8/3 - (cos (2θ)/4 + θ/2) | [-π/2, π/2]
8/3 - ((cos (π)/4 + π/4) - (cos (-π)/4 - π/4))
8/3 - π/2

And we are done.

Answer 2

Polars : x=rcos(?), y=rsin(?), dA=rdrd? : ? limits are [0,½?] for ? and [0.5] for r ?? = ? [r=0.5] ? [?=0,½?] rsin(?).exp{rcos(?)}.rdrd? ?? = ? [r=0.5] [?=0,½?] r²(?a million/r).exp{rcos(?}.d?dr = ?? [r=0.5] r{a million?exp(r) }.dr factors ? supplies ? ? r{a million?exp(r) }.dr = r{r?exp(r)} ? ? {r?exp(r)}.dr = ½r²+(a million?r)exp(r) ? ?? = ?½(25?0) ? {?4.exp(5)?a million} = 4exp(5)?23/2