Integrals in Polar Coordinates Π R Θ= 6

Polar coordinates

We describe points using the distance r from the origin and the angle θ anticlockwise from the x-axis.

(x,y)=(r cos(θ),r sin(θ)) π θ= 3 Integrals in polar coordinates π r θ= 6

Polar coordinates are related to ordinary (cartesian) coordinates by the formulae

x = r cos(θ) y = r sin(θ) p r = x 2 + y 2 θ = arctan(y/x).

(Care is needed to choose the right value of arctan(y/x).) In the diagram on the right above, we have divided a disk into small pieces using lines of constant θ and circles of constant r.To use this kind of subdivision for integration, we need to know the area of the small pieces.

The polar area element Disk integral of x2, again

RR 2 Consider a piece of angular width δθ, where the radius runs from r to r + δr. Consider again D x dA, where D is a disk of radius a around the origin. Here the appropriate limits are just 0 ≤ θ ≤ 2π and 0 ≤ r ≤ a. The integral is ZZ Z 2π Z a x 2 dA = r 2 cos2(θ) r dr dθ δθ δA r δθ D θ=0 r=0 Z 2π Z a = cos2(θ) r 3 dr dθ r θ=0 r=0 δr a4 Z 2π a4 Z 2π 1 + cos(2θ) = cos2(θ) dθ = dθ 4 4 2 Provided that δθ is small this will be approximately rectangular. If we measure θ=0 θ=0 2π angles in radians (as we always will) then the length of the curved side will be a4 1 1 πa4 = θ + sin(2θ) = r δθ, and the straight side has length δr, so the area is approximately 4 2 4 0 4 δA = r δr δθ. In the limit this becomes dA = r dr dθ, so we have the following as before. prescription: if D is a region that is conveniently described in polar coordinates, The following picture shows why R 2π cos2(θ) dθ = π: then 0 ZZ Z ··· Z ··· f (x, y) dA = f (r cos(θ), r sin(θ))r dr dθ, D θ=··· r=··· where the limits need to be ﬁlled in in accordance with the geometry of the region. π π 3π 2π 2 2

Each region has the same area, namely π/4. Integral over an inﬁnite wedge Integral over a semicircular disk

ZZ Z π/2 Z 10 I = x 2y 2 dA = (r 2 cos2(θ))(r 2 sin2(θ))r dr dθ r=1 θ=π/30 D θ=−π/2 r=0 10 D Z π/2 Z 10 = sin2(θ) cos2(θ)r 5 dr dθ θ=−π/2 r=0 θ=−π/30 If limits are just constants, then

Z b Z d Z b Z d f (u) g(v) dv du = f (u) du g(v) dv . u=a v=c u=a v=c ZZ 1 ZZ 1 Z π/30 Z ∞ 1 2 2 2 dA 2 2 2 dA = 4 r dr dθ D (x + y ) D (x + y ) θ=−π/30 r=1 r Z π/2 Z π/2 Z π/2 1 − cos(4θ) Z ∞ −2 ∞ 2 2 1 2 π −3 π r sin (θ) cos (θ) dθ = 4 sin (2θ) dθ = dθ = 2 r dr = θ=−π/2 θ=−π/2 θ=−π/2 8 30 r=1 15 −2 r=1 θ sin(4θ) π/2 π π 1 π = − = = 0 − = 8 4 × 8 8 15 −2 30 θ=−π/2 Z 10 r 6 10 1000000 r 5 dr = = r=0 6 r=0 6 π 1000000 (r = (x 2 + y 2)1/2 dA = r dr dθ) I = × ' 65449.84695. 8 6

A slotted rotor Area of a curved region

Moment of inertia I = RR (x 2 + y 2)dA. The picture shows the region D given in polar coor- D dinates by 0 ≤ r ≤ 2 + sin(2θ). We would like to ﬁnd the area of D, or in other words A = RR 1 dA. b D Here dA = r dr dθ as usual, and the relevant limits r=2+sin(2θ) a are 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 2 + sin(θ) θ

We ﬁrst use a simplifying trick. Let D0 be the region in the middle picture, and 0 RR 2 2 0 2π 2+sin(2θ) 2π 2 2+sin(2θ) put I = D0 (x + y )dA. As D is just obtained by turning D slightly, the Z Z Z r 0 0 A = r dr dθ = dθ moment of inertia will be the same, so I = I . On the other hand, 2I = I + I 2 is just the integral over the simpler region D00 shown on the right. We thus θ=0 r=0 θ=0 r=0 Z 2π Z 2π 1 RR 2 2 00 1 2 1 2 have I = 2 D00 (x + y )dA. For D the limits are just 0 ≤ θ ≤ 2π and = (2 + sin(2θ)) dθ = 4 + 4 sin(2θ) + sin (2θ) dθ a ≤ r ≤ b. The integrand is 2 θ=0 2 θ=0 2 2 2 2 2 1 Z 2π x + y = (r cos(θ)) + (r sin(θ)) = r , = 4 + 4 sin(2θ) + 1 − 1 cos(4θ) dθ. 2 2 2 and the area element is dA = r dr dθ. We thus have θ=0 Z 2π Z b Z 2π 4 4 The integral of sin(kθ) or cos(kθ) over a whole number of complete cycles is 1 3 1 b − a 1 I = 2 r dr dθ = 2 dθ zero. Thus, only the terms 4 and contribute to the integral, and we have θ=0 r=a θ=0 4 2 4 4 1 1 1 b − a 4 4 A = (2π.(4 + )) = 9π/2. = 2π = π(b − a )/4. 2 2 2 4 The Gaussian integral

It is an important fact (for the theory of the normal distribution in statistics, the analysis of heat ﬂow, the pricing of ﬁnancial derivatives, and other R ∞ −x2 √ applications) that −∞ e dx = π. We will explain one way to calculate R ∞ −x2 this. Put I = −∞ e dx. It obviously does not matter what we call the More general change of variables R ∞ −y 2 variable, so we also have I = −∞ e dy. We can now multiply these two expressions together to get ∞ ∞ Z Z 2 2 ZZ 2 2 I 2 = e−x −y dx dy = e−x −y dA. y=−∞ x=−∞ whole plane We can rewrite this using polar coordinates, noting that x 2 + y 2 = r 2 and dA = r dr dθ. We get

∞ 2π ∞ Z Z 2 Z 2 I 2 = r e−r dθ dr = 2π r e−r dr. r=0 θ=0 r=0 We now substitute u = r 2, so u also runs from 0 to ∞ and du = 2r dr. The integral becomes Z ∞ ∞ 2 −u 1 −u I = 2π e . 2 du = π − e = π((−0) − (−1)) = π u=0 u=0 √ so I = π as claimed.

Area of a parallelogram Change of variables in a double integral

a c Suppose x and y can be expressed in terms of some other variables u and v. The area of the parallelogram P is |ad − bc| = det . b d The Jacobian matrix: ∂x ∂x h a+c i ∂(x, y) ∂u ∂v xu xv b+d J = = ∂y ∂y = . c c ∂(u, v) ∂u ∂v yu yv [ d ] [ d ] For small changes δu and δv to u and v, resulting changes in x and y are P a d [ b ] δx ' xu δu + xv δv δy ' yu δu + yv δv. a [ b ] 0 u These equations can be combined as a single matrix equation: [ 0 ] [ 0 ] δx xu xv δu ∂(x, y) δu Indeed, P consists of the top triangle (shown in yellow) together with the = = . δy yu yv δv ∂(u, v) δv middle region (shown in green). The top triangle has the same area as the bottom one, so we may as well consider the parallelogram P0 consisting of the Now let the change in u vary between 0 and δu, and let the change in v vary bottom triangle together with the middle region. This parallelogram has a base x u between 0 and δv. The resulting changes in [ y ] then cover a small of length u and a perpendicular height of d, so the area is ud. Note that [ 0 ] is a c xu xv reached from [ b ] by moving in the opposite direction to the vector [ d ], so parallelogram spanned by δu and δv, and the area of this u a c yu yv [ 0 ] = [ b ] − t [ d ] for some t. By comparing the y-coordinates we see that parallelogram is |x y − x y |δu δv, or in other words det ∂(x,y) δu δv. t = b/d, and by looking at the x-coordinates we deduce that u = a − bc/d, so u v v u ∂(u,v) c a the area is ud = ad − bc. This works when [ d ] is anticlockwise from [ b ]. If ∂(x, y) c a Using this: dA = det du dv. [ d ] is clockwise from [ b ] it works out instead that ad − bc < 0 and the area is ∂(u, v) −(ad − bc). In all cases: the area is |ad − bc|. This is a key ingredient for double integrals by substitution. Area of an ellipse

We will ﬁnd the area of an ellipse E with equation x 2/a2 + y 2/b2 ≤ 1 (for some a, b > 0). For this it b is best to use a kind of distorted polar coordinates:

x = ar cos(θ) y = br sin(θ).

−a a Then x 2/a2 + y 2/b2 = r 2 cos2(θ) + r 2 sin2(θ) = r 2, so x 2/a2 + y 2/b2 ≤ 1 becomes 0 ≤ r ≤ 1. Partial derivatives: −b

xr = a cos(θ) xθ = −ar sin(θ) yr = b sin(θ) yθ = br cos(θ),

∂(x, y) a cos(θ) −ar sin(θ) J = = . ∂(r, θ) b sin(θ) br cos(θ) This means that the absolute value of the determinant is |det(J)| = |abr cos2(θ) − (−abr sin2(θ))| = |abr| = abr, so dA = abr dr dθ. We therefore have ZZ Z 2π Z 1 Z 2π r 2 1 Z 2π 1 area = 1 dA = abr dr dθ = ab dθ = ab dθ = πab. E θ=0 r=0 θ=0 2 r=0 θ=0 2