Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 5.4 Double Integrals in Polar Coordinates (12.4)
When we change variables in singe integral we also change the differential, e.g. dt = g' ( x ) dx in b
∫
f ( g ( x )) g' ( x ) dx
a
=
g( b )
∫ f (t ) dt t=g( x )
dt=g'( x )dx g( a )
When we change variables in double integral we often call it a change of coordinates. In this section we concentrate on polar coordinates, but the process can be generalized. Recall: The differential of a function f(x,y) is df = fx ( x, y ) dx + fy ( x, y ) dy






One understands the recent equations in sense of vectors, i.e. df = fx ( x, y ) dx + fy ( x, y ) dy where


 
dx,dy are a (non-­‐standard) basis vectors. Compare to the standard vectors i,j.





⎧ x = r cosθ ⎧⎪dx = cosθ dr − r sin θ dθ = cosθ ,− r sin θ
⇒ ⎨ 
We now define the differential ⎨
. 



⎩ y = r sin θ
dy
=
sin
θ
dr
+
r
cos
θ
d
θ
=
sin
θ
,
r
cos
θ
⎪⎩
b c
Since dxdy is associated with the area of the domain, i.e. ∫∫ f ( x, y ) dA = ∫ ∫ f ( x, y ) dy dx we look D
a c
at formula of the area in standard and polar bases: 
 

A = dx × dy = cosθ ,− r sin θ ,0 × sin θ , r cosθ ,0 = 0,0,r cos 2 θ + r sin 2 θ = r

 

A = dx × dy = i× j = k = 1
dr Finally we define dxdy = rdrdθ . Another way to get dxdy = rdrdθ : Find the area of polar “rectangle” [ ri−1 ,ri ] × ⎡⎣θ j−1 ,θ j ⎤⎦ , using difference between areas of sections thus with angle Δθ = θ j − θ j−1 , thus Δθ 2 Δθ 2 Δθ 2 2
Δθ
ri + ri−1
ΔAi =
ri −
ri−1 =
( ri − ri−1 ) = 2 ( ri + ri−1 )( ri − ri−1 ) = 2 ( ri − ri−1 ) Δθ = ri ΔrΔθ 2
2
2
Thm: If f is continuous on a polar “rectangle” R given by 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β where 0 ≤ β − α ≤ 2π , then β b
∫∫ f ( x, y ) dA = ∫ ∫ f ( r cosθ ,r sinθ ) r dr dθ α a
R
∫∫
{x<0,3≤x +y ≤5}
2
Ex 5.
=
3π /2
∫
2
3x + 2y dA =
3π /2 5
3π /2
π /2 3
π /2
∫ ∫ ( 3r cosθ + 2sinθ ) r dr dθ = ∫
5
r 3 cosθ + r 2 sin θ 3 dθ =
25 ( 5 cosθ + sin θ ) − 9 ( 3cosθ + sin θ ) dθ =
π /2
= ( 25 ( 5sin θ − cosθ ) − 9 ( 3sin θ − cosθ ) dθ )π /2 = −2 (125 − 27 ) = −2 ⋅ 98 = −196
3π /2
65 Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky Thm: If f is continuous of polar region of the form D = {( r,θ ) | α ≤ θ ≤ β ,h1 (θ ) ≤ r ≤ h2 (θ )} then β h2 (θ )
∫∫ f ( x, y ) dA = ∫ ∫ f ( r cosθ ,r sinθ ) r dr dθ α h1(θ )
D
Ex 6.
2
0
∫
∫
e
x=−2 y=− 4−y 2
dx dy = ∫∫ e
1−x 2 −y 2
R
3π 2
=−
Ex 7.
3π 2
dA =
2
∫ ∫
re1−r dr dθ =
2
θ =π 2 r=0
3π 2
2 2
1
1
π
e1−r dθ = −
e−3 − edθ = (e − e−3 )
∫
∫
0
2 θ =π 2
2 θ =π 2
4
∫∫ 3
D
π sin θ
x + y − 2 dA = ∫
2
2
0
π
∫ ( 3r − 2 ) r dr dθ = ∫ r
0
0
π
=−
1−x 2 −y 2
π
3
−r
-2 2 sinθ
0
π
dθ = ∫ sin 3 θ − sin 2 θ dθ =
0
1
1
2
π 4 π
⎛1
⎞
2 + sin 2 θ ) cosθ − ⎜ θ − sin 2θ ⎟ = − ( −1− 1) − = −
(
⎝
⎠
3
2
4
3
2 3 2
0
0
66