Chapter 17

Section 17.4The Double Integral as the Limit of a

Riemann Sum; Polar Coordinates

Polar CoordinatesAny point in the plane can be obtained by knowing r, the distance the point is from the origin and the angle the point makes with the positive x-axis. The values are called the polar coordinates of the point.

(𝑥 , 𝑦 )

y

x

𝑟

Conversions:

Polar to Rectangular:

Rectangular to Polar: 𝜃

Polar coordinates are used to describe curves that have complicated rectangular equations and make them simpler (at least from a calculus perspective (i.e. integrating)).Circles centered at the origin.

4 3 2 1 1 2 3 4

4 3 2 1

12

34

Cardioid. 3 leafed rose. Spiral.

1 2 3 4

2

1

1

2

3 2 1 1 2 3

4 3 2 1

1

2

2 1 1 2 3

2

1

1

2

Double Integrals and Polar CoordinatesA double integral in rectangular coordinates can be switched to a double integral in polar coordinates by replacing all x’s on the integrand by and all y’s in the integrand by . We need to multiply by the integrating factor .

∬Ω 𝑑𝑒𝑠𝑐𝑟𝑖𝑏𝑒𝑑

𝑖𝑛𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟𝑐𝑜𝑜𝑟𝑖𝑛𝑎𝑡𝑒𝑠

❑

𝑓 (𝑥 , 𝑦 )𝑑𝐴= ∬Ω 𝑑𝑒𝑠𝑐𝑟𝑖𝑏𝑒𝑑𝑖𝑛 𝑝𝑜𝑙𝑎𝑟

𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠

❑

𝑓 (𝑟 cos𝜃 ,𝑟 sin 𝜃 )𝑟 𝑑𝑟𝑑 𝜃Integrating factor (Do not forget!)

Polar Descriptions of RegionsThe region is no longer thought of as top and bottom or left and right curves. It must be thought of as outer and inner curve between a starting angle and an ending angle . The angles and for a (which is the most common) must continuously bound between the outer and inner curves.

∬Ω𝑑𝑒𝑠𝑐𝑟𝑖𝑏𝑒𝑑𝑖𝑛 𝑝𝑜𝑙𝑎𝑟

𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠

❑

h (𝑟 ,𝜃 )𝑑𝑟 𝑑𝜃= ∫𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔𝑎𝑛𝑔𝑙𝑒

𝑒𝑛𝑑𝑖𝑛𝑔𝑎𝑛𝑔𝑙𝑒

∫𝑖𝑛𝑛𝑒𝑟𝑐𝑢𝑟𝑣𝑒

𝑜𝑢𝑡𝑒𝑟𝑐𝑢𝑟𝑣𝑒

h (𝑟 ,𝜃 )𝑑𝑟 𝑑𝜃=∫𝜃 1

𝜃 2

∫𝜌1 (𝜃 )

𝜌2 (𝜃 )

h (𝑟 , 𝜃 )𝑑𝑟 𝑑𝜃

𝑟=𝜌2 (𝜃 )

x

y

𝜃1

𝜃2

𝑟=𝜌1 (𝜃 )Ω

ExampleFind the integral to the right where is the region in the first quadrant inside the circle , outside the circle and above the line .

0.5 1 .0 1 .5 2 .0

0 .5

1 .0

1 .5

2 .0

∬Ω

❑

8 𝑥2𝑑𝐴

First we graph and get the curves in polar form to determine the outside and inside curves.

𝑟=2

𝑟=1

𝜋4

𝜋2

Ω∬Ω

❑

8 𝑥2𝑑𝐴= ∬Ω 𝑖𝑛𝑝𝑜𝑙𝑎𝑟

❑

8𝑟 2cos2𝜃𝑟 𝑑𝑟 𝑑𝜃

¿∫𝜋4

𝜋2

∫1

2

8𝑟3 cos2𝜃𝑑𝑟 𝑑𝜃

¿∫𝜋4

𝜋2

2𝑟4 cos2𝜃|12𝑑𝜃=∫

𝜋4

𝜋2

32cos2𝜃−2cos2𝜃𝑑𝜃=∫𝜋4

𝜋2

30 cos2𝜃 𝑑𝜃

¿∫𝜋4

𝜋2

30 ( 1+cos (2𝜃 )2 )𝑑𝜃=∫

𝜋4

𝜋2

15+15cos (2𝜃 )𝑑𝜃=15𝜃+152sin (2𝜃 )|𝜋

4

𝜋2

¿ 15𝜋2−( 15𝜋4 + 15

2 )=15𝜋4 − 152

=152 ( 𝜋2 −1)

ExampleFind the integral to the right for the region inside the circle and outside the cardioid .

∬Ω

❑ 𝑥𝑥2+𝑦 2

𝑑𝐴

First graph the region and find the angles where the curves intersect.

4 3 2 1 1 2

2

1

1

2

and

You might be tempted to use the angles and , but this would get the wrong part of the region. Need to use the angles and . and

𝑟=2cos𝜃

𝑟=2−2cos𝜃𝜋3

−𝜋3

∬Ω

❑ 𝑥𝑥2+𝑦 2

𝑑𝐴=∫−𝜋3

𝜋3

∫2−2c os𝜃

2cos𝜃 𝑟 cos 𝜃𝑟2

𝑟 𝑑𝑟 𝑑𝜃=∫−𝜋3

𝜋3

∫2−2c os 𝜃

2cos𝜃

cos𝜃 𝑑𝑟 𝑑𝜃

¿ ∫−𝜋3

𝜋3

𝑟 cos𝜃|2−2cos𝜃2cos𝜃 𝑑𝜃=∫

−𝜋3

𝜋3

2cos2𝜃−2cos𝜃+2cos2𝜃𝑑𝜃=∫−𝜋3

𝜋3

4cos2𝜃−2cos𝜃 𝑑𝜃

¿ ∫−𝜋3

𝜋3

4 (1+cos (2 𝜃 )2 )−2cos𝜃 𝑑𝜃=∫

−𝜋3

𝜋3

2+2cos (2𝜃 )−2cos𝜃 𝑑𝜃=2𝜃+sin (2𝜃 )−2sin 𝜃|−𝜋3

𝜋3

¿ 2𝜋3

+ √32−√3−(−2𝜋3 − √3

2+√3)=4𝜋3 +√3−2√3=4𝜋3 −√3

ExampleSometimes you might need to convert a rectangular integral to polar like the integral to the right.

∫−1

1

∫0

√1−𝑦2

√𝑥2+𝑦 2𝑑𝑥𝑑𝑦

First graph the region.

0 .2 0 .4 0 .6 0 .8 1 .0

1 .0

0 .5

0 .5

1 .0

𝑟=1

𝜋2

−𝜋2

∫−1

1

∫0

√1−𝑦2

√𝑥2+𝑦 2𝑑𝑥𝑑𝑦=∫−𝜋2

𝜋2

∫0

1

𝑟 𝑟 𝑑𝑟 𝑑𝜃

¿ ∫−𝜋2

𝜋2

∫0

1

𝑟 2𝑑𝑟 𝑑𝜃

¿ ∫−𝜋2

𝜋213𝑟 3|

0

1

𝑑𝜃

¿ ∫−𝜋2

𝜋213𝑑𝜃

¿ 13𝜃|−𝜋

2

𝜋2

¿𝜋6 −

− 𝜋6 =

𝜋3