Double Integrals in Polar Coordinates - MATH 311,...

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Double Integrals in Polar CoordinatesMATH 311, Calculus III

J. Robert Buchanan

Department of Mathematics

Spring 2019

Area Elements

For the Riemann sum, development of the double integral useda rectangular area element with area

∆A = (∆x)(∆y)

and in differential form we write this as

dA = dx dy = dy dx .

Whenever a function f or region R exhibits a circular symmetryit may be more convenient to evaluate a double integral in polarcoordinates.

Question: what is the area element in polar coordinates?

Polar Area Element (1 of 2)

θ=θ1

θ=θ2

r=r2

r=r1

Δθ

x

y

Polar Area Element (1 of 2)

∆A =12

(∆θ)r22 −

12

(∆θ)r21

=12

(r22 − r2

1 )(∆θ)

=12

(r2 + r1)(r2 − r1)(∆θ)

=12

(r2 + r1)(∆r)(∆θ)

= r(∆r)(∆θ)

where r is the average radius of the area element.

In differential form then dA = r dr dθ.

Polar Area Element (1 of 2)

∆A =12

(∆θ)r22 −

12

(∆θ)r21

=12

(r22 − r2

1 )(∆θ)

=12

(r2 + r1)(r2 − r1)(∆θ)

=12

(r2 + r1)(∆r)(∆θ)

= r(∆r)(∆θ)

where r is the average radius of the area element.

In differential form then dA = r dr dθ.

Fubini’s Theorem

Theorem (Fubini’s Theorem)Suppose that f (r , θ) is continuous on the regionR = {(r , θ) |α ≤ θ ≤ β, g1(θ) ≤ r ≤ g2(θ)}, where g1(θ) ≤ g2(θ)for all θ in [α, β]. Then∫∫

Rf (r , θ) dA =

∫ β

α

∫ g2(θ)

g1(θ)f (r , θ)r dr dθ.

ExampleFind the area enclosed by one leaf of the 4-leafed roser = cos(2θ).

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

y

SolutionThe leaf pointing east is formed by the curve r = cos(2θ)between two angles for which r = 0.

0 = cos(2θ) =⇒ 2θ = ±π2⇐⇒ θ = ±π

4Thus the area of the leaf is

A =

∫∫R

1 dA

=

∫ π/4

−π/4

∫ cos(2θ)

0r dr dθ

=

∫ π/4

−π/4

12

cos2(2θ) dθ

=

∫ π/4

−π/4

14

(1 + cos(4θ)) dθ

=θ

4+

116

sin(4θ)

∣∣∣∣π/4

−π/4=π

8

Changing a Cartesian Double Integral to Polar

Find the volume of the solid region below the paraboloidz = 1− x2 − y2 and above the z = 0 plane.

Converting from Cartesian to Polar

V =

∫∫R

(1− x2 − y2) dA

=

∫ 1

−1

∫ √1−x2

−√

1−x2(1− x2 − y2) dy dx

=

∫ 2π

0

∫ 1

0(1− r2)r dr dθ

= 2π∫ 1

0(r − r3) dr

= 2π(

12

r2 − 14

r4)∣∣∣∣1

0

=π

2

Finding Volume Using Polar Coordinates (1 of 3)

Find the volume of the solid that lies under the paraboloidz = x2 + y2, above the z = 0 plane, and inside the cylinderx2 + y2 = 2x .

0.5 1.0 1.5 2.0x

-1.0

-0.5

0.5

1.0

y

Finding Volume Using Polar Coordinates (2 of 3)

If z = x2 + y2, then z = r2. The boundary of the cylinder isgiven in polar coordinates by

2x = x2 + y2

2r cos θ = r2

2 cos θ = r ,

for −π/2 ≤ θ ≤ π/2.

Finding Volume Using Polar Coordinates (3 of 3)

Thus the desired volume is

V =

∫∫R

z dA =

∫ π/2

−π/2

∫ 2 cos θ

0r2r dr dθ

=

∫ π/2

−π/2

14

(2 cos θ)4 dθ

=

∫ π/2

−π/2(1 + cos(2θ))2 dθ

= π +

∫ π/2

−π/2cos2(2θ) dθ

= π +12

∫ π/2

−π/2(1 + cos 4θ) dθ

=3π2

Homework

I Read Section 13.3.I Exercises: 1–45 odd