sin cos// cos sin

dr rdy dy d ddrdx dx d rd

The derivative of a polar curve is...

Integration in polar coordinates involves finding not the area underneath a curve but, rather, the area of a sector bounded by a curve. Consider the region bounded by the curve r = f (θ) and the two rays and , with . To derive a formula for the area, divide the region into N narrow sectors of angle

/ N , : 0 1 2 N

corresponding to a partition of the interval

Recall that a circular sector of angle Δθ and radius r has21 .

2K r

If Δθ is small, the jth narrow sector is nearly a circular sector of radius rj = f (θj), so its area is 2 .1

2 jK r

The total area is approximated by the sum:

2

1

2

1212

1N

jj

N

jj

K fr

This is a Riemann sum for the integral

22

1 1

1 12 2

N N

j jj j

K r f

212

.f d

If f (θ) is continuous, then the sum approaches the integral as

and we obtain the following formula.,N

THEOREM 1 Area in Polar Coordinates If f (θ) is a continuous function, then the area bounded by a curve in polar form r = f (θ) and the rays θ = and with is equal to

221 12 2r d f d

We know that r = R defines a circle of radius R. By THM 1, the area is equal to

THEOREM 1 Area in Polar Coordinates If f (θ) is a continuous function, then the area bounded by a curve in polar form r = f (θ) and the rays θ = and with is equal to

221 12 2r d f d

2

2 2 2

0

1 1 2 .2 2

R d R R

The equation r = 4 sin θ defines a circle of radius 2 tangent to the x-axis at the origin. The right semicircle is “swept out” as θ varies from 0 to

Use Theorem 1 to compute the area of the right semicircle with equation r = 4 sin θ.

/ 2.By THM 1, the area of the right semicircle is

221 1 .2 2r d f d

2 1sin 1 cos 2

2

/2 /22 2

0 0

/2/2

00 0

0

1 4sin 8 sin2

4 1 cos 2

2

4 2 cos

2 2 sin 2 2 0 0

d d

d udu

u

2 2u du d

Sketch r = sin 3θ and compute the area of one “petal.” Rectangular r f

r varies from 0 to 1 and back to 0 as θ varies from 0 to .

3

Polar Curve

r varies from 0 to -1 and back to 0 as θ varies from 2 to .

3 3

r varies from 0 to 1 and back to 0 as θ varies from 2 to .

3

212

K r d

/3 /32

0 0

/3 /3

0 0

22

00

1 1 1 cos6sin 32 2 2

1 1 cos 64 4

1 1cos sin12 24 12 24 12

d d

d d

udu u

2 1 cos 6sin 32

6 6u du d

/3 /3 /32 2 2

/3 /3 /3

2 /3 /32 /3 /3

2 /3 /32 /3 /3

1 1 12cos 1 4cos 1 2cos 2 12 2 2

1 1 1 1 2 3 2cos 32 3

sin2 2 2 2 2 3

d d d

udu d u

The area between two polar curves r = f1(θ) and r = f2(θ) with f2(θ) ≥ f1(θ), for , is equal to

2 22 1

12

f f d

Area Between Two Curves Find the area of the region inside the circle r = 2 cos θ but outside the circle r = 1.The two circles intersect at the points where (r, 2 cos θ) = (r, 1) orin other words, when 2 cos θ = 1.

,3 3

Region (I) is the difference of regions (II) and (III).

Area betweentwo polar curves

2 1cos 1 cos 22

2 2u du d

We close this section by deriving a formula for arc length in polar coordinates. Observe that a polar curve r = f (θ) has a parametrization with θ as a parameter:

by the product rule

' sin ' cosdxx f fd

' cos ' sindyy f fd

cos cos , sin sinx r f y r f

2 2Arc Length ' 'b

a

s x t y t dt 2 2 2 2Algebraically ' ' 'x y f f

2 2 Arc Length 's f f d

2 2

2 2 2 2

0 0

'

4 cos 4 sin 2 2 a

s f f d

a a d a d

Find the total length of the circle r = 2a cos θ for a > 0.

f (θ) = 2a cos θ ' 2 sinf a

Note that the upper limit of integration is π rather than 2π because the entire circle is traced out as θ varies from 0 to π. Rectangular Coordinates

Since x = r cos θ and y = r sin θ, we use the chain rule.

sin cos// cos sin

dr rdy dy d ddrdx dx d rd

To find the slope of a polar curve r = f (θ), remember that the curve is in the x-y plane, and so the slope is .dy

dx

sin cos// cos sin

dr rdy dy d ddrdx dx d rd

Find an equation of the line tangent to the polar curve r = sin 2θ when

4.3

3sin 1 the point will be in the 4th quadrant.2

r

2 22 02 22cos 2 sin cos 1

2cos 2 cos sin 2 22 02 2

rrr

r

2 212 2

y x

2cos2

2sin2

x r

y r

slope