Integration in polar coordinates involves finding not the area underneath a curve but, rather, the...

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Transcript of Integration in polar coordinates involves finding not the area underneath a curve but, rather, the...
sin cos/
/ cos sin
drrdy dy d d
drdx 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
.2
K 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
12
1
2
1N
jj
N
jj
K fr
This is a Riemann sum for the integral
22
1 1
1 1
2 2
N N
j jj j
K r f
21
2.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 1
2 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 1
2 2r d f d
2
2 2 2
0
1 12 .
2 2R d R R
The equation r = 4 sin θ defines a circle of radius 2 tangent to the xaxis 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 1
sin 1 cos 22
/2 /22 2
0 0
/2/2
00 0
0
14sin 8 sin
2
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
21
2K r d
/3 /32
0 0
/3 /3
0 0
22
00
1 1 1 cos 6sin 3
2 2 2
1 1cos 6
4 4
1 1cos sin
12 24 12 24 12
d d
d d
udu u
2 1 cos6sin 3
2
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 1
2 2 2
1 1 1 1 2 3 2cos
3
2 3sin
2 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 2
2 1
1
2f 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 between
two polar curves
2 1cos 1 cos 2
2
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 ' cosdx
x f fd
' cos ' sindy
y 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 2
Algebraically ' ' '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
drrdy dy d d
drdx dx d rd
To find the slope of a polar curve r = f (θ), remember that the curve is in the xy plane, and so the slope is .
dy
dx
sin cos/
/ cos sin
drrdy dy d d
drdx 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.
2r
2 22 0
2 22cos 2 sin cos1
2cos 2 cos sin 2 22 0
2 2
rr
rr
2 21
2 2y x
2cos
2
2sin
2
x r
y r
slope