• Double Integrals in Polar Coordinates


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    • Abstract: To aid in the use of (1), let us notice that if p is constant, then r = p is a circle ... p2: In polar coordinates, R is the region between r = 0 and r = p2 ...

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Polar Coordinates polar coordinate polar coordinates change of variable formula Double Integrals Double Integrals in Polar Coordinates
Part 1: The Area Di¤erential in Polar Coordinates
We can also apply the change of variable formula to the polar coordinate trans-
formation
x = r cos ( ) ; y = r sin ( )
However, due to the importance of polar coordinates, we derive its change of
variable formula more rigorously.
To begin with, the Jacobian determinant is
@ (x; y) cos ( ) sin ( )
= = r cos2 ( ) + r sin2 ( ) = r
@ (r; ) r sin ( ) r cos ( )
As a result, the area di¤erential for polar coordinates is
@ (x; y)
dA = drd = rdrd
@ (r; )
Let us consider now the polar region S de…ned by
= ; = ;r = g( );r = f ( )
where f ( ) and g ( ) are contained in [p; q] for all in [ ; ] : If 0 ; : : : ; m is
an h-…ne partition of [ ; ] and r0 ; : : : ; rn is an h-…ne partition of [p; q] ; then
the image of [ ; ] [p; q] is a partition of the image of the region with near
parallelograms whose areas are denoted by Ajk :
Since the area di¤erential is dA = rdrd , the area of the “near parallelogram”
is approximately
Ajk rj rj j
so that if xjk = rj cos ( k ) and yjk = rj sin ( k ) ; then
n m
XX n m
XX
lim (xjk ; yjk ) Ajk = lim (rj cos ( k ) ; rj sin ( k )) rj rj j
h!0 h!0
j=1 k=1 j=1 k=1
1
Writing each of these limits as double integrals results in the formula for
change of variable in polar coordinates:
ZZ Z Z f( )
(x; y) dA = (r cos ( ) ; r sin ( )) rdrd (1)
R g( )
To aid in the use of (1), let us notice that if p is constant, then r = p is a circle
of radius p centered at the origin in the xy-plane, while if is constant, then
= is a ray at angle beginning at the origin of the xy-plane.
Moreover, the origin corresponds to r = 0:
EXAMPLE 1 Use (1) to evaluate
Z Z p
1 2 x2
x2 + y 2 dydx
0 0
Solution: To do so, we transform the iterated integral into a double
integral
Z Z p ZZ
1 2 x2
x2 + y 2 dydx = x2 + y 2 dA
0 0 R
p
where R is a sector of a circle with radius 2: In polar coordinates,
p
R is the region between r = 0 and r = 2 for in [ =4; =2]:
Since r2 = x2 + y 2 ; the double integral thus becomes
ZZ Z Z p Z Z p
=2 2 =2 2
2 2 2
x +y dA = r rdrd = r3 drd
R =4 0 =4 0
2
and the resulting iterated integral is then easily evaluated:
p
ZZ Z =2 2 Z =2
2 2 r4
x +y dA = d = d =
R =4 4 0 =4 4
Check your Reading: What does y = x correspond to in polar coordinates?
Areas and Volumes in Polar Coordinates
If R is a region in the xy-plane bounded by = ; = ; r = g( ); r = f ( );
then (1) implies that
ZZ Z Z f( )
Area of R = dA = rdrd
R g( )
thus allowing us to …nd areas in polar coordinates.
p
EXAMPLE 2 Find the area of the region between x = 1, x = 2;
y = 0, and p
y= 2 x2
Solution: Since x = 1 corresponds to r cos ( ) = 1 or r = sec (p ;)
the region is between the line r = sec ( ) and a circle of radius 2
from = 0 to = =4:
Thus, the area of the region is
ZZ Z Z p
=4 2
Area = dA = rdrd
R 0 sec( )
3
and evaluation of the iterated integral leads to
p
Z =4 2
r2
Area = d
0 2 sec( )
Z =4
1
= 2 sec2 ( ) d
2 0
1 =4
= (2 tan ( ))j0
2
1 1
=
4 2
Moreover, we can use polar coordinates to …nd areas of regions enclosed by
graphs of polar functions.
EXAMPLE 3 What is the area of the region enclosed by the car-
dioid r = 1 + cos ( ) ; in [0; 2 ] :
Solution: Since the cardioid contains the origin, the lower boundary
is r = 0: Thus, its area is
Z 2 Z 1+cos( ) Z 2 1+cos( )
r2
Area = rdrd = d
0 0 0 2 0
Substituting and expanding leads to
Z
1 2
Area = 1 + 2 cos ( ) + cos2 ( ) d
2 0
Z
1 2 1 1
= 1 + 2 cos ( ) + + cos (2 ) d
2 0 2 2
2
1 3 1
= + 2 sin ( ) + sin (2 )
2 2 4 0
3
=
2
4
Polar coordinates can also be used to compute volumes. For example, the
equation of a sphere of radius R centered at the origin is
x2 + y 2 + z 2 = R2
Solving for z then yields shows us that the sphere can be considered the solid
between the graphs of the two functions
p p
g (x; y) = R2 x2 y 2 ; f (x; y) = R2 x2 y 2
over the circle x2 + y 2 = R2 in the xy-plane.
Since circle x2 + y 2 = R2 de…nes the type I region
p
x = R y = p R2 x2
x=R y = R2 x2
the volume of the sphere of radius R is given by the iterated integral
Z R Z pR2 x2 p
V = p 2 R2 x2 y 2 dydx (2)
R R2 x2
EXAMPLE 4 Use polar coordinates to evaluate
Z R Z pR2 x2 p
V = p 2 R2 x2 y 2 dydx
R R2 x2
Solution: To begin with, we rewrite the iterated integral as a double
integral over the interior of the circle of radius R centered at the
origin, which is often denoted by D:
Z Z p
V = 2 R2 (x2 + y 2 ) dA
D
In polar coordinates, the disc D of radius R is bounded by the curves
= 0; = 2 ; r = 0; r = R; so that
Z Z p
V = 2 R2 x2 y 2 dA
D
Z 2 Z R p
= 2 R2 r2 rdrd
0 0
5
Thus, if we let u = R2 r2 ; then du = 2rdr; u (0) = R2 ; u (R) = 0;
so that
Z 2 Z 0
V = u1=2 du d
0 R2
Z 2 0
u3=2
= d
0 3=2 R2
Z 2 2 3=2
2 R
= d
0 3
4 3
= R
3
Check your Reading: What is the volume of the unit sphere?
Independent Normal Distributions
In statistics, a normally distributed random variable with mean and standard
deviation has a Gaussian density, which is function of the form
1 (x )2 =(2 2
)
p1 (x) = p e (3)
2
It follows that the joint density for two independent, normally distributed events
is a function of two variables of the form
1 (x 1)
2
=(2 2
)e (x 1)
2
=(2 2
)
p (x; y) = p1 (x) p2 (y) = e 1 1
2 1 2
For simplicity, we will consider here only independent, normally distributed
events with mean = 0 in both and standard deviations = 1 = 2 : In such
cases, the joint density function is
1 2 2 2
p (x; y) = 2
e (x +y )=(2 )
2
EXAMPLE 5 Let (X; Y ) be the coordinates of the …nal resting
place of a ball which is released from a position on the z-axis toward
the xy-plane, and suppose the two coordinates are independently
normally distributed with a mean of 0 and a standard deviation of
6
3 feet.
What is the probability that the ball’ …nal resting place will be no
s
more than 5 feet from the origin?
Solution: Since = 3; the joint density function is
1 2 2
p (x; y) = e (x +y )=18
18
and we want to know the probability that (X; Y ) will be in a circle
with radius 5 centered at the origin. Since such a circle corresponds
to r = 0 to r = 5 for in [0; 2 ] ; the probability is
ZZ
1 2 2
P X 2 + Y 2 25 = e (x +y )=18 dA
R 18
Converting to polar coordinates then yields
Z 2 Z 5
1 r 2 =18
P X 2 + Y 2 25 = e rdrd
18 0 0
and if we now let u = r2 ; du = 2rdr; then u (0) = 0 and u (5) = 25
implies that
Z 2 Z 25
1
P X 2 + Y 2 25 = e u=18 dud
36 0 0
Z 2
1
= 18 18e 25=18 d
36 0
= 1 e 25=18
= 0:750648
7
Thus, there is about a 75% chance that the ball’ …nal resting place
s
will be no more than 5 feet from the origin.
Check your Reading: How exactly do we interpret P X 2 + Y 2 25 ?
An Important Result in Statistics
Finally, the value of the integral
Z 1
x2
I= e dx
0
is very important in statistical applications. To evaluate it, we …rst notice that
Z 1 Z 1 Z 1Z 1
2 2 2 2
I2 = e x dx e y dy = e x e y dydx
0 0 0 0
That is, I 2 is a type I iterated integral which can be converted to polar coordi-
nates.
EXAMPLE 6 Evaluate the integral
Z 1Z 1
2
y2
I2 = e x e dydx
0 0
Solution: To do so, let us notice that
Z Z
2 2
2
I = e (x +y ) dA
Quad I
However, in polar coordinates, the …rst quadrant is given by r = 0
to r = 1 for = 0 to = =2: Thus,
Z =2 Z 1
r2
I2 = e rdrd
0 0
As a result, we can write
Z " Z #
=2 R
2 r2
I = lim e rdr d
0 R!1 0
8
Thus, if we let u = r2 ; du = 2rdr; u (0) = 0; u (R) = R2 ; then
Z =2 " Z R2 #
2 1 u
I = lim e du d
2 0 R!1 0
Z =2 h i
1 2
= lim e0 e R d
2 0 R!1
Z =2
1
= d
2 0
=
4
p
Thus, I = =2; which implies both
Z 1 p Z 1
x2 x2
p
e dx = and e dx =
0 2 1
Exercises:
Evaluate the following iterated integrals by transforming to polar coordinates.
R 1 R p1 x2 p R 1 R p1 x2 y
1
1. 0 0
x2 + y 2 dydx 2. 0 0
tan x dydx
R 1 R p1 x2 x
R 1 R p1 x2 y
3. 0 0
p dydx 4. 0 0
p dydx
x2 +y 2 x2 +y 2
R 1 R p1 y 2 y dxdy R 1 R p1 y 2 x dxdy
5. 0 0 x2 +y 2 6. 0 0 x2 +y 2
R 2 R p4 2 p R 2 R p4 x2 x2 y 2
7. p x 9 x2 y 2 dydx 8.
0 4 x2 2 0 x2 +y 2 dydx
R 1 R p2 x2 x
R 1 R p2 x2 y
9. 0 x
p dydx 10. 0 x
p dydx
x2 +y 2 x2 +y 2
R1Rx x
R1Rx x
11. 0 0 x2 +y 2
dydx 12. 0 0
p dydx
x2 +y 2
R 1 R p2 y2 x
R 1 R p2 y2 y
13. 0 1 x2 +y 2 dxdy 14. 0 1 x2 +y 2 dxdy
R 1 R p1 x2 dydx
R1R1 dxdy
15. 0 1 x [x2 +y 2 ]3=2
16. 0 1 y [x2 +y 2 ]3=2
Each of the following polar curves encloses a region that contains the origin.
9
Find the area of the region the curve encloses.
17. r = 5; in [0; 2 ] 18. r = 3; in [ ; ]
19. r = sin ( ) ; in [0; ] 20. r = 4 cos ( ) ; in [0; ]
2
21. r = ; in [0; ] 22. r = j j + 1; in [ ; ]
23. r = sin (3 ) ; in [0; 2 =3] 24. r = 4 cos (3 ) ; in [0; 2 =3]
25. r = sin (5 ) ; in [0; 2 =5] 26. r = sin2 ( ) ; in [0; ]
27. r = 1 + cos (2 ) ; in [0; ] 28. r = 1 + sin (3 ) ; in [0; 2 =3]
29. r = sin ( ) + cos ( ) ; in [0; ] 30. r = 3 sin ( ) + 4 cos ( ) ; in [0; ]
31. Use polar coordinates to …nd the volume of a right circular cone with
height h and a circular base with radius R
(hint: the equation of the cone is
hp 2
z= x + y2
R
32. A right circular cone with a base of radius R is sliced by a plane of the
form
x+R
z = h1 + (h2 h1 )
2R
where h1 and h2 are positive. What is the shape of the solid between this plane
and the xy-plane, and what is its volume?
33. Recall that if 0 < " < 1 and p > 0; then
p
r=
1 " cos ( )
is an ellipse which encloses a region R. Evaluate
ZZ
dA
3=2
R [x2 + y 2 ]
34. Evaluate the double integral
ZZ
ydA
R
where R is the polar ellipse described in exercise 33.
35. In example 4, what is the probability that
10
1. (a) The …nal position of the ball is in the 1st quadrant and is no more
than 5 feet from the origin.
(b) The …nal position of the ball is between 3 and 7 feet from the origin.
(c) The …nal position of the ball is in the xy-plane.
36. After several throws at a dart board, a dart thrower …nds that both
the X and Y coordinates of his darts have a mean of 0 and a standard deviation
of 3 inches. What is the probability that a randomly selected dart throw from
all those he has thrown will be in the "bulls eye", if the bulls eye is a circle of
radius one inch centered at the origin?
37. Suppose an airplane has two rocket engines whose time of ignition with
respect to a “time zero” is normally distributed with a standard deviation of
= 0:01 seconds. If the rockets’ignitions are independent events, what is the
probability that the sum of the squares of the …ring times is less than 0.01?
38. Inp exercise 37, what is the probability that the left engine will …re no
more than 3 times later than the right engine?
39. The antennae lengths of a sample of 32 woodlice were measured and
found to have a mean of 4 mm and standard deviation of 2.37 mm. Assuming
the antennae lengths are normally distributed, what is the probability of one of
the antennae of a woodlice being twice as long as the other? (Hint: substitute
to translate the means to 0).
40. Acme sheet metal produces several hundred rectangular sheets of metal
each day. If errors in the lengths and widths of the rectangular sheets are
independent random variables with mean of 0 and a standard deviation of s
= 0.1 inches, then what is the probability that the error in the area of the
rectangular sheets exceeds 0.1 inches?
41. Use the method in the discussion preceding example 6 to evaluate
Z 1
2
J= x2 e x dx
0
42. Find the area and the centroid of a cardioid of the form
r = 1 + cos ( )
43. Write to Learn: A freezer produces ice cubes with normally distrib-
uted temperatures with a mean of 0 F and a standard deviation of 2 F: Write
a short essay in which you compute and explain the probability that two ice
cubes chosen at random will have temperatures that di¤er by no more than
3 F; assuming the temperatures are independent.
44. Try it out! Drop a ball several times (i.e., 20-30 times) from a position
directly above an “origin” in an xy-plane you create. (Hint: to avoid any bias,
you might want to secure the ball with a thread and then release the ball by
cutting the thread). Suppose that
(x1 ; y1 ) ; (x2 ; y2 ) ; : : : ; (xn ; yn )
11
denotes the …nal stopping points of the ball. The sample means of both the x’s
and the y’ should be practically zero. The sample standard deviation for the
s
x’ is
s sP
n
j=1 (xj x)
x =
n
and the sample standard deviation y for the y’ is similar. Show that x
s y
and then repeat example 5 using the sample standard deviations as the value
for :
12


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