Download March 13, 2009 Lecture

Math 1272 -- Lecture for 13 March 2009
We started looking at problems involving calculus for polar curves when we
worked out the slope of the tangent line to a point on such a curve. Now we will see
how to compute arclength along a polar curve and the area enclosed by one, along with
the places where we must be cautious in such calculations.
For arclength, we return the parametric version of our generic integral:
θB
B
2
∫
s =
dx + dy
A
2
=
∫
θA
2
( ) +( )
dx
dθ
dy
dθ
2
dθ
.
This time, however, we have specific relationships between the Cartesian coordinates
( x, y ) and the polar coordinates ( r, θ ) :
€ x = r(θ ) cos θ ⇒
y = r(θ ) sin θ ⇒
€
2
( ) ( dθ )
dx
dθ
+ dy
dy
= ddrθ sin θ + r cos θ
dθ
,
=
2
(
=
dr
dθ
cos θ − r sin θ
2
) (
+ ddrθ sin θ + r cos θ
)
2
[( ddrθ ) cos θ − 2r ( ddrθ ) sin θ cos θ + r sin θ ]
+ [( ddrθ ) sin θ + 2r ( ddrθ ) sin θ cos θ + r cos θ ]
2
€
= ddrθ cos θ − r sin θ ,
making the radical in the integrand
€
€
dx
dθ
2
2
2
2
2
2
2
( ddrθ ) 2 ⋅ [cos 2 θ + sin 2 θ ] + r 2 ⋅ [sin 2 θ + cos 2 θ ] =
=
We will therefore work with the arclength integral for polar curves,
θB
€
s =
∫
θA
€
( ddrθ ) 2 + [r(θ )]2 dθ .
( ddrθ ) 2 + [r(θ )]2
.
Section 10.4, Problem 47: Find the arclength of the polar curve r =
θ2 for 0 ≤ θ ≤ 2π .
We are given the range of integration for this curve, so we only need to
determine the derivative of the curve function in order to proceed:
dr
dθ
s =
€
∫
2π
0
= 2θ ⇒
(2 θ ) 2 + (θ 2 ) 2 dθ =
0
4 θ 2 + θ 4 dθ =
This integral can be solved using the substitution u =
4π 2 + 4
€
∫
2π
→
∫
4
u ⋅ ( 12 du) =
1
⋅ ( 23 u 3 / 2 )
2
∫
2π
0
θ 4 + θ 2 dθ .
θ2 + 4 ⇒ du = 2θ dθ :
4π 2 + 4
4
= 13 ⋅ [(4 π 2 + 4) 3 / 2 − 4 3 / 2 ] = 83 ⋅ [(π 2 +1) 3 / 2 − 1] ≈ 93 .
€
€
Problem 45 (modified): Find the arclength of the polar curve r = 2 sin
θ for 0 ≤ θ ≤ π/3 .
Here, the derivative is
dr
dθ
= 2 cos θ , so the arclength integral is
π /3
s =
∫
π /3
(2 cos θ ) 2 + (2 sin θ ) 2 dθ =
€
0
€
∫ 2 dθ
(cos θ ) 2 + (sin θ ) 2 dθ
0
π /3
=
∫2
= 2θ
π /3
0
0
= 23π
.
We could go on to find the circumference of this circle, which we might
reasonably expect that we could calculate by going the full angle of 2π radians around
the circle. When we do this, we find
s = 2θ
2π
0
= 4 π . But something’s wrong with
this€
number: we know that this circle has a diameter of 2 , so its circumference should
be C = πd = 2π !
This brings up a matter
€ with polar curves to be wary of: it is important to know
how the angle parameter θ behaves for the curve we are working with. In this problem,
the circle is traced once completely as θ runs from zero to π , so if we overlook this
and integrate the arclength from zero to 2π , we will have followed the circle twice, and
so it is not surprisingly that we get double the circumference. We do note that the
π
circumference is correctly found from s = 2 θ 0 = 2π = C . [Another problem that
can come up is discussed in Addendum 1.]
Because the origin is a€special point in the polar coordinate system, finding the
area within a curve is not found in the same fashion that we calculate it using rectangular
coordinates. The reference is not the x-axis, so we cannot use
∫
b
y dx . Instead
we must measure from the origin and advance the integration in angle θ .
A =
a
€
If we use an infinitesimal wedge dθ in the direction
θ0 , the radius of the curve,
r( θ0 ) , hardly changes and the bit of arclength is found from the angle-arclength
relation for a circle, s = R · θ . Because it is so short, this infinitesimal arclength looks
very nearly straight. So the wedge is quite close to being a narrow triangle with a height
of r( θ0 ) and a base of ds = r( θ0 ) · dθ , which makes its area dA = ½ h · b
= ½ r(
θ0 ) · r( θ0 ) · dθ . The integral needed for finding areas of polar curves is then
θ2
A =
1
2
∫
θ1
[ r (θ )]2 dθ .
The same caution applies for finding areas of polar curves as for finding their
arclengths: we must be sure that we understand how the angle parameter behaves
along the curve. So if we calculate the area of the circle we have just discussed, we
should use the integral
€
A =
∫
π
0
1
[2 sin θ ]2
2
dθ = 12 ⋅ 4
= ( θ − 12 sin 2θ )
π
0
∫
π
0
π
sin 2 θ dθ = 2/ ∫ 0
1
(1 − cos 2θ )
2/
= ( π − 12 sin 2π ) − (0 − 12 sin 0) = π
dθ
,
which is the result we expect for a circle of diameter 2 ( A = ¼ π · d2 = ¼ π · 22 = π ). Had
we integrated the area from zero to 2π , we again would have traced over the circle twice
and gotten twice the correct area.
€
€
To avoid the pitfall of retracing a curve, we can sometimes take advantage of the
symmetry of the curve to evaluate arclength or area over a portion of the complete
figure and then multiply the result by the appropriate factor.
Problem 13: Find the area enclosed by the curve r = 2 cos 3θ .
We have a three-petaled rosette, which is a curve which is traced once completely as θ
runs from zero to π . But we can sidestep this issue by noticing that this curve has a
three-fold symmetry and that each petal is bilaterally symmetric. We can then start our
area integration at θ = 0 and continue to the first angle at which the radius of the curve
becomes zero, which occurs at θ = π/6 . Since this section (marked in orange) makes
up one-sixth of the full area, we can multiply our result by 6 to obtain the desired value:
π /6
A = 6
∫
0
€
1
(2 cos 3θ ) 2
2
π /6
dθ = 3
∫
0
π /6
4 cos2 3θ dθ = 12
∫
0
1
(1+ cos 6 θ ) dθ
2
= 6 ⋅ ( θ + 16 sin 6 θ )
π /6
0
= 6 ⋅ [( π6 + 16 sin 6 ⋅ π6 ) − (0 + 16 sin 0)]
= 6 ⋅ ( π6 + 0 − 0 − 0) = π .
€
€
As with other area problems we have done before, we may need to work out the
limits of integration, when they are not already given to us.
Problem 21: Find the area of the inner loop of the limaçon r = 1 + 2 sin
θ.
A plot of the curve function versus θ shows us that there is a section of the curve
which has “negative radius”. Inspecting the graph, or solving for the angles where the
radius falls to zero:
r = 1 + 2 sin θ = 0 ⇒ sin θ = − 12 ⇒ θ = 76π , 116π
,
tells us that we need to carry out the area integration over this range:
€ A =
11π / 6
∫
7π / 6
=
€
1
2
1
1
2
(1+
2
sin
θ
)
d
θ
=
2
2
11π / 6
∫
7π / 6
(1+ 4 sin θ +
€
€
€
€
∫
(1+ 4 sin θ + 4 sin 2 θ ) dθ
7π / 6
4[ 12 {1 − cos
= ( 23 θ − 2 cos θ − 12 sin 2θ )
€
11π / 6
2θ }]) dθ =
1
2
11π / 6
∫
( 3 + 4 sin θ − 2 cos 2θ ) dθ
7π / 6
11π / 6
7π / 6
= ( 23 ⋅ 116π − 2 cos 116π − 12 sin 113π ) − ( 23 ⋅ 76π − 2 cos 76π − 12 sin 73π )
= ( 114π − 2/ ⋅ [ 2/3 ] − 12 ⋅ [− 23 ] − 74π + 2/ ⋅ [− 2/3 ] + 12 ⋅ [ 23 ]) /
= 44/π + 3 ⋅ ( −1 + 14 −1 + 14 ) = π − 23 3 .
The special nature of the origin also affects the way in which we calculate the
area between two polar curves. In rectangular coordinates, we can simply subtract the
area under the “lower curve” from that of the “upper curve” :
A =
∫
b
a
f (x) − g(x) dx .
€
For polar curves, we must instead subtract the area of the “inner wedge” from that of
the “outer wedge”, giving us
θ2
A =
∫
θ1
1
[
2
f (θ )]2 − 12 [ g(θ )]2 dθ .
We can proceed once we know the limits of integration. Often this will require
us to find the intersection points between the two curves. When we did this with
functions in Cartesian coordinates, it was sufficient to set the two functions equal to
€( f(x) = g(x) ) and solve the equation for the values of x . We can do the
one another
same thing for two polar curves, but that doesn’t necessarily cover everything.
Problem 37: Find all points of intersection of the curves r = 1 + sin
θ and r = 3 sin θ .
By setting the equations for the two curves equal, we find
1 + sin θ = 3 sin θ ⇒ 2 sin θ = 1 ⇒ sin θ = 12 ⇒ θ = π6 , 56π
€
.
The circle and the cardioid certainly intersect at these two angles. However, we
can see in the figure above that they also meet at the origin. It is not possible to
discover these angles by the preceding method, since we are asking to find when both
functions are zero, which is equivalent to asking when 0 = 0 . We must take the
additional step of finding the angles at which each curve function equals zero
individually:
circle --
3 sin θ = 0 ⇒ θ = 0, π
cardioid --
;
1 + sin θ = 0 ⇒ sin θ = −1 ⇒ θ = 32π
.
Why are these values different? Remember that the cardioid is only swept out once as
€ from zero to 2π , while the circle is traced twice in this same interval. It can
θ runs
also be the case with two curves that the angles at which r = 0 occur at differing points
€
in their cycles.
So this becomes another consideration when we intend to integrate the area
between two curves. If we were asked to find the area within the circle but outside of
the cardioid (see the first of the pair of figures above), we could simply integrate
between the two angles of intersection we found; we can also exploit the symmetry of
the two curves about the y-axis and write
5π / 6
A =
∫
π /6
1
1
2
2
(
3
sin
θ
)
−
(1
+
sin
θ
)
dθ
2
2
π /2
= 2
∫
π /6
€
or
1
( 3 sin θ ) 2 − 12 (1 + sin θ ) 2 dθ 2
.
However, if we instead needed to find the area within the cardioid but outside the circle
(second figure in the pair above), we would have to take into account the fact that the
origin is reached at different angles for the two curves. So the area of the cardioid must
€
be integrated
over a range of angles different from that for the circle; again using the
symmetry of the curves, we have
 π /6 1
A = 2 ⋅  ∫ 2 (1 + sin θ ) 2 dθ −
 −π / 2
π /6
∫
0

1
2
( 3 sin θ ) dθ 2

.
€
end lecture for 13 March
Addendum 1 -I received an e-mail from a student a few days after this lecture, concerning
Example 4 in Section 10.4 (pp. 652-53). We are asked to find the arclength of the
cardioid r = 1 + sin θ , shown in the figure below.
It is straightforward to set up the arclength integral, as is done in the book, to arrive at
the integral
L =
∫
2π
0
2 + 2 sin θ dθ . The integral is evaluated by using the
“conjugate factor” method, in which we multiply the numerator and denominator to
obtain
€
L =
∫
2 + 2 sin θ ⋅ 2 − 2 sin θ
dθ =
2 − 2 sin θ
2π
0
1− sin 2 θ
2π
= 2 ∫0
€
€
2 − 2 sin θ
2π
dθ = 2 ∫ 0
∫
2π
0
4 − 4 sin 2 θ
dθ
2 − 2 sin θ
cos2 θ
dθ .
2 − 2 sin θ
Up to this point, everything is fine. In setting up the integrand, we began with an
expression under the radical which is the sum of two squares, so it is guaranteed to
provide a positive value for the arclength. We have observed proper use of algebra and
the Pythagorean Identity, so our latest version of the integral is correct.
However, if we now take the typical step of writing
2π
2 ∫0
2π
cos2 θ
dθ = 2 ∫ 0
2 − 2 sin θ
cos θ
dθ 2 − 2 sin θ
,
and solve the integral using the simple substitution u = 2 − 2 sin
zero!
€
θ , our result will be
There is a pitfall awaiting careless use of the square root, but it is
understandable that this could happen because the choice of integration limits obscures
the source of the problem. The step we took should properly be written as
2π
2 ∫0
2π
cos2 θ
dθ = 2 ∫ 0
2 − 2 sin θ
cos θ
2 − 2 sin θ
dθ .
Perhaps not enough of the point is made in algebra classes, but it can be critical to keep
x 2 = x , and not merely x . If we use the symmetry of the cardioid
(refer
€ to the figure above) and start the integration on the y-axis at θ = −π/2 , we notice
that the value of cos θ on the right-hand side of the y-axis has the opposite sign of the
in mind that
value it has on the left side. This means that the cardioid consists of two halves which
will €
contribute exactly cancelling terms in the integral, giving us a false zero value for
the arclength.
It would be better to exploit the symmetry of the figure to integrate over, say,
only the right-hand half of the cardioid. We would then write our arclength integral as
 π /2
L = 2 ⋅ 2 ∫
 −π / 2
π /2

dθ = 4 ∫
2 − 2 sin θ

−π / 2
cos θ
cos
cos θ
dθ
2 − 2 sin θ
.
θ ≥ 0 in this range
Now, when we make the substitution described above, we find
€
π /2
4
cos θ
dθ
2 − 2 sin θ
∫
−π / 2
u = 2 − 2 sin
θ ⇒ du = −2 cos θ
θ:
u = 2 − 2 sin
€
→
∫
0
4
4 du
(−2 du)
= 2 ∫0
= 2 ⋅ (2 u1/ 2 )
u
u
thereby confirming the arclength given in the textbook.
€
θ:
4
0
−π/2
4
π/2
0
= 2 ⋅ 2 ⋅ 41/ 2 = 8 ,
Addendum 2 -Here’s another little example of how the determination of intersection points
between two polar curves can get complicated.
Problem 31: Find the area inside both the curves r = cos 2θ and r = sin 2θ .
If we set out to find the intersection points between these curves for the purpose
of finding the limits of integration for an area integral, we might start by setting
cos 2θ = sin 2θ ⇒ tan 2θ = 1
⇒ 2θ = π4 , 54π , 94π , 134π
€
€
⇒
θ = π8 , 58π , 98π , 138π
,
following the standard procedure for solving such a trigonometric equation. We can see
from the graph, however, that this gets us only half of the eight intersection points at
the tips of the “leaves” for the area of interest (we won’t even concern ourselves here
with the values that θ has for each curve at r = 0 ). What about the other four?
This returns us to the issue of the behavior of the angle parameter for curves
such as rosettes. In each of our two curves, two of the petals represent intervals in θ
π 3π
) and ( 54π , 74π ) ,
where the radius is negative. For cos 2θ , this occurs for ( ,
4 4
π
3π
, 2π ) for sin 2θ . A
while the intervals of “negative radius” are ( , π ) and (
2
2
consequence of this is that the remaining four intersection points represent different
values of θ for each curve, and thus cannot be found by solving an equation of the two
€
curve functions. An instance of this is marked on the graph above, where the point
3π
11π
marked is reached at θ =
on €
r = sin 2θ , but at θ =
on r = cos 2θ .
8
8
There is really no easily-described method for dealing with a situation such as
this; it is probably safest to investigate the curves and the behavior of the angle
parameter by€plotting them in such cases. €
For the purpose of solving the original
problem, we can evade this complication by noting the eight-fold symmetry of the
pattern for the area and the bilateral symmetry of the “leaves” in it. We can then simply
use the first intersection point that we found (giving us the sector marked in gold),
allowing us to write the integral for the area of the complete figure as
π /8
A = 8⋅2
∫
0
1
2
(sin
2
θ
)
dθ = 8
2
π /8
∫ sin
2
2θ dθ
,
0
from which we can go on to compute the total area of
π
−1 .
2
€
€
Credit: In the interests of saving some time in preparing these notes, most of the
figures have been taken and adapted from the textbook (Stewart, 6th ed.) and the
Complete Solutions Manual for that text.
-- G. Ruffa
19 - 21 March 2009