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:

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s = dx 2 + dy 2A

B

∫ = dxdθ( )

2+ dy

dθ( )2

θ A

θ B

∫ dθ .

This time, however, we have specific relationships between the Cartesian coordinates ( x, y ) and the polar coordinates ( r, θ ) :

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x = r(θ) cos θ ⇒ dxdθ = dr

dθ cosθ − r sin θ ,

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y = r(θ) sin θ ⇒ dydθ = dr

dθ sin θ + r cosθ , making the radical in the integrand

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dxdθ( )

2+ dy

dθ( )2

= drdθ cos θ − r sinθ( )

2+ dr

dθ sin θ + r cos θ( )2

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=( drdθ )

2 cos2θ − 2r ( drdθ ) sin θ cosθ + r2 sin2θ[ ]+ ( drdθ )

2 sin2θ + 2r ( drdθ ) sin θ cos θ + r2 cos2θ[ ]

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= ( drdθ )2 ⋅ cos2θ + sin2θ[ ] + r2 ⋅ sin2θ + cos2θ[ ] = ( drdθ )

2 + [r(θ)]2 .

We will therefore work with the arclength integral for polar curves,

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s = ( drdθ )2 + [r(θ)]2

θ A

θ B

∫ dθ .

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:

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drdθ = 2θ ⇒

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s = (2θ)2 + (θ 2)2 dθ0

2π∫ = 4θ 2 + θ 4 dθ

0

2π∫ = θ 4 + θ 2 dθ

0

2π∫ .

This integral can be solved using the substitution u = θ2 + 4 ⇒ du = 2θ dθ :

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→ u4

4π 2 + 4

∫ ⋅ ( 12 du) = 12 ⋅ (

23 u

3 / 2) 44π 2 + 4

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= 13 ⋅ [(4π

2 + 4)3 / 2 − 43/ 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

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drdθ = 2 cosθ , so the arclength integral is

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s = (2 cos θ)2 + (2 sinθ)2 dθ0

π / 3

∫ = 2 (cosθ)2 + (sinθ)2 dθ0

π / 3

∫

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= 2 dθ0

π / 3

∫ = 2θ 0π / 3 = 2π

3 .

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

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s = 2θ 02π = 4π . 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

circumference is correctly found from

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s = 2θ 0π = 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

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A = y dxa

b∫ . Instead

we must measure from the origin and advance the integration in angle θ .

If we use an infinitesimal wedge dθ in the direction θ

0 , 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( θ

0 ) and a base of ds = r( θ

0 ) · dθ , which makes its area dA = ½ h · b

= ½ r( θ0 ) · r( θ

0 ) · dθ . The integral needed for finding areas of polar curves is then

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A = 12

θ1

θ 2

∫ [ 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

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A = 12 [2 sinθ]2 dθ

0

π

∫ = 12 ⋅ 4 sin2θ dθ

0

π

∫ = / 2 1/ 2

(1− cos 2θ ) dθ0

π

∫

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= (θ − 12 sin 2θ ) 0π = (π − 12 sin 2π ) − (0 −

12 sin 0) = π ,

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:

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A = 6 12 (2 cos 3θ )

2

0

π / 6

∫ dθ = 3 4 cos2 3θ0

π / 6

∫ dθ = 12 12 (1+ cos 6θ )

0

π / 6

∫ dθ

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= 6 ⋅ (θ + 16 sin 6θ ) 0

π / 6 = 6 ⋅ [( π6 + 16 sin 6 ⋅

π6 ) − (0 + 1

6 sin 0)]

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= 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:

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r = 1+ 2 sinθ = 0 ⇒ sin θ = − 12 ⇒ θ = 7π6 ,

11π6 ,

tells us that we need to carry out the area integration over this range:

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A = 12

7π / 6

11π / 6

∫ (1+ 2 sin θ)2 dθ = 127π / 6

11π / 6

∫ (1+ 4 sin θ + 4 sin2θ) dθ

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= 127π / 6

11π / 6

∫ (1+ 4 sin θ + 4[ 12{1− cos 2θ}]) dθ = 127π / 6

11π / 6

∫ ( 3+ 4 sin θ − 2 cos 2θ ) dθ

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= ( 32θ − 2 cosθ −12 sin 2θ ) 7π / 6

11π / 6

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= ( 32 ⋅11π6 − 2 cos 11π6 − 12 sin

11π3 ) − (

32 ⋅7π6 − 2 cos 7π6 − 12 sin

7π3 )

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= ( 11π4 − / 2 ⋅ [ 3

/ 2 ] − 12 ⋅ [−

32 ] − 7π

4 + / 2 ⋅ [− 3/ 2 ] + 1

2 ⋅ [3

2 ])

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= / 4 π/ 4 + 3 ⋅ ( −1+ 1

4 −1+ 14 ) = π − 3

2 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” :

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A = f (x) − g(x) dxa

b∫ .

For polar curves, we must instead subtract the area of the “inner wedge” from that of the “outer wedge”, giving us

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A = 12

θ1

θ 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 one another ( f(x) = g(x) ) and solve the equation for the values of x . We can do the 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

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1+ sinθ = 3 sin θ ⇒ 2 sin θ = 1 ⇒ sin θ = 12 ⇒ θ = π

6 ,5π6 .

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 --

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3 sinθ = 0 ⇒ θ = 0, π ; cardioid --

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1+ sinθ = 0 ⇒ sin θ = −1 ⇒ θ = 3π2 .

Why are these values different? Remember that the cardioid is only swept out once as θ runs from zero to 2π , while the circle is traced twice in this same interval. It can 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

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A = 12 (3 sin θ )

2 − 12 (1+ sin θ )2π / 6

5π / 6

∫ dθ or

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= 2 12 (3 sinθ )

2 − 12 (1+ sin θ )2π / 6

π / 2

∫ dθ .

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

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A = 2 ⋅ 12 (1+ sin θ )2

−π / 2

π / 6

∫ dθ − 12 (3 sinθ )

2

0

π / 6

∫ dθ

.

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

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L = 2 + 2 sin θ dθ0

2π∫ . The integral is evaluated by using the

“conjugate factor” method, in which we multiply the numerator and denominator to obtain

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L =2 + 2 sin θ ⋅ 2 − 2 sin θ

2 − 2 sinθdθ

0

2π∫ =

4 − 4 sin2θ2 − 2 sin θ

dθ0

2π∫

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= 21− sin2θ2 − 2 sin θ

dθ0

2π∫ = 2 cos2θ

2 − 2 sin θdθ

0

2π∫ .

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

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2 cos2θ2 − 2 sin θ

dθ0

2π∫ = 2 cos θ

2 − 2 sin θdθ

0

2π∫ ,

and solve the integral using the simple substitution u = 2 − 2 sin θ , our result will be zero! 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

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2 cos2θ2 − 2 sin θ

dθ0

2π∫ = 2

cos θ2 − 2 sin θ

dθ0

2π∫ .

Perhaps not enough of the point is made in algebra classes, but it can be critical to keep

in mind that

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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 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

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L = 2 ⋅ 2cosθ2 − 2 sinθ

dθ−π / 2

π / 2

∫

= 4 cos θ

2 − 2 sin θdθ

−π / 2

π / 2

∫ .

cos θ ≥ 0 in this range

Now, when we make the substitution described above, we find

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4 cos θ2 − 2 sin θ

dθ−π / 2

π / 2

∫ u = 2 − 2 sin θ ⇒ du = −2 cos θ

θ : −π/2 π/2 u = 2 − 2 sin θ : 4 0

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→(−2 du)

u4

0∫ = 2 du

u0

4∫ = 2 ⋅ (2u1/ 2) 0

4 = 2 ⋅ 2 ⋅ 41/ 2 = 8 , thereby confirming the arclength given in the textbook.

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

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cos 2θ = sin 2θ ⇒ tan 2θ = 1

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⇒ 2θ = π4 ,

5π4 , 9π4 ,

13π4 ⇒ θ = π

8 ,5π8 , 9π8 ,

13π8 ,

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

where the radius is negative. For cos 2θ , this occurs for

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( π4 ,3π4 ) and (

5π4 ,

7π4 ) ,

while the intervals of “negative radius” are

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( π2 , π ) and (3π2 , 2π ) for sin 2θ . A

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

marked is reached at

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θ = 3π8 on r = sin 2θ , but at

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θ = 11π8 on r = cos 2θ .

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

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A = 8 ⋅ 2 12

0

π / 8

∫ (sin 2θ )2 dθ = 8 sin2 2θ dθ0

π / 8

∫ ,

from which we can go on to compute the total area of

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π2 −1 .

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