Calculus-Double integrals in polar coordinates (r,phi)

7/30/2019 Calculus-Double integrals in polar coordinates (r,phi)

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- r

P

y

x

I.e., this is the same point as given by (r, +). We note that in some books

r is not allowed to be negative. This can actually make a difference in a problem.To conclude this brief review, we plot a couple of graphs.

Example 1. Sketch r = .

Answer. We mean the curve in the (x, y) plane, since in the (r, ) plane r = is a

straight line! We start at = 0 and increase . When = 0 we have r = 0 and weare at the origin. When = /2, r = /2 and as increases so does r. This means

that the curve spirals away from the origin.

( , ) (2 , 2)

( / 2, /2)

y

x

We must not forget what happens for < 0. We also get a spiral as follows:96


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x

y

(2 , 2)

( / 2, / 2)

( , )

So the final picture is two spirals:

x

y

( , )

Example 2. Sketch r = cos .

Answer. We mean the curve in the (x, y) plane not in the (r, ) plane. Again we

start at = 0 and note

= 0 r = 1

r

=

2r = 0.

Note that r means here we are moving towards the origin. So we obtain Fig a.97


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0=

2= r = 0 , r = 1,

x

y

Fig. a

We now keep on increasing :

=

2, r = 0

, r

= , r = 1,

and obtain Fig. b.

= , r = - 1

y

x

Fig. b

Observe that here r means we are moving away from the origin and the point isin the fourth quadrant. By the properties of the cosine (i.e., periodicity), as keeps

on increasing, we shall retrace these curves over and over again, and the same will98


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happen for negative. So the final total picture is

x

y

.

We can see that this is a circle as follows:

r = cos = r2 = r cos ,

i.e., a point on r = cos must satisfy (in (x, y) coordinates)

x2 + y2 = x

or

x 1

2

2+ y2 =

1

4.

I.e., lie on a circle centered at ( 12 , 0) of radius12 .

We shall practice sketching more graphs with the examples that follow.

Let us now suppose we have a pie shaped region R as shown.99


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r

R

Since the area of a circle is r2, and there are 2 radians all the way around, then

Area of R = (r2) 2

=r2

2.

Consider now the next region as shown, with r, dr, d positive.

r

R

We find

dA =(d)(r + dr)2

2 d

2r2

=d

2

2rdr + (dr)2

= d

r dr +

(dr)2

2

.

This formula is exact, but in cases of interest to us we shall let dr 0. We canneglect the term (dr)2/2 in comparison with dr and thus shall state:

dA = r dr d .

An easy way to remember this is to note100


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r

d

dA

dr

r d

and treat dA as for a rectangle, dA = rddr. Let us return to our problem.

Consider a region D, bounded by r = h(), r = g(), = 1, = 2 as shown.

1

2

r = h ( )

r = g ( )

x

y

D

Our problem is to find

D

f(x, y) dA.

We now decompose the region into circular caps of area dA = rdrd. These

replace dxdy. We then have

f(x, y) dA = f(r cos , r sin ) rdrd.

Now assemble the caps into spikes (these replace the strips).101


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1

2

r = h ( )

r = g ( )

x

y

D

The spike between and + d contributes:

r=f() (spike ends)r=g() (spike starts)

f(r cos , r sin ) rdrd.

Finally, assembling all the spikes gives:

D

f(x, y) dA =

=2=1

r=h()r=g()

f(r cos , r sin ) rdrd.

We shall illustrate this below. Before, note that if the boundary of D is given by

= h(r), = g(r), r = r1, r = r2, then we assemble the caps the other way aroundand get

D

f(x, y) dA =

r=r2r=r1

=h(r)=g(r)

f(r cos , r sin )rdrd.

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= h ( r )

= g ( r )

r = r 1

r = r 2

x

y

One more remark: We found dA = rdrd under the assumption that r, dr,

d are all positive. Since r can be negative, this formula may not hold (if r < 0,

dr > 0, d > 0, then dA < 0!). We have several ways to deal with this problem and

some are as follows:

(1) If r = f() changes sign, break the problem into r > 0 and r < 0 regions.

(2) Where possible, use symmetry. This is the easiest.

(3) If r = f() is negative, rewrite the curve as r = g() with g positive.

(4) If r = f() is negative, then take both r and dr negative.

We illustrate the above four ways with the following example.

Example 3. Find the area enclosed by the curve r = cos .

Answer. As we saw in Example 2 this is the circle

x 1

2

2+ y2 =

1

4,

So we know the answer: 14

.103


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x

y

We however ignore this and use the example to practice the four ways indi-

cated. Note that r = cos changes sign at = /2, so we split the problem by

considering first between 0 and /2 and later between /2 and .

Now as varies between and /2 we get a semicircle.

= 0

= / 2

r = cos

r = 0

The area is

/2=0

cos r=0

rdrd =

/20

cos2

2d =

1

4

/20

[1 + cos(2)] d

= 14 + sin(2)2

/2

0= 8 .

A not unexpected result. Since the figure is symmetric (its a circle, or from the

properties of the cosine), we use this fact, and obtain that the total area is 2 (/8) = /4 as we know. We now practice the other ways for the purposes of

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illustration. Keep in mind that we are trying to calculate the area enclosed by

r = cos , r = 0, =

2, = .

The next way (changing to positive r) gives as follows: A point P = (r, ) is

on r = cos for some between /2 and iff P is also given by (r, + ). I.e., ifwe plot ( cos , + ) with /2 < < , we will get the same curve.

P ( r , )

- r+

y

x

Put = +. Then our curve is also given by r = cos() with 3/2 < < 2,and now r is positive.

= 2

= 3 / 2 x

y

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So we get

A = 2

= 32

cos()r=0 rdrd =

1

2 2

3

2cos

2

( ) d

=1

2

23

2

1 + cos(2( ))2

d =1

2

23

2

1 + cos(2)

2

=1

4

1 +

sin(2)

2

23

2

=1

4

2 3

2

=

8.

For the final way: We now keep between /2 and . Suppose we increase

by d and decrease r (remember r is negative!!) by dr.

d

- r

dA

x- dr

d

y

We then obtain in the fourth quadrant a dA given by

dA = (r d)(dr) = rddr.

This looks the same as before, except now both r and dr are negative! We calculate

the area of a spike, from the biggest (least negative) r to the smallest (most negative)

rsince our dr are negativeand get

cos r=0

rdrd.

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So that, since d is positive, we get

A =

2

cos

r=0

r drd =

2

cos2

2

d =1

4

2

(1 + cos(2))d =

8

.

It is obvious that symmetry (when it exists) yields the easiest approach! We

now consider more examples.

Example 4. Evaluate

D(x + 3y) dA if D is the circle centered at the origin of

radius 2.

Answer. Note that the boundary of D is given by x2 + y2 = 4. This is the

motivation for changing to polar coordinates, since we can describe D in the much

simpler form r = 2!

x

y

We then have:

2=0

2r=0

(cos + 3 sin )r2 dr d =2

=0

(cos + 3 sin ) r3

32

0

d

=

20

(cos + 3 sin )8

3d

=8

3[sin 3cos

20

= 0.

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Imagine we tried to do this problem using (x, y).

4 - xy = 2

y = - 4 - x 2

- 2 2x

y

We would have: 2x=2

4x2y=

4x2

(x + 3y) dy dx.

This can be done, but the integrals are more complicated.

Example 5. Find D(x2 + y2) dA where D is the cardioid r = 1 + cos .Answer. Since the boundary ofD is given in polar coordinates, we use this system.

Can you imagine how complicated this would become if we tried to switch to ( x, y)?

We first need to sketch the cardioid.

= 0 r = 2

r =

2r = 1

r

= r = 0108


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r

=3

2r = 1

= 2 r = 2.

We only plot the graph for 0 2 since cos is periodic.

= / 2 , r = 1

= , r = 0

= 0 , r = 2

y

x

We now write the integral in polar coordinates:

x = r cos , y = r sin

and so

x2 + y2 = r2.

We then have

(x2 + y2)dA = (r2)(rdrd)

and so

D

(x2 + y2) dA =

2=0

1+cos r=0

r2r dr d

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=

20

r4

4

1+cos

0

d =1

4

20

(1 + cos )4 d.

Note that

(1 + cos )4 =

(1 + cos )22

= (1 + 2 cos + cos2 )2

= 1 + 4 cos + 6 cos2 + 4 cos3 + cos4

= 1 + 4 cos + 6

1 + cos 2

2

+ 4 cos3 +

1 + cos 2

2

2

= 1 + 4 cos + 3(1 + cos 2) + 4 cos3

+

1

4 [1 + 2 cos 2 + cos2

2]

= 41

4+ 4 cos + 3

1

2cos2 + 4 cos3 +

1

4cos2 2

= 41

4+ 4 cos + 3

1

2cos2 + 4 cos3 +

1

4

1 + cos 4

2

= 43

8+ 4 cos + 3

1

2cos2 + 4 cos3 +

1

8cos4.

So, finally,

1

4

20

(1 + cos )4 =1

4

20

4

3

8+ 4 cos + 3

1

2cos2 +

1

8cos4 + 4 cos3

d

=1

4

4

3

8 2 + 4

20

cos3 d

=1

4

4

3

8 2 + 4

20

cos2 cos d

=

1

44 38 2 + 4 2

0 (1 sin2

)cos d put u=sin , du=cos d

=

1

4

4

3

8 2 + 4

sin sin

3

3

20

=1

4

4

3

8

2.

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Example 6. Find the area of one loop ofr = cos 2.

Answer. Again we start with a sketch.

= 0 r = 1

r

=

4r = 0

r

= 2 r = 1

r

=3

4r = 0

r

= r = 1

r

=5

4r = 0

r

=3

2r = 1

=7

4r = 0

= 2 r = 1.

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x

y

= / 4

= / 2

= , r = 1

= 3 / 4 = 3 / 2 , r = -1

= 0

, r = - 1

Note that the graph has four equal and symmetric loops, and that r changes from

positive to negative and back again. What I feel is the easiest way is to calculate

the area for between 0 and /4 and multiply by 2. We get

Area of a loop = 2

/4=0

cos2r=0

r dr d

=

/4=0

cos2 2

d

=

/40

1 + cos 4

2d =

1

2

+

sin4

4

/4

0

= 124 = 8 .

This seems the best way. To practice, let us calculate the area for between /4

and /2 and again multiply by 2.112


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= / 2

= / 4

- r

dr

Here r is negative and we choose dr < 0. Then

Area = 2

/2/4

cos 20

rdrd =

/2/4

cos2 2 d

=1

2

/2/4

[1 + cos 4] d =1

2

+

sin4

4

/2/4

=

8

as before.

Example 7. Find the volume outside the cone z =

x2 + y2 and inside the

sphere x2 + y2 + z2 = 1.

Answer. Note that it is easiest to figure out just the volume common to the cone

and sphere, and subtract this from the volume of the sphere.

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z = r

2 2z + r = 1

We can set up the problem using (x, y) and then switch to (r, ) or start right away

with (r, ). Let us start with (r, ) directly. If we look down on our solid we see a

circle coming from the intersection of the sphere and cone, i.e.,

z2 + r2 = 1, z = r

or 2r2 = 1 or r = 1/

2.

2r = 1 /

y

x

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We now cut the disc into regions of size dA = rdrd and obtain that the volume of

a column over dA = (height)(cross-sectional area)

= (

1 r2 from the

sphere

rfrom the

cone

)rdrd.

Hence

volume over a spike =

1/2r=0

(

1 r2 r)rdrd

and volume V, common to sphere and cone, is:

V =

20

1/2r=0

(

1 r2 r)rdrd

=

20

1/2r=0

1 r2r dr

1/2

r=0

r2 dr

d

=

20

1

3(1 r2) 32 r

3

3

1/2r=0

d

= 2

0 1

31 11

23/2

+1

23

2 d=

20

1

3

1 2

23

2

d =

1

3

1 1

2

2.

So the final answer is

Volume of sphere V = 43

23

1 1

2

=

4

3

1

2+

1

2

2

.

Example 8. Find the volume bounded above by the cone z = r and below by

y = 0, y = x, x = 1 and the xy plane.

Answer. Since the altitude is z = r (simple), we do this problem using (r, ), even

though the boundary is given by y = f(x). We rewrite the boundary of the base115


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triangle as:

y = 0 = 0y = x r sin = r cos tan = 1 =

4

x = 1 r cos = 1 r = sec .

r = sec

x

y

z

= / 4

= 0

y

x

So,

Volume over dA = altitude cross sectional area

= (r)(r dr)d.

And we get

V =

/4=0

sec r=0

r2 dr d =1

3

/40

sec3 d.

Now to evaluate:

/40

sec3 d =

/40

sec2

u(sec

v)d

(u = tan , v = sec tan )

= tan sec

/40

/4

0

tan2 sec d

= tan sec

/40

/4

0

(sec2 1) sec d116


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= tan sec

/4

0

/4

0

sec3 d +

/40

sec d,

or

2

/40

sec3 d = tan sec

/40

+ ln | sec + tan |/4

0

= tan

4sec

4+ ln

sec 4

+ tan

4

=

22

+ ln

2

2+ 1

.

And, finally,

V =1

3

1

2+

1

2ln

2

2+ 1

.

Example 9. Find the volume under the plane 4x + 6y + z = 12 and above the

region in the (x, y) plane with boundary x2 + y2 = x.

Answer. Note that x2 + y2 = x (or (x

12 )

2 + y2 = 14 ) is a circle. We change it to

polar coordinates as follows: r2 cos2 + r2 sin2 = r cos or r = cos . Observe the

values of which are needed to trace the circle, and the fact that r goes negative!

x

= 0

= / 2

= / 2

y

We could split the problem up into regions where r is positive and where its neg-117


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ative, but the best way is probably to instead have range over /2 /2.This also covers the circle. Next note that the plane does not cross the circle. In-

deed, in the circle x 1 and y 1/2 and so z = 12 4x 6y 12 4 6 12 > 0.

If the plane crossed, we would have to split the problem, since the altitude would

sometimes be z and sometime z. So we get

(12 4x 6y)dA = (12 4r cos 6r sin )rdd

and

Volume =/2

=/2

cos r=0

(12 4r cos 6r sin )rdrd

=

/2/2

6r2 4r

3

3cos 2r3 sin

cos 0

d

=

/2/2

6cos2 4

3cos4 2cos3 sin

d.

Now

/2/2

6cos2 d =/2/2

3(1 + cos 2)d = 3

+sin2

2

/2/2

= 3/2/2

43

cos4 d =

/2/2

4

3

1 + cos 2

2

2d

=

/2/2

1

3

[1 + 2 cos 2 + cos2 2] d

=/2/2

13

1 + 2 cos 2 + 1

2(1 + cos 4)

d

= 13

+ sin 2 +

1

2

+

sin4

4

/2/2

= 13 3

2 =

2.

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Finally,

/2

/2 2cos3 sin d =

cos4

2 /2

/2= 0.

(u = cos , du = sin d)

So

V = 3 2

=5

2.

Example 10. Convert to polar coordinates and evaluate

aa

a2x20

(x2 + y2)1/2dy dx.

Answer. We begin by sketching the region D. Note that x varies from a to a, andfor a given x between a and a, y goes from 0 to a2 x2. But if y = a2 x2,

then y2

+ x2

= a2

(i.e., on the circle) and y 0. So that D is:

x + y = a222

D

x

y

- a a

Now

(x2 + y2)1/2 = r and x2 + y2 = a2 r = a.119


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So we get

=

=0 a

r=0

(r)(rdrd) =

0

a3

3d =

a3

3.

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Further Exercises:

1) Evaluate A y2 dA if A is the area enclosed by r = 4(1 cos ).

2) Change to polar coordinates and evaluate

22

4y2

4y2e(x

2+y2) dxdy.

3) Find the total area enclosed by r2 = 2 cos(2).

4) Find by means of polar coordinates the area in the circular cap bounded above

by x2 + y2 = 1 and below by y = 1/2.

5) Find the area of the region common to the cardioid r = 1 + cos and the

circle r = sin .

6) Change to polar coordinates and evaluate

1

z=0

1z2

y=0

sin(y2 + z2) dy dz.

7) Calculate the area bounded by the curves:

= 0, r =

4, r =

2, =

r 3

8

2.

8) Evaluate

B

x2 + y2 dxdy by means of polar coordinates ifB is the region

bounded by the circle r =

sin .

Note: the answer is positive.

9) Change to (x, y) coordinates and evaluate:

2

=4

csc r=0

r5 cos2 sin2 drd.

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10) Evaluate

D(1+x2 +y2)10 dxdy where D is the region bounded by the curves

y = 0, y = x, x2 + y2 = 1.

11) Evaluate D x2 + y2 dy dx where D is the triangle with vertices (0, 0),(1, 0), (1, 1).

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Calculus-Double integrals in polar coordinates (r,phi)

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