Topic 32 Notes Jeremy Orlo - MIT - Massachusetts...

Topic 32 NotesJeremy Orloff

32 Polar coordinates and double integrals

32.1 Polar Coordinates

x

y

r y

(x, y) = (r cos(θ), r sin(θ))

xθ

Standard x, y, r, θ triangle

Polar coordinates are just standard trigonometric relations. In the figure above, we knowthat

x = r cos(θ), y = r sin(θ) r =√x2 + y2, θ = ‘ tan−1(y/x).′

θ is a bit tricky, ‘tan−1(y/x)’ is in quotes to indicate that it is not perfectly well-defined.We will examine this in more detail below.

We call (r, θ) the polar coordinate representation of (x, y).

The following example shows is more than one way to represent any point in polar coordi-nates.

Example 32.1. In the table below, the (x, y) coordinates are at the top of each columnand various possible r,θ representations are below.

(x, y) (1, 0) (0, 1) (2, 0) (1, 1) (−1, 1) (−1,−1) (0, 0)

(r, θ) (1, 0) (1, π/2) (2, 0) (√

2, π/4) (√

2, 3π/4) (√

2, 5π/4) (0, π/2)

(r, θ) (1, 2π) (√

2, 9π/4) (√

2,−3π/4) (0,−7.2)(r, θ) (1, 4π)

32.2 Double integrals in polar coordinates

Recall that a double integral

∫ ∫Rf(x, y) dA is a ‘sum’ . Here dA is the area of a tiny region

around the point (x, y). In rectangular coordinates we put a tiny rectangle at (x, y). Tocompute double integrals using polar coordinates we will put a ’polar rectangle’ at (x, y).

1

32 POLAR COORDINATES AND DOUBLE INTEGRALS 2

x

y

r

rr∆θ

∆θ

∆r

∆A

θ(r, θ)

(r + ∆r, θ)

(r + ∆r, θ + ∆θ)

Starting at the point with polar coordinates (r, θ) we form a curvy rectangle by changing rby ∆r and θ by ∆θ. That is, with vertices (r, θ), (r + ∆, θ), (r, θ + ∆θ), (r + ∆r, θ + ∆θ).The curvy sides are small arcs of circles and the straight sides are radial segments.

If ∆r and ∆θ are very small, the curvy rectangle is approximately a rectangle with sides∆r and r∆θ. So, it’s area is ∆A ≈ r∆θ∆r. In the limit we get an exact formula

Area in polar coordinates = dA = r dr dθ.

Example 32.2. The circular sector R shown below has angle π/3 and radius 2. Assumeits density is δ(x, y) = xy. Use integration with polar coordinates to find its mass.

x

y

2

π/3

r = 2

answer: As a double integral the mass is

∫∫Rδ(x, y) dA.

In polar coordinates: δ(x, y) = xy = r2 cos(θ) sin(θ).

To find the limits of integration we cover R with radial lines.:

Inner limits (fix θ): 0 < r < 2, i.e. the radial lines run from the origin to the circle r = 2.

Outer limits: 0 < θ < π/3, i.e. we have a radial line for each θ in this range.

Thus the mass is

M =

∫ ∫Rδ(x, y) dA =

∫ π/3

θ=0

∫ 2

r=0r2 cos(θ) sin(θ) r dr dθ.

Inner integral:

∫ 2

0r3 cos(θ) sin(θ) dr =

r4

4cos(θ) sin(θ)

∣∣∣∣20

= 4 cos(θ) sin(θ)

Outer integral: M =

∫ π/3

04 cos(θ) sin(θ) dθ = 2 sin2 θ

∣∣π/30

=3

2.

Example 32.3. Compute I =

∫ 2

1

∫ x

0

1

(x2 + y2)3/2dy dx by switching to polar coordinates.

answer: Step 1 is to draw the region.

The integral as given has limits:inner limits: 0 ≤ y ≤ x, outer limits: 1 ≤ x ≤ 2.


In rectangular coordinates this means we’ve covered the region with vertical lines. This isshown in the figure on the left.

x

y

y = x

x = 1

x = 2

1 2x

y

θ = π/4

r = sec(θ) r = 2 sec(θ)

1 2

Region using vertical lines (left) and radial lines (right)

In polar coordinates, the vertical line x = a can be written r cos(θ) = a. That is, r =a sec(θ).

So covering the region with radial lines gives limits (see above figure on right) :

Inner limits: r from sec(θ) to 2 sec(θ); outer limits: θ from 0 to π/4.

Integral =

∫ π/4

0

∫ 2 sec(θ)

sec(θ)

1

r3r dr dθ.

Inner integral:

∫ 2 sec(θ)

sec(θ)

1

r3r dr =

1

2cos(θ). Outer integral:

∫ π/4

0

1

2cos(θ) dθ =

1

2sin(θ)

∣∣∣∣π/40

=

√2

4.

Example 32.4. Find the volume of the region above the xy-plane and below the graph ofz = 1− x2 − y2.answer: As always, we start with a picture;

x

y

z

11

1

R

As a double integral the volume is

∫∫Rz dA, where R is the unit disk in the xy plane below

the graph and above R.

Since R is the disk and, in polar coordinates, z = 1− r2 we will compute this integral inpolar coordinates. For R we have the following limits.

Inner limits: 0 ≤ r ≤ 1; outer limits 0 ≤ θ ≤ 2π.

volume =

∫ 2π

0

∫ 1

0(1− r2) rdr dθ.


Inner integral:

∫ 1

0(1− r2) rdr =

1

2− 1

4=

1

4.

Outer integral: V =

∫ 2π

0

1

4dθ =

π

2.

32.3 Graphing with polar coordinates

We’ll explain what it means to graph a function r = f(θ) with an example.

Example 32.5. (Cardiod.) Plot the graph of r = 1 + cos(θ).

answer: This means to plot every point (r, θ) in polar coordinates, where r = 1 + cos(θ).To get a sense we make a table of r for various θ between −π and π. Since cos(θ) is periodicthere is no need to look outside this range. Also, since cos(−θ) = cos(θ) we get the samevalue of r for ±θ.

θ 0 ±π/6 ±π/4 ±π/3 ±π/2 ±2π/3 ±3π/4 ±πr 2 1 +

√3/2 1 +

√2/2 3/2 1 1/2 1−

√2/2 0

In the figure, the labels at the end of radial segments indicate the polar angle θ of thesegment. Of course, the length of the segment is r.

x

y

a 2a

π/12

−π/12

π/6

−π/6

−π/4

π/4

−π/3

π/3

−5π/12

5π/12

π/2

−π/2

2π/3

−2π/3

This figure is called a cardioid

In words, at θ = 0, r has its maximum value of 2. As θ increases to π, r decreases to 0. Youcan see this in the top half of the cardioid, i.e. as the angle increases the radial segmentgets shorter. The symmetry means the bottom half (for negative θ) is the mirror image ofthe top.

32.4 Gallery of polar functions

Lines and circles can be written described in polar coordinates. We give formulas in polarcoordinates for four of these. In the formulas we use specific constants, the general caseshould be clear. We also give a brief derivation of each formula.

• Circle of radius 2 centered at the origin: r = 2.Derivation: this is the definition of such a circle.


• Ray at angle π/3 from the origin: θ = π/3.Derivation: see the figure below.

• Vertical line x = 2: r = 2 sec(θ).Derivation: x = r cos(θ) = 2 ⇒ r = 2 sec(θ).

• Horizontal line y = 2: r = 2 csc(θ).Derivation: y = r sin(θ) = 2 ⇒ r = 2 csc(θ).

Here are plots of each of the above.

x

y

π/3

ray θ = π/3

x

y

2

2

2

Circle r = 2.

x

y

2

line x = 2; r = 2 sec(θ)

x

y

2

line y = 2; r = 2 csc(θ)

32.4.1 Circles through the origin

We’ll state the conclusions up front and then derive them.

The graph of r = 2a cos(θ) for −π/2 ≤ θ ≤ π/2 is a circle of radius a and center (a, 0).That is, it is a circle through the origin with center on the x-axis.

Likewise, the graph of r = 2a sin(θ) for 0 ≤ θ ≤ π is a circle of radius a and center (0, a).That is, it is a circle through the origin with center on the y-axis.

x

y

a 2a

Circle r = 2a cos(θ)−π/2 ≤ θ ≤ π/2.

x

y

a

2a

Circle r = 2a sin(θ).0 ≤ θ ≤ π.

Let’s prove this for r = 2a cos(θ). We can convert this equation to rectangular coordinatesby first multiplying it by r to get: r2 = 2ar cos(θ). Then, since r2 = x2+y2 and x = r cos(θ)we can write this in rectangular coordinates as x2 + y2 = 2ax. Now a little algebraicmanipulation and completing the square gives

x2 + y2 − 2ax = 0 ⇒ x2 − 2ax+ a2 + y2 = a2 ⇒ (x− a)2 + y2 = a2.


This last formula shows we have a circle of radius a and center on the x-axis at (a, 0), justas claimed. We still have to show that we cover the entire circle exactly once as θ goes from−π/2 to π/2.

To do this, we’ll first check the endpoints:1. at θ = −π/2 we have r = 2a cos(−π/2) = 0,2. at θ = π/2 we have r = 2a cos(π/2) = 0.

So, as θ goes through the range from −π/2 to π/2, the graph starts and ends at the origin.That is basically enough to show what we want, but let’s go through some other values ofθ.

x

y

a 2a

Circle r = 2a cos(θ)−π/2 ≤ θ ≤ π/2.

−π/4

π/4

−π/3

π/3

−π/6

π/6

−π/12

π/12

In the figure the angle θ is marked at the end of each radial segment. It illustrates that as θgoes from −π/2 to π/2 the point (r, θ) with r = 2a cos(θ) starts at 0 and goes once aroundthe circle in a counterclockwise direction.

The picture for r = 2a sin(θ) is similar. Since the circle is above the x-axis, the angle θ runsfrom 0 to π.

32.4.2 Plotting negative values of r.

When plotting, we can allow r to be negative. If the formula produces a negative value forr, we just draw the radial line segment backwards. For example, if r = −1 when θ = π/4,then we plot this at the point (x, y) = (−

√2/2,−

√2).

x

y

θ = π/4

1

1

r = −1, θ = π/4

Warning. While this works for plots, you should never use it in integration. Bad thingscan happen if you let r be negative in an integral. It is better to make use of symmetryand only integrate over regions where r is positive.


Example 32.6. A limacon is a curve with equation r = a(1 + b cos(θ)), where b > 1. Plotthe limacon r = 1 + 2 cos(θ).

answer: We start with a table of values. Notice that at θ = 2π/3, r = 0 and for 2π/3 <θ ≤ π, r is negative. As with the cardioid, we know that the limacon is symmetric acrossthe x axis.

θ 0 ±π/6 ±π/4 ±π/3 ±π/2 ±2π/3 ±3π/4 ±πr 3 1 +

√3 1 +

√2 2 1 0 1−

√2 ≈ −0.41 −1

x

y

π

π/6

−π/6

−π/4

π/4

−π/3

π/3

π/2

−π/2

−3π/4

3π/4

2π/3−2π/3

−3π/4

3π/4

Limacon, r = 1 + 2 cos(θ)

32.5 More polar plots

Here are two more nice plots. We show them without any comment.

x

y

Four leaved rose r = a sin(2θ).

x

y

Lemniscate: r2 = 2a2 cos(2θ)

32.6 Ellipse

Let E be the ellipse given by the equationx2

a2+y2

b2= 1. This is an ellipse centered at 0 and

oriented with the coordinate axes.


x

y

a

b

(x, y) = (r cos(θ), r sin(θ))

Ellipsex2

a2+y2

b2= 1.

Claim. In polar coordinates, E has the equation

r2 =a2b2

b2 cos2(θ) + a2 sin2(θ).

Proof. First, rewrite the formula for E as b2x2 + a2y2 = a2b2. Now, use x2 = r2 cos2(θ)and y2 = r2 sin2(θ). So,

b2x2+a2y2 = a2b2 ⇒ b2r2 cos2(θ)+a2r2 sin2(θ) = a2b2 ⇒ r2 =a2b2

b2 cos2(θ) + a2 sin2(θ). QED.

Topic 32 Notes Jeremy Orlo - MIT - Massachusetts...

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