Integral Calculus 201-SN3-RE (201-NYB) • CEGEP, Montreal

Revision sheet: polar coordinates, 201-SN3-RE (201-NYB)

Q: Can the polar radius be negative?

Yes. A negative radius means the point is plotted in the direction opposite to the given angle. This is what allows a rose with an even coefficient to have twice as many petals as its coefficient, half of them being traced with negative values of the radius.

Q: Why does the system of equations not give all the intersections?

Because it assumes that the two curves pass through the same point at the same angle. But the pole has infinitely many names, one for each angle, so two curves can pass through it at different angles. To test the pole, you must therefore solve separately for each radius being zero.

Q: Where does the factor one half in the polar area come from?

From the area of a circular sector, which is half the square of the radius times the angle. The elementary slice of a polar region is a very thin sector, not a rectangle. The immediate check is to apply the formula to a circle of constant radius, which must give back pi times the radius squared.

Q: How do you find the bounds of a polar area?

By looking for the angles where the radius vanishes, because that is where the curve comes back to the pole and closes a loop. For an area between two curves, the bounds are the angles of intersection. Integrating beyond them counts regions already traced again and doubles or quadruples the answer.

Q: How many petals does a polar rose have?

If the coefficient of the angle is odd, the rose has exactly that number of petals. If it is even, it has twice as many. The reason is the sign of the radius: with an odd coefficient, the second half of the turn retraces the same petals, while with an even coefficient it traces new ones.

This sheet does not explain again how to convert between the two coordinate systems: that fits in two lines. It deals with what decides the mark in 201-SN3-RE (formerly 201-NYB), that is, the fact that a polar representation is not unique and what that does to intersections, then the three factors that the area formula loses most often.

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

In polar coordinates, $r$ can be NEGATIVE and the angle is read modulo $2\pi$ : one point has infinitely many representations. That is where every lost mark in the chapter comes from, starting with the intersections that the system of equations never gives.

This chapter is part of 201-SN3-RE (201-NYB), Integral Calculus

The essentials

One point, infinitely many names

• Conversion: $x=r\cos\theta$ , $y=r\sin\theta$ ; conversely $r^{2}=x^{2}+y^{2}$ and $\tan\theta=\frac{y}{x}$ , watching the QUADRANT.
• $\left(r,\theta\right)$ and $\left(r,\theta+2k\pi\right)$ designate the same point.
• So do $\left(-r,\theta\right)$ and $\left(r,\theta+\pi\right)$ : a NEGATIVE radius is read in the opposite direction.
• The POLE, the origin of the system, is written $\left(0,\theta\right)$ for any $\theta$ : it is the most ambiguous point of the plane.

The rose

r=\cos 2\theta

has FOUR petals: two are traced by positive values of

r

, two by negative values, in the direction opposite to the angle.

This lack of uniqueness is not a curiosity: it is the direct cause of the two most costly pitfalls of the chapter, missed intersections and loops counted twice.

The formulas of the chapter, with their factors

• AREA of a sector: $\mathcal{A}=\frac{1}{2}\int_{\alpha}^{\beta}\left[r(\theta)\right]^{2}d\theta$ . The $\frac{1}{2}$ AND the square are both required.
• Area BETWEEN two polar curves: $\frac{1}{2}\int\left(r_{\text{out}}^{2}-r_{\text{in}}^{2}\right)d\theta$ , a difference of squares.
• ARC LENGTH: $L=\int_{\alpha}^{\beta}\sqrt{\left[r(\theta)\right]^{2}+\left[r'(\theta)\right]^{2}}\,d\theta$ .
• SLOPE of the tangent line: $\frac{dy}{dx}=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$ , obtained by parametrizing by $\theta$ .

The area element is a SECTOR with angle

d\theta

and radius

r

, of area

\frac{1}{2}r^{2}\,d\theta

: hence the one half and the square.

The $\frac{1}{2}$ in the area comes from the area of a circular sector, $\frac{1}{2}r^{2}\Delta\theta$ : the slice is a thin triangle, not a rectangle.

The curves to recognize at a glance

• $r=a$ : circle centred at the pole. $\theta=\alpha$ : line through the pole.
• $r=2a\cos\theta$ : circle of radius $a$ centred at $(a,0)$ ; $r=2a\sin\theta$ : circle centred at $(0,a)$ .
• $r=a\left(1\pm\cos\theta\right)$ or $a\left(1\pm\sin\theta\right)$ : CARDIOID.
• $r=a\cos(n\theta)$ : rose with $n$ petals if $n$ is ODD, with $2n$ petals if $n$ is EVEN.
• $r^{2}=a^{2}\cos 2\theta$ : lemniscate, shaped like a figure eight lying on its side.

Four shapes to recognize without computing: a circle centred at the pole, a circle through the pole, a cardioid, a rose.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Rejecting a negative radius

the whole plotting question, since half the petals disappear

What not to write

“ $r=\cos 2\theta$ becomes negative between $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ , so the curve does not exist there.”

What to write

“ $r<0$ means the point is plotted in the direction OPPOSITE to $\theta$ : these values trace two more petals.”

Why: The polar radius is a signed coordinate, not a distance. A four-petal rose drawn with only two petals signals exactly this mistake, and it shows at first glance on the test.

2. Forgetting the pole in an intersection

the whole question, and a wrong area if the splitting depends on it

What not to write

“ $1+\cos\theta=1-\cos\theta$ gives $\theta=\frac{\pi}{2}$ and $\frac{3\pi}{2}$ : two points of intersection.”

What to write

“There are THREE: the two from the system, plus the POLE, reached by the first curve at $\theta=\pi$ and by the second at $\theta=0$ .”

The two cardioids cross at three visible points, but the system gives only two: the POLE is reached by each one at a different angle, so it escapes the equation.

Why: The system of equations assumes that the two curves pass through the same point FOR THE SAME angle. At the pole this is not necessary, since $\left(0,\theta\right)$ is the same point for every $\theta$ . You must therefore always test the pole SEPARATELY, by solving $r=0$ for each curve.

3. Forgetting the factor one half in the area

the whole question, since the answer is twice too large

What not to write

$\mathcal{A}=\int_{\alpha}^{\beta}r^{2}\,d\theta$

What to write

$\mathcal{A}=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}\,d\theta$

Why: The elementary slice is a circular sector of area $\frac{1}{2}r^{2}\Delta\theta$ , not a rectangle. Immediate check on the circle $r=a$ : the formula must give back $\pi a^{2}$ , which only the version with $\frac{1}{2}$ does.

4. Forgetting the square in the area

the whole question

What not to write

$\mathcal{A}=\frac{1}{2}\int r\,d\theta$

What to write

$\mathcal{A}=\frac{1}{2}\int r^{2}\,d\theta$

Why: Without the square, the expression does not even have the dimension of an area: $r\,d\theta$ is an arc length, not a surface. The dimensional check is enough to rule this version out, and it costs nothing.

5. Integrating over a full turn when one loop is enough

the whole question, since the result counts all four petals

What not to write

“Area of one petal of $r=\cos 2\theta$ : $\frac{1}{2}\int_{0}^{2\pi}\cos^{2}2\theta\,d\theta$ .”

What to write

“One petal is traced for $\theta\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$ : $\frac{1}{2}\int_{-\pi/4}^{\pi/4}\cos^{2}2\theta\,d\theta=\frac{\pi}{8}$ .”

One petal of

r=\cos 2\theta

is traced for

\theta\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]

: that is where, and only where, you integrate.

Why: The bounds of a polar area are found by looking for where $r$ vanishes, that is, where the curve comes back to the pole: those angles delimit a loop. Integrating beyond them counts regions already traced again.

6. Getting the number of petals wrong

1 to 2 marks on the graph

What not to write

“ $r=\cos 2\theta$ has two petals, since the coefficient is $2$ .”

What to write

“An EVEN coefficient gives $2n$ petals, an ODD coefficient gives $n$ : here $n=2$ , so FOUR petals.”

\cos 2\theta

(even coefficient) gives four petals,

\cos 3\theta

(odd) gives three: parity decides.

Why: The rule comes from the sign of $r$ : when $n$ is odd, the second half of the turn retraces the same petals with $r<0$ ; when $n$ is even, it traces new ones. Counting the zeros of $r$ over one turn gives the answer without having to memorize the rule.

7. Confusing the slope of the tangent with the derivative of the radius

the whole question

What not to write

“The tangent line has slope $r'(\theta)$ .”

What to write

“ $\frac{dy}{dx}=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$ , obtained by parametrizing $x=r\cos\theta$ and $y=r\sin\theta$ .”

Why: $r'(\theta)$ measures how fast the point moves away from the pole, not the inclination of the curve. The slope is obtained by treating the curve as parametrized by $\theta$ , which brings in two terms on top and two at the bottom through the product rule.

8. Losing the quadrant when converting to polar

the whole question, since the point obtained is its mirror image through the pole

What not to write

“ $x=-1$ , $y=-1$ gives $\tan\theta=1$ , so $\theta=\frac{\pi}{4}$ .”

What to write

“The point is in the THIRD quadrant: $\theta=\frac{5\pi}{4}$ , and $r=\sqrt{2}$ .”

\arctan\frac{y}{x}=\frac{\pi}{4}

points to

(1, 1)

(red); the point

(-1, -1)

is in the third quadrant, at

\theta=\frac{5\pi}{4}

(green).

Why: The arctangent function only returns angles between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ : it cannot tell the first quadrant from the third, nor the second from the fourth. A rough sketch of the point, before any computation, settles the question.

Which method to choose

How to find ALL the intersections of two polar curves

The system of equations is never enough. Three checks are needed, in this order.

If you are looking for intersections “at the same angle” → solve $r_{1}(\theta)=r_{2}(\theta)$ over a full turn

Example: $1+\cos\theta=1-\cos\theta$ gives $\theta=\frac{\pi}{2}$ and $\frac{3\pi}{2}$
If you are checking whether the POLE is common → solve $r_{1}=0$ and $r_{2}=0$ SEPARATELY; if both have a solution, the pole is an intersection

Example: $1+\cos\theta=0$ at $\pi$ and $1-\cos\theta=0$ at $0$ : the pole is common
If the two curves can meet with opposite radii → also solve $r_{1}(\theta)=-r_{2}(\theta+\pi)$

Example: useful for roses and lemniscates
If a doubt remains → SKETCH both curves: the drawing gives the number of points, the algebra gives their coordinates

Example: three visible points for two solutions of the system signals a forgotten pole

The drawing is not a proof but a counter: it tells you how many points to look for, which is enough to know whether the system missed some.

Which symmetries to use

A symmetry test takes ten seconds and halves the plotting work, or cuts it to a quarter.

If replacing $\theta$ with $-\theta$ does not change $r$ → symmetry about the POLAR AXIS, the x-axis

Example: $r=1+\cos\theta$ , since cosine is even
If replacing $\theta$ with $\pi-\theta$ does not change $r$ → symmetry about the line $\theta=\frac{\pi}{2}$ , the y-axis

Example: $r=1+\sin\theta$
If replacing $r$ with $-r$ , or $\theta$ with $\theta+\pi$ , does not change the curve → symmetry about the POLE

Example: $r^{2}=\cos 2\theta$
If no test works → plot point by point over one turn, tabulating $r$ for the standard angles

Example: every $\frac{\pi}{12}$ is enough for an exam curve

A symmetry test that fails does not prove there is no symmetry: these tests are sufficient, not necessary. So you can try all three without worrying.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

A polar area, written for the grader

When to use it: Any question asking for the area of a loop, a petal or a region between two polar curves.

1 Find the BOUNDS by solving $r(\theta)=0$ , or by finding the angles of intersection.
2 Justify that these bounds delimit exactly ONE loop, traced only once.
3 Write the complete formula with the $\frac{1}{2}$ and the square.
4 Rewrite the squared trigonometric term with the double-angle formulas before integrating.
5 Compute and conclude with the area unit.

Concluding sentence

The area of the cardioid $r=1+\cos\theta$ is $\frac{1}{2}\int_{0}^{2\pi}\left(1+\cos\theta\right)^{2}d\theta=\frac{3\pi}{2}$ square units.

The trap: Integrating $\cos^{2}\theta$ without reducing the power. The table does not give this antiderivative directly: you must go through $\cos^{2}\theta=\frac{1+\cos 2\theta}{2}$ , and this line is expected on the test.

Marking: Usually 1 mark for the justified bounds, 1 mark for the complete formula, 1 mark for the power reduction, 1 mark for the final computation.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Checking a polar area on a circle

Apply the formula to the constant $r=a$ : it must give back $\pi a^{2}$ . This is the check that catches the missing factor $\frac{1}{2}$ .

$\frac{1}{2}\int_{0}^{2\pi}a^{2}d\theta=\frac{1}{2}a^{2}(2\pi)=\pi a^{2}$ ✓. Without the $\frac{1}{2}$ , you would get $2\pi a^{2}$ , twice as much.
Counting the points of intersection on the sketch

Sketch the two curves roughly and count the visible crossings, then compare with the number of solutions of the system.

Three visible crossings for two algebraic solutions: the pole was forgotten, as with the two opposite cardioids.
Checking the number of petals with the zeros of $r$

Count the values of $\theta$ over one turn where $r$ vanishes: each pair of consecutive zeros delimits a loop.

$\cos 2\theta$ vanishes four times on $\left[0, 2\pi\right]$ , at $\frac{\pi}{4}$ , $\frac{3\pi}{4}$ , $\frac{5\pi}{4}$ and $\frac{7\pi}{4}$ : these four zeros cut the turn into four arcs, one per petal, so four petals ✓.
Converting a point you found and placing it again

Once you have $\left(r,\theta\right)$ , compute $x=r\cos\theta$ and $y=r\sin\theta$ and check that the point falls in the right quadrant.

$\left(\sqrt{2},\frac{5\pi}{4}\right)$ gives $(-1, -1)$ ✓, third quadrant; $\left(\sqrt{2},\frac{\pi}{4}\right)$ would have given $(1, 1)$ .

The typical problem, taken apart

The area of a cardioid, from the bounds to the power reduction

Compute the area of the region inside the cardioid $r=1+\cos\theta$ .

The cardioid

r=1+\cos\theta

reaches its maximum

2

\theta=0

and comes back to the pole at

\theta=\pi

: one full turn traces it exactly once.

Step 1

Bounds: $r=1+\cos\theta$ vanishes only at $\theta=\pi$ , and the curve closes up after one full turn. So we integrate from $0$ to $2\pi$ .

Why

Justifying that the turn is traced only ONCE is essential: on a rose, the same reasoning would lead to much narrower bounds, and integrating over $2\pi$ would count all four petals instead of one.

Step 2

Symmetry: $\cos(-\theta)=\cos\theta$ , so $r(-\theta)=r(\theta)$ : the curve is symmetric about the polar axis. We can integrate from $0$ to $\pi$ and DOUBLE.

Why

The test takes ten seconds and halves the work. It is optional here but essential on curves where the full integral is heavy.

Step 3

$\mathcal{A}=\frac{1}{2}\int_{0}^{2\pi}\left(1+\cos\theta\right)^{2}d\theta=\frac{1}{2}\int_{0}^{2\pi}\left(1+2\cos\theta+\cos^{2}\theta\right)d\theta$ .

Why

The $\frac{1}{2}$ and the square are written before any expansion: it is the only protection against the two most frequent omissions of the chapter.

Step 4

Power reduction: $\cos^{2}\theta=\frac{1+\cos 2\theta}{2}$ , so the integrand becomes $\frac{3}{2}+2\cos\theta+\frac{\cos 2\theta}{2}$ .

Why

No table gives the antiderivative of $\cos^{2}$ directly. The double-angle formula is the only way, and it is the line the marking scheme explicitly expects.

Step 5

On $\left[0, 2\pi\right]$ , $\int\cos\theta\,d\theta=0$ and $\int\cos 2\theta\,d\theta=0$ : only $\frac{3}{2}\times 2\pi=3\pi$ remains. So $\mathcal{A}=\frac{1}{2}(3\pi)=\frac{3\pi}{2}$ .

Why

Both cosine terms vanish over a full turn, which makes the computation immediate. Noticing it before integrating saves two lines.

Step 6

Check: the cardioid lies inside the disk of radius $2$ , of area $4\pi\approx 12.57$ ; the answer $\frac{3\pi}{2}\approx 4.71$ is a little more than a third of it, which matches the drawing ✓.

Why

Bounding by a simple disk catches the missing factor $2$ : without the $\frac{1}{2}$ , you would have found $3\pi\approx 9.42$ , almost the whole area of the disk, which is visibly wrong.

The conclusion, written out

The area inside the cardioid $r=1+\cos\theta$ is $\frac{3\pi}{2}\approx 4.71$ square units, one and a half times the area of the unit disk.

The classic mistake on this problem: Integrating $\left(1+\cos\theta\right)^{2}$ without reducing $\cos^{2}\theta$ , or worse, writing $\left(1+\cos\theta\right)^{2}=1+\cos^{2}\theta$ . The cross term $2\cos\theta$ does not disappear, even if it ends up integrating to zero over a full turn.

Learn by heart

• $x=r\cos\theta$ , $y=r\sin\theta$ ; and for the converse, watch the QUADRANT.
• $r$ can be NEGATIVE: the point is plotted in the direction opposite to $\theta$ .
• The pole is written $\left(0,\theta\right)$ for every $\theta$ : it always escapes the intersection system.
• Area: $\frac{1}{2}\int r^{2}\,d\theta$ . The $\frac{1}{2}$ and the square, both of them.
• Bounds of a loop: the angles where $r$ vanishes.
• Rose $r=a\cos(n\theta)$ : $n$ petals if $n$ is odd, $2n$ if $n$ is even.
• Length: $\int\sqrt{r^{2}+\left[r'\right]^{2}}\,d\theta$ .
• Reduce $\cos^{2}$ and $\sin^{2}$ before integrating: the table does not give their antiderivatives.

Frequently asked questions

Can the polar radius be negative?

Yes. A negative radius means the point is plotted in the direction opposite to the given angle. This is what allows a rose with an even coefficient to have twice as many petals as its coefficient, half of them being traced with negative values of the radius.

Why does the system of equations not give all the intersections?

Because it assumes that the two curves pass through the same point at the same angle. But the pole has infinitely many names, one for each angle, so two curves can pass through it at different angles. To test the pole, you must therefore solve separately for each radius being zero.

Where does the factor one half in the polar area come from?

From the area of a circular sector, which is half the square of the radius times the angle. The elementary slice of a polar region is a very thin sector, not a rectangle. The immediate check is to apply the formula to a circle of constant radius, which must give back pi times the radius squared.

How do you find the bounds of a polar area?

By looking for the angles where the radius vanishes, because that is where the curve comes back to the pole and closes a loop. For an area between two curves, the bounds are the angles of intersection. Integrating beyond them counts regions already traced again and doubles or quadruples the answer.

How many petals does a polar rose have?

If the coefficient of the angle is odd, the rose has exactly that number of petals. If it is even, it has twice as many. The reason is the sign of the radius: with an odd coefficient, the second half of the turn retraces the same petals, while with an even coefficient it traces new ones.

Practise it

Corrected exercises: Polar coordinates, 201-SN3-RE (201-NYB)

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-polar-coordinates. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: polar coordinates, 201-SN3-RE (201-NYB)

The essentials

One point, infinitely many names

The formulas of the chapter, with their factors

The curves to recognize at a glance

The mistakes that cost marks

1. Rejecting a negative radius

2. Forgetting the pole in an intersection

3. Forgetting the factor one half in the area

4. Forgetting the square in the area

5. Integrating over a full turn when one loop is enough

6. Getting the number of petals wrong

7. Confusing the slope of the tangent with the derivative of the radius

8. Losing the quadrant when converting to polar

Which method to choose

How to find ALL the intersections of two polar curves

Which symmetries to use

How the answer is expected to be written

A polar area, written for the grader

Check before you hand in

The typical problem, taken apart

The area of a cardioid, from the bounds to the power reduction

Learn by heart

Frequently asked questions

Can the polar radius be negative?

Why does the system of equations not give all the intersections?

Where does the factor one half in the polar area come from?

How do you find the bounds of a polar area?

How many petals does a polar rose have?

Corrected exercises: Polar coordinates, 201-SN3-RE (201-NYB)

See also

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