CP-7 Polar coordinates — revision question pack

3 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section CP-7. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

CP-7.1 · Understand and use polar coordinates and be able to convert between polar and Cartesian coordinates.

Explanation

Polar coordinates $(r,\theta)$ locate a point at directed distance $r$ from the pole, with $\theta$ measured anticlockwise from the initial line.
Convert using $x=r\cos\theta$ and $y=r\sin\theta$ ; in reverse, use $r^2=x^2+y^2$ and determine $\theta$ from $\tan\theta=y/x$ with a quadrant check.
For a polar curve, multiplying by $r$ can expose the substitutions $r\cos\theta=x$ , $r\sin\theta=y$ and $r^2=x^2+y^2$ .
Coordinates are not unique: $(r,\theta)$ , $(r,\theta+2\pi)$ and $(-r,\theta+\pi)$ represent the same point.
Examiners expect an angle in the requested interval and a Cartesian equation simplified enough to identify the locus.

A polar point resolved into its Cartesian horizontal and vertical components.

Worked example

Convert $r=6\cos\theta-4\sin\theta$ to Cartesian form and identify the curve.

1.Multiply by $r$ : $r^2=6r\cos\theta-4r\sin\theta$ .
2.Substitute to obtain $x^2+y^2=6x-4y$ .
3.Complete the squares: $(x-3)^2+(y+2)^2=13$ .

Answer: The curve is a circle with centre $(3,-2)$ and radius $\sqrt{13}$ .

Common mistakes

Don't use $\arctan(y/x)$ without correcting the angle to the point's quadrant.
Don't replace $r$ by $x^2+y^2$ instead of replacing $r^2$ by $x^2+y^2$ .
Don't treat a negative value of $r$ as invalid rather than reversing the direction.

Exam tip

When an interval for $\theta$ is given, state the quadrant before selecting the inverse-tangent value.

Tier 1 · Easy

1.
Convert the polar coordinates $(4,\pi/6)$ to Cartesian coordinates.
(2)
(Total for Question 1 is 2 marks)
2.
The line $l$ has Cartesian equation $x+y=3$ . Find a polar equation of $l$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Express the Cartesian point $(-3,3\sqrt3)$ in polar form, taking $r>0$ and $0\leq\theta<2\pi$ .
(3)
(Total for Question 1 is 3 marks)
2.
The curve $C$ has equation $(x-2)(y+1)=6$ . Show that a polar equation of $C$ is $r^2\sin\theta\cos\theta+r\cos\theta-2r\sin\theta=8$ .
(4)
(Total for Question 2 is 4 marks)
3.
The point $P$ has Cartesian coordinates $(-5,-5)$ . (a) Find polar coordinates of $P$ with $r>0$ and $0\leq\theta<2\pi$ . (b) Find the polar coordinates of the reflection of $P$ in the initial line, again with $r>0$ and $0\leq\theta<2\pi$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Convert the polar curve $r=4\cos\theta+2\sin\theta$ to Cartesian form and identify the curve.
(5)
(Total for Question 1 is 5 marks)
2.
The curve $C$ has Cartesian equation $x^2-y^2=9$ . Show that a polar equation of $C$ is $r^2\cos2\theta=9$ , where $0\leq\theta<2\pi$ , and state the values of $\theta$ for which $C$ has no points.
(4)
(Total for Question 2 is 4 marks)
3.
The curve $C$ has polar equation $r^2\sin^2\theta-8r\cos\theta=16$ . Convert the equation to Cartesian form and identify the curve. Taking $r\geq0$ , state the polar coordinates of any point of $C$ on the initial line and on the half-line $\theta=\pi$ .
(5)
(Total for Question 3 is 5 marks)
4.
The polar curve has equation $r\cos(\theta-\pi/3)=4$ . Find its Cartesian equation. Hence find polar coordinates, with $r>0$ and $0\leq\theta<2\pi$ , of the point on the curve closest to the pole.
(5)
(Total for Question 4 is 5 marks)

CP-7.2 · Sketch curves with r given as a function of theta, including use of trigonometric functions.

Explanation

A polar sketch should be built from zeros and extreme values of $r$ , symmetry, and values on the initial line.
A negative radius places the point in the direction opposite to $\theta$ , so discarding negative values can remove a loop or petal.
The specification includes straight lines such as $r=p\sec(\alpha-\theta)$ , circles, spirals, cardioids, limacons, rose curves and $r^2=a^2\cos2\theta$ .
For tangents, write $x=r\cos\theta$ and $y=r\sin\theta$ : a tangent parallel to the initial line normally has $dy/d\theta=0$ , while a perpendicular tangent normally has $dx/d\theta=0$ .
Examiners expect key angles, pole crossings, symmetry and maximum radii to be labelled rather than an unsupported calculator sketch.

The four-petalled shape of the rose curve

r=a\cos4\theta

(under

r\geq0

), with its symmetry and angular directions labelled.

Worked example

Sketch $r=3(1+\cos\theta)$ for $0\leq\theta\leq2\pi$ , labelling the pole crossing and maximum radius.

1.The equation is unchanged by $\theta\mapsto-\theta$ , so the curve is symmetric about the initial line.
2. $r=0$ at $\theta=\pi$ , while $r=6$ at $\theta=0$ .
3.At $\theta=\pi/2$ and $3\pi/2$ , $r=3$ ; join the points as one cardioid with its cusp at the pole.

Answer: A right-facing cardioid, symmetric about the initial line, with cusp at the pole and furthest point $(6,0)$ .

Common mistakes

Don't plot a negative radius in the direction $\theta$ instead of the opposite direction.
Don't call $dy/d\theta=0$ sufficient for a horizontal tangent without checking that $dx/d\theta\ne0$ .
Don't sketch a rose petal at every zero of $r$ rather than using extrema to locate petal axes.

Exam tip

For a sketch question, find zeros and extrema of $r$ and label them on the curve before joining smoothly.

Tier 1 · Easy

1.
Sketch the polar curve $r=3$ , labelling its key geometric features.
(2)
(Total for Question 1 is 2 marks)
2.
Sketch the curve with polar equation $r=3+2\cos\theta$ for $0\leq\theta<2\pi$ , showing the value of $r$ on the initial line and on the line $\theta=\pi/2$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Sketch the polar curve $r=2\cos\theta$ . State its Cartesian equation, centre and radius.
(4)
(Total for Question 1 is 4 marks)
2.
Sketch the curve with polar equation $r=3\sin3\theta$ , $0\leq\theta\leq\pi/3$ , showing the values of $\theta$ at which the curve meets the pole, the value of $\theta$ at which $r$ is greatest, and the line of symmetry of the petal.
(4)
(Total for Question 2 is 4 marks)
3.
Taking $r\geq0$ , sketch the curve with polar equation $r=1+2\sin\theta$ . State the values of $\theta$ , $0\leq\theta<2\pi$ , for which $r\geq0$ , and label the pole crossings, the line of symmetry and the greatest value of $r$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
For $0\leq\theta<2\pi$ and taking $r\geq0$ , sketch the complete polar curve $r=4\sin2\theta$ , stating the intervals of $\theta$ on which $r\geq0$ and labelling the directions and lengths of all petals and the values of $\theta$ at which the curve passes through the pole.
(5)
(Total for Question 1 is 5 marks)
2.
Sketch the complete curve $r=4+2\sin2\theta$ for $0\leq\theta<2\pi$ , showing the value of $r$ on the initial line, its maximum and minimum radii, and its lines of symmetry.
(5)
(Total for Question 2 is 5 marks)
3.
Taking $r\geq0$ , sketch the complete curve $r^2=18\cos2\theta$ for $0\leq\theta<2\pi$ . Label the directions and lengths of its loops and state its lines of symmetry. Find the exact value of $r$ when $\theta=\dfrac\pi6$ .
(5)
(Total for Question 3 is 5 marks)
4.
On the same polar diagram, sketch $C:r=\theta$ and $K:r=\pi$ for $0\leq\theta\leq2\pi$ , taking $r\geq0$ . Hence find their point of intersection and the intervals for which $C$ lies inside and outside $K$ .
(5)
(Total for Question 4 is 5 marks)
5.
Taking $r\geq0$ , show that the curve $r=4/(2+\sin\theta)$ has Cartesian equation $4x^2+3y^2+8y=16$ . Sketch the complete curve for $0\leq\theta<2\pi$ , labelling its four axis points and its line of symmetry.
(5)
(Total for Question 5 is 5 marks)

CP-7.3 · Find the area enclosed by a polar curve.

Explanation

The polar area swept while $\theta$ runs from $\alpha$ to $\beta$ is $A=\dfrac12\int_\alpha^\beta r^2\,d\theta$ . The limits must describe the required region exactly; find them from pole crossings, intersections or symmetry, because one complete loop may occupy only part of a full revolution.
For a region between curves over the same angles, use $\dfrac12\int(r_{\text{outer}}^2-r_{\text{inner}}^2)\,d\theta$ , checking which radius is larger throughout.
Trigonometric identities are often needed before integration.
Examiners expect a sketch or clear limit justification, the factor $\tfrac12$ , correct squaring of $r$ , and an exact value unless a decimal is requested.
A symmetric multiplier is valid only when the chosen sector is not already the whole region.

Worked example

Find the exact area of one petal of $r=4\cos(2\theta)$ .

1.The petal centred on the initial line is traced from $\theta=-\pi/4$ to $\theta=\pi/4$ .
2. $A=\dfrac12\int_{-\pi/4}^{\pi/4}16\cos^2(2\theta)\,d\theta$ .
3.Using $\cos^2(2\theta)=\dfrac12(1+\cos4\theta)$ gives $A=4[\theta+\tfrac14\sin4\theta]_{-\pi/4}^{\pi/4}=2\pi$ .

Answer: The area of one petal is $2\pi$ square units.

Common mistakes

Don't integrate $r$ rather than $r^2$ in the polar area formula.
Don't omit the factor $\tfrac12$ from $A=\tfrac12\int r^2\,d\theta$ .
Don't use $0\leq\theta\leq2\pi$ for one loop, because this counts the same region repeatedly.

Exam tip

State how the intersection or pole-crossing equations produce the angular limits before evaluating the integral.

Tier 1 · Easy

1.
Find the exact area of the sector enclosed by $r=2$ and the rays $\theta=0$ and $\theta=\pi/3$ .
(3)
(Total for Question 1 is 3 marks)
2.
The curve $C$ has polar equation $r=2\theta$ , $0\leq\theta\leq\pi/2$ . Find the exact area of the region enclosed by $C$ and the line $\theta=\pi/2$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Determine the exact area enclosed by the loop $r=3\sin\theta$ .
(5)
(Total for Question 1 is 5 marks)
2.
For the curve $r=2+\cos\theta$ , where $0\leq\theta<2\pi$ , find the exact values of $\theta$ at which the tangent is (a) parallel to the initial line and (b) perpendicular to the initial line.
(6)
(Total for Question 2 is 6 marks)
3.
The curve $C$ has equation $r=4\sqrt{\cos\theta}$ for $-\dfrac\pi2\leq\theta\leq\dfrac\pi2$ , taking $r\geq0$ . Sketch $C$ , state its line of symmetry and find the exact area it encloses.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
The curves $r=2$ and $r=4\cos\theta$ enclose a region that lies inside $r=4\cos\theta$ but outside $r=2$ . Find its exact area.
(7)
(Total for Question 1 is 7 marks)
2.
Curves $C_1:r=3+\cos\theta$ and $C_2:r=3+\sin\theta$ are defined for $0\leq\theta<2\pi$ . Find the exact area of the region which lies inside $C_1$ and outside $C_2$ .
(8)
(Total for Question 2 is 8 marks)
3.
The curve $C$ has polar equation $r=3+3\cos2\theta$ , $0\leq\theta<2\pi$ . Taking $r\geq0$ , the parts of $C$ above the initial line enclose two separate regions with the initial line and the half-line $\theta=\pi$ . Find the exact area of each region and their total area.
(8)
(Total for Question 3 is 8 marks)
4.
The curves $C_1:r=5+3\cos\theta$ and $C_2:r=5$ are defined for $-\pi\leq\theta\leq\pi$ . Find the exact area of the region which lies inside $C_1$ and outside $C_2$ .
(8)
(Total for Question 4 is 8 marks)
5.
For $0\leq\theta<2\pi$ , the polar equation of $C$ is $r=a+b\cos\theta$ , where $a>b>0$ . The greatest distance of $C$ from the pole is $8$ , and the exact area enclosed by $C$ is $38\pi$ . Find the values of $a$ and $b$ , and state the least distance of $C$ from the pole.
(8)
(Total for Question 5 is 8 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

CP-7.1 · Understand and use polar coordinates and be able to convert between polar and Cartesian coordinates.

Tier 1 · Easy

Mark scheme for CP-7.1 Tier 1 · Easy
Question	Scheme	Marks
1	$(x,y)=(2\sqrt3,2)$	2
(2 marks)		2
Notes
$x=4\cos(\pi/6)=4(\sqrt3/2)=2\sqrt3$ and $y=4\sin(\pi/6)=4(1/2)=2$ .
2	$x=r\cos\theta$ and $y=r\sin\theta$ $r(\cos\theta+\sin\theta)=3$ $r=\dfrac{3}{\cos\theta+\sin\theta}$ , or equivalently $r=\dfrac{3}{\sqrt2}\sec\left(\theta-\dfrac\pi4\right)$	3
(3 marks)		3
Notes
Substitute $x=r\cos\theta$ and $y=r\sin\theta$ into $x+y=3$ . This gives $r(\cos\theta+\sin\theta)=3$ , or $r=3/(\cos\theta+\sin\theta)$ .

Tier 2 · Standard

Mark scheme for CP-7.1 Tier 2 · Standard
Question	Scheme	Marks
1	$(r,\theta)=(6,2\pi/3)$	3
(3 marks)		3
Notes
$r=\sqrt{(-3)^2+(3\sqrt3)^2}=\sqrt{9+27}=6$ . The point lies in quadrant II and $\tan\theta=(3\sqrt3)/(-3)=-\sqrt3$ , so $\theta=2\pi/3$ .
2	$xy+x-2y=8$ $x=r\cos\theta$ and $y=r\sin\theta$ $xy=r^2\sin\theta\cos\theta$ $r^2\sin\theta\cos\theta+r\cos\theta-2r\sin\theta=8$	4
(4 marks)		4
Notes
Expanding gives $xy+x-2y=8$ . Substitute $x=r\cos\theta$ and $y=r\sin\theta$ , so $xy=r^2\sin\theta\cos\theta$ . This gives the required polar equation.
3	$r=\sqrt{(-5)^2+(-5)^2}=5\sqrt2$ $P$ is in quadrant III, so $\theta=\dfrac{5\pi}{4}$ . Reflection in the initial line gives the Cartesian point $(-5,5)$ . Its polar coordinates are $\left(5\sqrt2,\dfrac{3\pi}{4}\right)$ .	4
(4 marks)		4
Notes
The radius is $\sqrt{(-5)^2+(-5)^2}=5\sqrt2$ . Since $P$ is in quadrant III with reference angle $\pi/4$ , its angle in $0\leq\theta<2\pi$ is $5\pi/4$ . Reflection in the initial line negates the angle, giving $-5\pi/4$ ; normalising into $0\leq\theta<2\pi$ gives $3\pi/4$ . The reflected point is therefore $(-5,5)$ with polar coordinates $(5\sqrt2,3\pi/4)$ .

Tier 3 · Hard

Mark scheme for CP-7.1 Tier 3 · Hard
Question	Scheme	Marks
1	$x^2+y^2=4x+2y$ $(x-2)^2+(y-1)^2=5$ A circle with centre $(2,1)$ and radius $\sqrt5$	5
(5 marks)		5
Notes
Multiply by $r$ : $r^2=4r\cos\theta+2r\sin\theta$ . Hence $x^2+y^2=4x+2y$ . Completing the square gives $(x-2)^2+(y-1)^2=5$ , so the curve is the stated circle.
2	$x^2-y^2=r^2(\cos^2\theta-\sin^2\theta)$ $\cos^2\theta-\sin^2\theta=\cos2\theta$ $r^2\cos2\theta=9$ No points for $\dfrac\pi4\leq\theta\leq\dfrac{3\pi}4$ or $\dfrac{5\pi}4\leq\theta\leq\dfrac{7\pi}4$ .	4
(4 marks)		4
Notes
Substitute $x=r\cos\theta$ and $y=r\sin\theta$ . Then $x^2-y^2=r^2(\cos^2\theta-\sin^2\theta)=r^2\cos2\theta$ , giving $r^2\cos2\theta=9$ . Since $r^2\geq0$ , points require $\cos2\theta\geq0$ ; and where $\cos2\theta=0$ the equation becomes $0=9$ , so no point exists there either, so points exist only when $\cos2\theta>0$ .
3	$r^2\sin^2\theta=y^2$ and $r\cos\theta=x$ . $y^2=8(x+2)$ $C$ is a parabola with vertex $(-2,0)$ , axis the $x$ -axis, opening to the right. At $\theta=0$ , the equation would require $r=-2$ , so there is no point of $C$ on the initial line when $r\geq0$ . At $\theta=\pi$ , $r=2$ , so the point on that half-line has polar coordinates $(2,\pi)$ .	5
(5 marks)		5
Notes
Use $r\sin\theta=y$ and $r\cos\theta=x$ to give $y^2-8x=16$ , or $y^2=8(x+2)$ . This is a right-opening parabola with vertex $(-2,0)$ . On the initial line, $\theta=0$ gives $-8r=16$ , which has no non-negative solution. On the half-line $\theta=\pi$ , the equation gives $8r=16$ , so $r=2$ and the polar coordinates are $(2,\pi)$ .
4	$\cos(\theta-\pi/3)=\dfrac12\cos\theta+\dfrac{\sqrt3}{2}\sin\theta$ $\dfrac12x+\dfrac{\sqrt3}{2}y=4$ $x+\sqrt3y=8$ The perpendicular distance from the pole is $8/\sqrt{1+3}=4$ , in the direction $(1,\sqrt3)$ . The closest point is $(2,2\sqrt3)$ , with polar coordinates $(4,\pi/3)$ .	5
(5 marks)		5
Notes
Expanding the shifted cosine and using $r\cos\theta=x$ and $r\sin\theta=y$ gives the line $x+\sqrt3y=8$ . Its normal $(1,\sqrt3)$ points at angle $\pi/3$ and has magnitude $2$ . The perpendicular from the pole therefore has length $8/2=4$ and meets the line at $(2,2\sqrt3)$ , giving polar coordinates $(4,\pi/3)$ .

CP-7.2 · Sketch curves with r given as a function of theta, including use of trigonometric functions.

Tier 1 · Easy

Mark scheme for CP-7.2 Tier 1 · Easy
Question	Scheme	Marks
1	A circle centred at the pole with radius $3$ . It passes through $(3,0)$ at $\theta=0$ and $(-3,0)$ at $\theta=\pi$ .	2
(2 marks)		2
Notes
Every point has constant distance $3$ from the pole while $\theta$ ranges through a complete turn. The locus is therefore the circle $x^2+y^2=9$ , centred at the pole with radius $3$ ; it meets the positive initial line at $(3,0)$ and its negative continuation at $(-3,0)$ .
2	$r=5$ at $\theta=0$ . $r=3$ at $\theta=\pi/2$ . A closed limacon symmetric about the initial line, drawn with $r\geq0$ .	3
(3 marks)		3
Notes
At $\theta=0$ , $r=5$ , and at $\theta=\pi/2$ , $r=3$ . Since $r(-\theta)=r(\theta)$ , the curve is symmetric about the initial line. Draw the closed limacon smoothly using non-negative radii.

Tier 2 · Standard

Mark scheme for CP-7.2 Tier 2 · Standard
Question	Scheme	Marks
1	$x^2+y^2=2x$ $(x-1)^2+y^2=1$ Centre $(1,0)$ , radius $1$	4
(4 marks)		4
Notes
Multiply by $r$ to obtain $r^2=2r\cos\theta$ , so $x^2+y^2=2x$ . Completing the square gives $(x-1)^2+y^2=1$ . Sketch this circle through the pole and $(2,0)$ , symmetric about the initial line.
2	The greatest radius is $r=3$ at $\theta=\pi/6$ . $r=0$ at $\theta=0$ and $\theta=\pi/3$ . The curve is one petal of length $3$ . The petal is symmetric about $\theta=\pi/6$ .	4
(4 marks)		4
Notes
On $0\leq\theta\leq\pi/3$ , the radius is non-negative. The table $(\theta,r)=(0,0),(\pi/6,3),(\pi/3,0)$ traces one petal. Since $r(\pi/3-\theta)=r(\theta)$ , it is symmetric about $\theta=\pi/6$ .
3	$r\geq0$ for $0\leq\theta\leq\dfrac{7\pi}{6}$ or $\dfrac{11\pi}{6}\leq\theta<2\pi$ , with pole crossings at the two included endpoints. It is symmetric about the line $\theta=\dfrac\pi2$ . The greatest radius is $3$ at $\theta=\dfrac\pi2$ . The sketch is the outer loop of an upward-facing limacon, passing through radius $1$ at $\theta=0$ and $\theta=\pi$ .	4
(4 marks)		4
Notes
The condition $1+2\sin\theta\geq0$ is equivalent to $\sin\theta\geq-1/2$ , giving the stated intervals and pole crossings. Since $\sin(\pi-\theta)=\sin\theta$ , the curve is symmetric about $\theta=\pi/2$ . Its maximum radius is $3$ when $\sin\theta=1$ . The points at $\theta=0,\pi$ both have radius $1$ , fixing the outer-loop sketch.

Tier 3 · Hard

Mark scheme for CP-7.2 Tier 3 · Hard
Question	Scheme	Marks
1	$r\geq0$ only for $\theta\in[0,\pi/2]\cup[\pi,3\pi/2]$ . A two-petalled rose. Petals along $\theta=\pi/4$ and $5\pi/4$ , in the first and third quadrants. Each petal has length $4$ . The curve passes through the pole at $\theta=0,\pi/2,\pi,3\pi/2$ .	5
(5 marks)		5
Notes
Under the $r\geq0$ convention, the curve exists only where $\sin2\theta\geq0$ , namely $\theta\in[0,\pi/2]\cup[\pi,3\pi/2]$ . It reaches the pole at $\theta=0,\pi/2,\pi,3\pi/2$ . The maximum radius $4$ occurs at $\theta=\pi/4$ and $5\pi/4$ , so the curve has two petals of length $4$ , in the first and third quadrants.
2	$r=4$ at $\theta=0,\pi/2,\pi,3\pi/2$ . The maximum radius is $6$ at $\theta=\pi/4$ and $5\pi/4$ . The minimum radius is $2$ at $\theta=3\pi/4$ and $7\pi/4$ , so there is no inner loop or pole contact. A closed two-lobed curve, with lobes along $\theta=\pi/4$ and $5\pi/4$ . Symmetric about $\theta=\pi/4$ and $\theta=3\pi/4$ , hence half-turn symmetry about the pole.	5
(5 marks)		5
Notes
The radius is $4$ on both coordinate axes. Since $-1\leq\sin2\theta\leq1$ , the maximum radius is $6$ along $\theta=\pi/4,5\pi/4$ and the minimum is $2$ along $\theta=3\pi/4,7\pi/4$ . Thus $r$ is always positive. Also $r(\theta+\pi)=r(\theta)$ and $r(\pi/2-\theta)=r(\theta)$ , giving the stated symmetries and two-lobed shape.
3	The maximum radius is $3\sqrt2$ at $\theta=0$ and $\theta=\pi$ . There are two equal loops directed along the initial line and the half-line $\theta=\pi$ . The curve is symmetric about both the initial line and the line $\theta=\pi/2$ . At $\theta=\dfrac\pi6$ , $r^2=18\cos\dfrac\pi3=9$ . Since $r\geq0$ , the exact value is $r=3$ .	5
(5 marks)		5
Notes
The maximum $r^2=18$ occurs when $\cos2\theta=1$ , at $\theta=0,\pi$ , so the horizontal lemniscate has two loops of length $3\sqrt2$ . The equation is unchanged by reflection in either coordinate axis, giving symmetry about the initial line and the line $\theta=\pi/2$ . At $\theta=\pi/6$ , $r^2=18\cos(\pi/3)=9$ ; the stated convention selects $r=3$ .
4	$C$ starts at the pole, makes one anticlockwise turn with steadily increasing radius and ends at radius $2\pi$ on the initial line. $K$ is the circle centred at the pole with radius $\pi$ ; both curves are shown on the same diagram. They meet where $\theta=\pi$ , at the point with polar coordinates $(\pi,\pi)$ . $C$ lies inside $K$ for $0\leq\theta<\pi$ . $C$ lies outside $K$ for $\pi<\theta\leq2\pi$ .	5
(5 marks)		5
Notes
The radius of $C$ increases continuously from $0$ to $2\pi$ during one complete anticlockwise turn, while $K$ has constant radius $\pi$ . Equality of the radii gives $\theta=\pi$ and therefore the intersection $(\pi,\pi)$ in polar form. Comparing $\theta$ with $\pi$ gives the two stated intervals.
5	$2r+r\sin\theta=4$ , so $2r+y=4$ . $4(x^2+y^2)=(4-y)^2$ $4x^2+3y^2+8y=16$ The axis points, in polar form, are $(2,0)$ , $(2,\pi)$ , $(4/3,\pi/2)$ and $(4,3\pi/2)$ . The sketch is a closed oval through these points, elongated below the pole and symmetric about the line $\theta=\pi/2$ .	5
(5 marks)		5
Notes
Multiplying by the denominator gives $2r+y=4$ . Squaring $2r=4-y$ and using $r^2=x^2+y^2$ gives the stated Cartesian equation. Evaluating the polar equation at $0$ , $\pi$ , $\pi/2$ and $3\pi/2$ gives the four labelled polar points. Since $\sin(\pi-\theta)=\sin\theta$ , the curve is symmetric about $\theta=\pi/2$ .

CP-7.3 · Find the area enclosed by a polar curve.

Tier 1 · Easy

Mark scheme for CP-7.3 Tier 1 · Easy
Question	Scheme	Marks
1	Area $=\dfrac{2\pi}{3}$	3
(3 marks)		3
Notes
$A=\frac12\int_0^{\pi/3}2^2\,d\theta=2[\theta]_0^{\pi/3}=2\pi/3$ .
2	$A=\dfrac12\int_0^{\pi/2}4\theta^2\,d\theta$ $=2\left[\dfrac{\theta^3}{3}\right]_0^{\pi/2}$ $A=\dfrac{\pi^3}{12}$	3
(3 marks)		3
Notes
The curve starts at the pole when $\theta=0$ and meets the boundary line when $\theta=\pi/2$ . Hence $A=\tfrac12\int_0^{\pi/2}(2\theta)^2\,d\theta=2[\theta^3/3]_0^{\pi/2}=\pi^3/12$ .

Tier 2 · Standard

Mark scheme for CP-7.3 Tier 2 · Standard
Question	Scheme	Marks
1	Area $=\dfrac{9\pi}{4}$	5
(5 marks)		5
Notes
The loop is traced once for $0\leq\theta\leq\pi$ . Hence $A=\frac12\int_0^\pi9\sin^2\theta\,d\theta$ . Since $\int_0^\pi\sin^2\theta\,d\theta=\pi/2$ , the area is $9\pi/4$ .
2	$\dfrac{dx}{d\theta}=-2\sin\theta(1+\cos\theta)$ . $\dfrac{dy}{d\theta}=2\cos^2\theta+2\cos\theta-1$ . $dy/d\theta=0$ gives $\cos\theta=(-1\pm\sqrt3)/2$ ; reject $(-1-\sqrt3)/2<-1$ . Parallel: $\theta=\arccos\left(\dfrac{\sqrt3-1}{2}\right)$ or $2\pi-\arccos\left(\dfrac{\sqrt3-1}{2}\right)$ , and $dx/d\theta\ne0$ there. $dx/d\theta=0$ gives $\theta=0$ or $\pi$ . Perpendicular: $\theta=0$ or $\pi$ , and $dy/d\theta=3$ or $-1$ respectively.	6
(6 marks)		6
Notes
Using $x=(2+\cos\theta)\cos\theta$ and $y=(2+\cos\theta)\sin\theta$ gives $dx/d\theta=-2\sin\theta(1+\cos\theta)$ and $dy/d\theta=2\cos^2\theta+2\cos\theta-1$ . A tangent parallel to the initial line has $dy/d\theta=0$ , so $\cos\theta=(\sqrt3-1)/2$ . A perpendicular tangent has $dx/d\theta=0$ , giving $\theta=0,\pi$ . The other derivatives are non-zero at these values.
3	$r=0$ at $\theta=\pm\dfrac\pi2$ and its greatest value is $4$ at $\theta=0$ , fixing the closed-loop sketch. Since $r(-\theta)=r(\theta)$ , $C$ is symmetric about the initial line. $A=\dfrac12\int_{-\pi/2}^{\pi/2}16\cos\theta\,d\theta$ $A=16$ square units.	4
(4 marks)		4
Notes
Cosine is non-negative on the stated interval, so $r$ is real and non-negative throughout, with $r=0$ at both endpoints and $r=4$ at $\theta=0$ . Since cosine is even, the curve is symmetric about the initial line. Its area is $\tfrac12\int_{-\pi/2}^{\pi/2}16\cos\theta\,d\theta=8[\sin\theta]_{-\pi/2}^{\pi/2}=16$ .

Tier 3 · Hard

Mark scheme for CP-7.3 Tier 3 · Hard
Question	Scheme	Marks
1	Area $=\dfrac{4\pi}{3}+2\sqrt3$	7
(7 marks)		7
Notes
The curves meet when $2=4\cos\theta$ , so $\theta=\pm\pi/3$ for the required region. Thus $A=\frac12\int_{-\pi/3}^{\pi/3}(16\cos^2\theta-4)\,d\theta$ . Using $\cos^2\theta=(1+\cos2\theta)/2$ , the integral simplifies to $\frac12(16\pi/3+4\sqrt3-8\pi/3)=4\pi/3+2\sqrt3$ .
2	$C_1=C_2$ when $\cos\theta=\sin\theta$ . The intersections are at $\theta=\pi/4$ and $5\pi/4$ . $C_1\geq C_2$ on $[0,\pi/4]\cup[5\pi/4,2\pi]$ . $A=\dfrac12\left(\int_0^{\pi/4}+\int_{5\pi/4}^{2\pi}\right)\left((3+\cos\theta)^2-(3+\sin\theta)^2\right)d\theta$ The integrand is $\cos(2\theta)+6\cos\theta-6\sin\theta$ . An antiderivative is $\dfrac12\sin(2\theta)+6\sin\theta+6\cos\theta$ . $\int_0^{\pi/4}+\int_{5\pi/4}^{2\pi}$ of the squared-radius difference $=12\sqrt2$ . $A=6\sqrt2$	8
(8 marks)		8
Notes
The curves meet where $\cos\theta=\sin\theta$ , giving $\theta=\pi/4,5\pi/4$ . Testing $\theta=0$ and $\theta=\pi$ shows that $C_1$ is outside $C_2$ on $[0,\pi/4]\cup[5\pi/4,2\pi]$ . The squared-radius difference has antiderivative $\tfrac12\sin2\theta+6\sin\theta+6\cos\theta$ ; evaluating the two integrals gives $12\sqrt2$ before the factor $1/2$ , hence $A=6\sqrt2$ .
3	$r=6\cos^2\theta$ , so $r\geq0$ throughout. $r=0$ at $\theta=\pi/2$ , separating the two upper regions. The right region is traced for $0\leq\theta\leq\pi/2$ and the left for $\pi/2\leq\theta\leq\pi$ . $A_{\rm right}=\dfrac12\int_0^{\pi/2}36\cos^4\theta\,d\theta$ $\cos^4\theta=\dfrac{3+4\cos2\theta+\cos4\theta}{8}$ $A_{\rm right}=\dfrac{27\pi}{8}$ By symmetry, $A_{\rm left}=\dfrac{27\pi}{8}$ . The total area is $\dfrac{27\pi}{4}$ square units.	8
(8 marks)		8
Notes
Since $1+\cos2\theta=2\cos^2\theta$ , the curve has $r=6\cos^2\theta$ and reaches the pole at $\theta=\pi/2$ . This splits the part above the initial line into the intervals $[0,\pi/2]$ and $[\pi/2,\pi]$ . For the right region, $A=18\int_0^{\pi/2}\cos^4\theta\,d\theta$ . Using the power-reduction identity gives $\int_0^{\pi/2}\cos^4\theta\,d\theta=3\pi/16$ , so its area is $27\pi/8$ . Symmetry gives the same left area and total $27\pi/4$ .
4	$5+3\cos\theta=5$ at the intersections. The intersections occur at $\theta=-\pi/2$ and $\theta=\pi/2$ . $C_1$ is outside $C_2$ for $-\pi/2\leq\theta\leq\pi/2$ . $A=\dfrac12\int_{-\pi/2}^{\pi/2}\left((5+3\cos\theta)^2-25\right)d\theta$ The squared-radius difference is $30\cos\theta+9\cos^2\theta$ . $\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta=2$ $\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta=\dfrac\pi2$ $A=30+\dfrac{9\pi}{4}$ square units.	8
(8 marks)		8
Notes
The curves meet when $3\cos\theta=0$ , at $\theta=\pm\pi/2$ . Between these angles $C_1$ is the outer curve. The squared-radius difference is $30\cos\theta+9\cos^2\theta$ . Its two required integrals are $2$ and $\pi/2$ , so applying the factor $1/2$ gives $30+9\pi/4$ .
5	The greatest radius occurs at $\theta=0$ , so $a+b=8$ . $38\pi=\dfrac12\int_0^{2\pi}(a+b\cos\theta)^2\,d\theta$ $(a+b\cos\theta)^2=a^2+2ab\cos\theta+b^2\cos^2\theta$ $\dfrac12\int_0^{2\pi}(a+b\cos\theta)^2\,d\theta=\pi a^2+\dfrac{\pi b^2}{2}$ $a^2+\dfrac{b^2}{2}=38$ Substituting $b=8-a$ gives $3a^2-16a-12=0$ . The roots are $a=6$ and $a=-\dfrac23$ ; since $a>b>0$ , $a=6$ and $b=2$ . The least radius is $a-b=4$ , attained at $\theta=\pi$ .	8
(8 marks)		8
Notes
Because $a>b>0$ , the greatest radius is $a+b$ at $\theta=0$ , giving $a+b=8$ . Expanding the polar area integral over a complete turn gives $A=\pi a^2+\pi b^2/2$ . Equating this to $38\pi$ and substituting $b=8-a$ gives $3a^2-16a-12=0$ . Its roots are $6$ and $-2/3$ , so the stated inequalities select $a=6$ , $b=2$ . The least radius is then $a-b=4$ at $\theta=\pi$ .