What are the steps for finding an area between two polar curves?

Sketch both curves and find the angle at which they meet. Draw the ray at that angle, then split the region into a sum or a difference of two polar areas.

True or False? An area between two polar curves is always the difference of two integrals.

False. It depends on the region. Where one curve bounds it on one side of the ray and the other bounds it on the other side, the area is a sum instead.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 9: Parametric Equations, Vector-Valued Functions & Polar CoordinatesPolar Coordinates

Polar Coordinates (College Board AP® Calculus BC): Flashcards

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FrontDefining Polar Coordinates
Define polar coordinates.

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Cards in this collection (27)

Define polar coordinates.
A way of locating a point by a distance $r$ from a fixed origin called the pole, and an angle $θ$ measured in radians from an initial line.
They are written $(r, θ)$ .
How do you convert polar coordinates into Cartesian coordinates?
Use $x = r \cos θ$ and $y = r \sin θ$ .
These come from the right triangle with hypotenuse $r$ and angle $θ$ at the pole.
A point has polar coordinates $(r, θ)$ and Cartesian coordinates $(x, y)$ . Making $r^{2}$ and $θ$ the subject gives
$r^{2} = x^{2} + y^{2}$ and $\tan θ =______$
The completed relationship is $\tan θ = \frac{y}{x}$ .
Dividing $y = r \sin θ$ by $x = r \cos θ$ cancels the $r$ and leaves $\tan θ$ .
True or False?
A point in the plane has exactly one pair of polar coordinates.
False.
Adding $2 π$ to $θ$ describes the same point, and allowing $r$ to be negative describes it again.
Every point has infinitely many polar coordinate pairs.
What does a negative value of $r$ mean?
The point is plotted in the opposite direction to $θ$ .
So $(- r, θ)$ is the same point as $(r, θ + π)$ .
Convert the polar equation $r = 4 \cos θ$ into $x$ and $y$ , and say what curve it is.
Multiplying by $r$ gives $r^{2} = 4 r \cos θ$ , so $x^{2} + y^{2} = 4 x$ .
Completing the square gives ${(x - 2)}^{2} + y^{2} = 2^{2}$ , a circle of radius $2$ centred at $(2, 0)$ .
What are the polar equations of a circle of radius $a$ centred at the pole, and of the line $x = k$ ?
The circle is simply $r = a$ .
The vertical line becomes $r \cos θ = k$ , which rearranges to $r = k \sec θ$ .
A question asks for the average distance from the origin to a point on $r (θ)$ . What do you use?
The ordinary average value formula, applied to $r$ with respect to $θ$ .
That gives $\frac{1}{β - α} \int_{α}^{β} r (θ) d θ$ , so nothing new has to be learned.
What does the sign of $\frac{d r}{d θ}$ tell you about a polar curve?
A positive value means points on the curve are moving further from the origin as $θ$ increases.
A negative value means they are moving closer to it.
Why does finding $\frac{d x}{d θ}$ from $x = r \cos θ$ need the product rule?
Because $r$ is not a constant: it varies with $θ$ .
Differentiating the product gives $\frac{d x}{d θ} = \frac{d r}{d θ} \cos θ - r \sin θ$ .
For a polar curve $r (θ)$ , differentiating $y = r \sin θ$ by the product rule gives
$\frac{d y}{d θ} = \frac{d r}{d θ} \sin θ +______\cos θ$
The completed derivative is $\frac{d y}{d θ} = \frac{d r}{d θ} \sin θ + r \cos θ$ .
The product rule gives one term from differentiating $r$ and one from differentiating $\sin θ$ , and the second keeps $r$ as it stands.
What is the slope $\frac{d y}{d x}$ of a polar curve?
It is $\frac{\frac{d r}{d θ} \sin θ + r \cos θ}{\frac{d r}{d θ} \cos θ - r \sin θ}$ , which is $\frac{d y}{d θ}$ divided by $\frac{d x}{d θ}$ .
True or False?
The second derivative formula for a polar curve has the same shape as the one for a parametric curve.
True.
It is $\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d θ} (\frac{d y}{d x})}{\frac{d x}{d θ}}$ , which is the parametric formula with $θ$ in place of $t$ .
A polar curve is a parametric curve whose parameter is the angle.
Find $\frac{d y}{d x}$ for $r = e^{- θ}$ and evaluate it at $θ = \frac{π}{4}$ .
With $\frac{d r}{d θ} = - e^{- θ}$ the factor $- e^{- θ}$ cancels, leaving $\frac{\sin θ - \cos θ}{\cos θ + \sin θ}$ .
At $θ = \frac{π}{4}$ the numerator is zero, so the slope is $0$ .
A particle moves on a polar curve with $\frac{d θ}{d t} = 3$ . How do you find $\frac{d y}{d t}$ ?
Use the chain rule, $\frac{d y}{d t} = \frac{d y}{d θ} \cdot \frac{d θ}{d t}$ .
Work out $\frac{d y}{d θ}$ from the curve, then multiply by the given rate.
The area bounded by the polar curve $r = f (θ)$ and the rays $θ = α$ and $θ = β$ is
$A = \int_{α}^{β} \frac{1}{2}______d θ$
The completed formula is $A = \int_{α}^{β} \frac{1}{2} r^{2} d θ$ .
The integrand is half the square of the distance from the pole, and the limits are the two angles bounding the region.
Define a ray, as the term is used in a polar area question.
A straight line extending from the pole in a fixed direction $θ = α$ .
Two rays bound the region, and the area is swept out anticlockwise between them.
True or False?
The polar area formula fails on parts of a curve where $r$ is negative.
False.
The formula still works, because $r$ is squared in the integrand.
A loop traced out with negative $r$ has its area found in exactly the same way.
The curve $r = \sin 3 θ$ has three loops. How do you find the rays bounding one of them?
A loop starts and ends where $r = 0$ , so solve $\sin 3 θ = 0$ .
That gives $θ = 0$ , $\frac{π}{3}$ and $\frac{2 π}{3}$ , and consecutive pairs bound the loops.
Write down the integral for the area of the loop of $r = \sin 3 θ$ between $θ = 0$ and $θ = \frac{π}{3}$ .
It is $\int_{0}^{\frac{π}{3}} \frac{1}{2} {(\sin 3 θ)}^{2} d θ$ , which comes to $0.262$ to 3 decimal places.
How can symmetry shorten a polar area calculation?
A larger area can often be written as a multiple of a smaller one.
The total area enclosed by $r = \sin 3 θ$ is three times one loop, since the three loops are congruent.
What are the steps for finding an area between two polar curves?
Sketch both curves and find the angle at which they meet.
Draw the ray at that angle, then split the region into a sum or a difference of two polar areas.
True or False?
An area between two polar curves is always the difference of two integrals.
False.
It depends on the region.
Where one curve bounds it on one side of the ray and the other bounds it on the other side, the area is a sum instead.
How do you tell which of two polar curves is the outer one?
Substitute a value of $θ$ from the range into both and compare the two values of $r$ .
At $θ = 0$ , $r = 3 \cos θ$ gives $3$ while $r = 1 + \cos θ$ gives $2$ , so the first lies outside.
Two polar areas over the same limits are subtracted. Because the limits match, they can be written as one integral:
$\int_{0}^{\frac{π}{3}} \frac{1}{2} [{(3 \cos θ)}^{2} -______] d θ$
The completed integral is $\int_{0}^{\frac{π}{3}} \frac{1}{2} [{(3 \cos θ)}^{2} - {(1 + \cos θ)}^{2}] d θ$ .
The two integrands are combined inside one set of brackets, with the outer curve squared first.
A region runs from $θ = 0$ to $\frac{π}{3}$ along $r = 1 + \cos θ$ , then from $\frac{π}{3}$ to $\frac{π}{2}$ along $r = 3 \cos θ$ . Write its area.
It is $\int_{0}^{\frac{π}{3}} \frac{1}{2} {(1 + \cos θ)}^{2} d θ + \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{1}{2} {(3 \cos θ)}^{2} d θ$ , which comes to $1.963$ to 3 decimal places.
When can you avoid integrating for one of the two polar areas?
When that part is a sector of the circle $r = a$ .
The whole circle has area $π a^{2}$ , so taking a known fraction of it is quicker than an integral.

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