Area of a polar region

Learning goals

Identifies the angular interval tracing the desired region once.
Sets up the polar-area integral 1/2∫r² dθ.
Uses symmetry to simplify the calculation when appropriate.
Evaluates the integral and reports nonnegative geometric area.

Difficulty: 1 – 5

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Area of a polar region 1

A polar region has area \(7\). A new curve is formed by multiplying every radius of its boundary by \(3\) while keeping the same angular interval. What is the area of the new region?

Hints

- Consider how the area formula changes when the radius is scaled. - Compare the square of the new radius with the square of the original radius.

Solution

1. Polar area depends on \(r^2\). 2. Multiplying every radius by \(3\) multiplies every squared radius by \(9\). 3. Therefore, the new area is \(9\cdot7=63\).

Answer

\(63\)

Area of a polar region 2

The shaded part of the circle \(r=2\) is shown. What interval of polar angles sweeps the shaded region exactly once?

Hints

- Identify the rays that bound the shaded region. - Translate those two coordinate-axis rays into polar angles. - The interval should sweep the shaded region once without continuing into an unshaded quadrant.

Solution

1. The shaded region starts on the positive x-axis, which corresponds to \(\theta=0\). 2. It ends on the positive y-axis, which corresponds to \(\theta=\frac{\pi}{2}\). 3. Therefore, the region is swept exactly once for \(0\le\theta\le\frac{\pi}{2}\).

Answer

\(0\le\theta\le\frac{\pi}{2}\)

Area of a polar region 3

A closed polar curve is traced exactly once for \(0\le\theta\le2\pi\) and satisfies \(r^2=5-\cos\theta\). Set up, but do not evaluate, an integral for its enclosed area.

Hints

- Compare the quantity given in the equation with the quantity that determines a small polar sector's area. - The stated interval already represents one complete tracing. - Do not introduce an unnecessary square root when building the area model.

Solution

1. Polar area is accumulated from \(\frac{1}{2}r^2\,d\theta\). 2. Because \(r^2\) is already given, substitute it directly. 3. The required setup is \(A=\frac{1}{2}\int_{0}^{2\pi}(5-\cos\theta)\,d\theta\).

Answer

\(A=\frac{1}{2}\int_{0}^{2\pi}(5-\cos\theta)\,d\theta\)

Area of a polar region 4

The marked arc and its two radius segments bound what fraction of the full disk: one-quarter, one-half, or three-quarters?

Hints

- Compare the sector angle with a full turn. - The radius stays the same around the arc.

Solution

1. The two radii point along the positive x- and y-axes, making a right angle. 2. A right angle is one-quarter of a full turn, so the sector is one-quarter of the disk.

Answer

One-quarter.

Area of a polar region 5

For \(r=2\sin\theta\) on \(0\le\theta\le\pi\), does the point trace one complete loop or two? Explain using the radius and the interval.

Hints

- What would mark the start and end of a loop at the pole? - Could two different angles in this interval represent the same point away from the pole?

Solution

1. The radius begins at zero, becomes positive, and returns to zero without becoming negative on this interval. 2. The curve forms one circle and visits that loop once.

Answer

One complete loop. The radius is zero at \(\theta=0\) and \(\theta=\pi\) and positive between them. As the angle increases over this interval, the point goes around the circle once and returns to the pole.

Area of a polar region 6

Find the polar-sector area for the constant radius \(r=2\) and angles \(0\le\theta\le\frac{\pi}{3}\).

Hints

- Sector area is proportional to its angle in radians. - The radius does not vary over this sector.

Solution

1. A sector of constant radius \(r\) and angle \(\Delta\theta\) has area \(\frac{1}{2}r^2\Delta\theta\). 2. Substitute to get \(\frac{1}{2}\cdot4\cdot\frac{\pi}{3}=\frac{2\pi}{3}\).

Answer

\(\frac{2\pi}{3}\) square units.

Area of a polar region 7

At an angle where a polar curve has \(r=-2\), does the small swept area \(\frac{1}{2}r^2\,d\theta\) have a positive or negative coefficient? Explain.

Hints

- Does reversing the radius direction reverse the area of a small sector? - For an increasing angle, consider what determines the sign of the sector-area coefficient.

Solution

1. Squaring the signed radius gives \((-2)^2=4\). 2. The area coefficient is \(\frac{1}{2}\cdot4=2\), which is positive.

Answer

Positive; squaring the radius gives \(r^2=(-2)^2=4\), so the area coefficient is \(\frac{1}{2}r^2=2\).

Area of a polar region 8

Find the exact area of the sector traced by the polar curve \(r=2\) over \(0\le\theta\le\frac{\pi}{3}\).

Hints

- Think of the region as being built from many thin polar sectors as the angle changes. - Decide how the radius of each small sector affects its area contribution. - Keep the exact angular width in radians so the final answer remains exact.

Solution

1. Use the polar area formula \(A=\frac{1}{2}\int_{0}^{\pi/3}r^{2}\,d\theta\). 2. Substituting \(r=2\) gives \(A=\frac{1}{2}\int_{0}^{\pi/3}4\,d\theta\). 3. Evaluating the integral gives \(A=\frac{2\pi}{3}\).

Answer

\(A=\frac{2\pi}{3}\)

Area of a polar region 9

For \(r=3\cos(\theta)\) and \(-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\), write and evaluate the integral for the area of the circle traced once.

Hints

- Does the interval trace the circle just once? - Use the polar sector-area formula with the varying radius. - Which identity will help integrate a squared cosine?

Solution

1. The given interval traces the circle once, so \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}r^{2}\,d\theta\). 2. Substitute \(r=3\cos(\theta)\): \(A=\frac{9}{2}\int_{-\pi/2}^{\pi/2}\cos^{2}(\theta)\,d\theta\). 3. The squared-cosine integral is \(\frac{\pi}{2}\), giving \(A=\frac{9\pi}{4}\).

Answer

\(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}(3\cos(\theta))^{2}\,d\theta=\frac{9\pi}{4}\)

Area of a polar region 10

For \(r=4\sin(\theta)\), \(0\leq\theta\leq\pi\), write and evaluate the polar area integral for the entire circle.

Hints

- Check whether the given angle interval covers the circle once. - How does the polar area formula use the radius? - A squared-sine identity makes the integral easier.

Solution

1. The curve traces its circle once on the stated interval. 2. \(A=\frac{1}{2}\int_{0}^{\pi}(4\sin(\theta))^{2}\,d\theta=8\int_{0}^{\pi}\sin^{2}(\theta)\,d\theta\). 3. Since \(\int_{0}^{\pi}\sin^{2}(\theta)\,d\theta=\frac{\pi}{2}\), the area is \(4\pi\).

Answer

\(A=\frac{1}{2}\int_{0}^{\pi}(4\sin(\theta))^{2}\,d\theta=4\pi\)

Area of a polar region 11

Find the exact area of the cardioid \(r=1+\cos(\theta)\) over \(0\le\theta\le2\pi\).

Hints

- Check first whether the stated angle interval traces the cardioid exactly once. - Think of the region as an accumulation of narrow polar sectors whose radii depend on \(\theta\). - After setting up the area, an exact trigonometric identity can simplify the squared term.

Solution

1. The area is \(A=\frac{1}{2}\int_{0}^{2\pi}(1+\cos(\theta))^{2}\,d\theta\). 2. Expanding gives \(1+2\cos(\theta)+\cos^{2}(\theta)\). 3. Over a full period, the cosine term integrates to \(0\), while the other terms contribute \(2\pi\) and \(\pi\). Therefore, \(A=\frac{3\pi}{2}\).

Answer

\(A=\frac{3\pi}{2}\)

Area of a polar region 12

Find the exact area of the cardioid \(r=2+2\sin(\theta)\) over \(0\le\theta\le2\pi\).

Hints

- Decide what angular interval represents one complete tracing of this cardioid. - Look for a useful common factor in the radial expression before doing any algebra. - Once the area integral is set up, use full-period trigonometric behavior to keep the evaluation exact.

Solution

1. The area is \(A=\frac{1}{2}\int_{0}^{2\pi}(2+2\sin(\theta))^{2}\,d\theta\). 2. Factoring gives \(A=2\int_{0}^{2\pi}(1+\sin(\theta))^{2}\,d\theta\). 3. The constant, sine, and squared-sine terms contribute \(2\pi\), \(0\), and \(\pi\), respectively, so \(A=6\pi\).

Answer

\(A=6\pi\)

Area of a polar region 13

Find the exact area inside the limaçon \(r=3+\cos(\theta)\) over \(0\le\theta\le2\pi\).

Hints

- Verify that the interval captures the entire limaçon without repeated tracing. - Relate the radius at each angle to the area of a narrow polar sector. - Keep the evaluation exact by using full-period properties of the trigonometric terms that appear after simplification.

Solution

1. The area is \(A=\frac{1}{2}\int_{0}^{2\pi}(3+\cos(\theta))^{2}\,d\theta\). 2. Expanding gives \(9+6\cos(\theta)+\cos^{2}(\theta)\). 3. Over \([0,2\pi]\), these terms integrate to \(18\pi\), \(0\), and \(\pi\), so \(A=\frac{19\pi}{2}\).

Answer

\(A=\frac{19\pi}{2}\)

Area of a polar region 14

For \(r=2\cos(\theta)\), \(0\leq\theta\leq\frac{\pi}{2}\), write and evaluate the polar area integral for the part of the circle in the first quadrant.

Hints

- Which part of the circle corresponds to the angle interval? - Use the squared radius in the polar area formula. - Evaluate the squared cosine exactly.

Solution

1. The interval traces the upper half of the circle, which lies in the first quadrant. 2. \(A=\frac{1}{2}\int_{0}^{\pi/2}(2\cos(\theta))^{2}\,d\theta=2\int_{0}^{\pi/2}\cos^{2}(\theta)\,d\theta\). 3. The squared-cosine integral is \(\frac{\pi}{4}\), so \(A=\frac{\pi}{2}\).

Answer

\(A=\frac{1}{2}\int_{0}^{\pi/2}(2\cos(\theta))^{2}\,d\theta=\frac{\pi}{2}\)

Area of a polar region 15

For \(r=4\sin(\theta)\), \(0\leq\theta\leq\frac{\pi}{2}\), write and evaluate the polar area integral for the first-quadrant region.

Hints

- Determine the region traced by this angle interval. - Each narrow sector contributes area based on the square of its radius. - What identity simplifies squared sine?

Solution

1. The stated interval traces the right half of the circle, lying in the first quadrant. 2. The area is \(A=\frac{1}{2}\int_{0}^{\pi/2}(4\sin(\theta))^{2}\,d\theta=8\int_{0}^{\pi/2}\sin^{2}(\theta)\,d\theta\). 3. Since \(\int_{0}^{\pi/2}\sin^{2}(\theta)\,d\theta=\frac{\pi}{4}\), \(A=2\pi\).

Answer

\(A=\frac{1}{2}\int_{0}^{\pi/2}(4\sin(\theta))^{2}\,d\theta=2\pi\)

Area of a polar region 16

Set up, but do not evaluate, an integral for the area of the upper half of the cardioid \(r=1-\cos(\theta)\).

Hints

- Translate “upper half” into the directions of the rays that sweep that part of the plane. - Check that your chosen angle interval traces the desired half of the cardioid once. - Your final expression should represent area accumulation and remain unevaluated.

Solution

1. The upper half of the cardioid is traced once as \(\theta\) runs from \(0\) to \(\pi\). 2. Therefore, the required setup is \(A=\frac{1}{2}\int_{0}^{\pi}(1-\cos(\theta))^{2}\,d\theta\).

Answer

\(A=\frac{1}{2}\int_{0}^{\pi}(1-\cos(\theta))^{2}\,d\theta\)

Area of a polar region 17

Set up, but do not evaluate, an integral for the area swept from the positive x-axis to the positive y-axis by \(r=2+\sin(\theta)\).

Hints

- Translate the two named coordinate-axis rays into polar angles. - Check how the radius behaves while the angle sweeps between those rays. - Stop after writing an integral that represents the swept area; do not evaluate it.

Solution

1. The rays from the positive x-axis to the positive y-axis correspond to \(0\le\theta\le\frac{\pi}{2}\). 2. Therefore, the required setup is \(A=\frac{1}{2}\int_{0}^{\pi/2}(2+\sin(\theta))^{2}\,d\theta\).

Answer

\(A=\frac{1}{2}\int_{0}^{\pi/2}(2+\sin(\theta))^{2}\,d\theta\)

Area of a polar region 18

Find the exact area swept by \(r=\sqrt{\theta+1}\) from \(\theta=0\) to \(\theta=2\).

Hints

- Which power of the radius appears in the area formula? - Simplify the square of the square root before integrating. - An exact result is possible here.

Solution

1. \(A=\frac{1}{2}\int_{0}^{2}(\sqrt{\theta+1})^{2}\,d\theta\). 2. Simplify the square before integrating: \(A=\frac{1}{2}\int_{0}^{2}(\theta+1)\,d\theta\). 3. Therefore, \(A=\frac{1}{2}\left[\frac{\theta^{2}}{2}+\theta\right]_{0}^{2}=2\).

Answer

\(2\)

Area of a polar region 19

Find the exact area swept by the polar spiral \(r=\theta\) from \(\theta=0\) to \(\theta=2\).

Hints

- The spiral sweeps out narrow sectors as the angle increases. - Which expression should be squared in the area formula? - Evaluate the integral at the two stated angles.

Solution

1. The polar-area formula gives \(A=\frac{1}{2}\int_{0}^{2}r^{2}\,d\theta\). 2. Substitute \(r=\theta\): \(A=\frac{1}{2}\int_{0}^{2}\theta^{2}\,d\theta\). 3. Thus, \(A=\frac{1}{2}\left[\frac{\theta^{3}}{3}\right]_{0}^{2}=\frac{4}{3}\).

Answer

\(\frac{4}{3}\)

Area of a polar region 20

A closed polar curve is traced once by the nonnegative radius \(r=\sqrt{4+2\cos\theta}\) for \(0\le\theta\le2\pi\). Find its exact enclosed area.

Hints

- The polar-area formula uses the square of the radius. - What happens to the sine term after integrating over a full revolution? - Keep the calculation exact; you do not need to expand the square root.

Solution

1. Since \(r^{2}=4+2\cos\theta\), the polar-area formula gives \(A=\frac{1}{2}\int_{0}^{2\pi}(4+2\cos\theta)\,d\theta\). 2. Integrate: \(A=\frac{1}{2}[4\theta+2\sin\theta]_{0}^{2\pi}\). 3. Therefore, \(A=4\pi\).

Answer

\(4\pi\)

Area of a polar region 21

Find the exact area swept by \(r=e^{\theta}\) from \(\theta=0\) to \(\theta=\ln(2)\).

Hints

- Polar area uses the square of the radius. - How do you integrate an exponential with an angle in its exponent? - Use the relationship between the exponential and natural logarithm at the upper limit.

Solution

1. Substitute the radius in the polar-area formula: \(A=\frac{1}{2}\int_{0}^{\ln(2)}e^{2\theta}\,d\theta\). 2. Integrate: \(A=\frac{1}{4}\left[e^{2\theta}\right]_{0}^{\ln(2)}\). 3. Using \(e^{2\ln(2)}=4\), obtain \(A=\frac{1}{4}(4-1)=\frac{3}{4}\).

Answer

\(\frac{3}{4}\)

Area of a polar region 22

A polar region is traced with \(r=1\) for \(0\le\theta\le\pi\) and \(r=2\) for \(\pi<\theta\le\frac{3\pi}{2}\). Find its total area.

Hints

- Notice where the rule defining the radius changes as the angle increases. - Represent the swept region on each angular portion before combining the contributions. - Check that no angular interval is omitted or counted twice.

Solution

1. Split the area integral where the radial rule changes. 2. The first part is \(\frac{1}{2}\int_{0}^{\pi}1\,d\theta=\frac{\pi}{2}\). 3. The second part is \(\frac{1}{2}\int_{\pi}^{3\pi/2}4\,d\theta=\pi\). 4. The total area is \(\frac{3\pi}{2}\).

Answer

\(\frac{3\pi}{2}\)

Area of a polar region 23

For \(r=\sec\theta\) on \(-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}\), write the polar area integral for the swept region and evaluate it exactly.

Hints

- Model the area swept as the angle moves across the symmetric interval. - After the area model is established, inspect the resulting trigonometric expression for a familiar derivative relationship. - Use the symmetry of the endpoint angles to check the exact result.

Solution

1. Use the polar-area formula: \(A=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\sec^2\theta\,d\theta\). 2. An antiderivative of \(\sec^2\theta\) is \(\tan\theta\). 3. Therefore, \(A=\frac{1}{2}\left(\tan\frac{\pi}{4}-\tan\left(-\frac{\pi}{4}\right)\right)=1\).

Answer

\(A=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\sec^2\theta\,d\theta=1\)

Area of a polar region 24

Let \(A(\theta)\) be the area swept by the radius vector of \(r=2+\theta\), starting at \(\theta=0\). Find \(\frac{dA}{d\theta}\) at \(\theta=1\).

Hints

- Think of the swept area as an accumulation function. - Differentiate the area integral with respect to its upper limit. - Substitute the requested angle only after finding the rate expression.

Solution

1. The accumulated area is \(A(\theta)=\frac{1}{2}\int_{0}^{\theta}(2+u)^2\,du\). 2. By the Fundamental Theorem of Calculus, \(\frac{dA}{d\theta}=\frac{1}{2}(2+\theta)^2\). 3. At \(\theta=1\), \(\frac{dA}{d\theta}=\frac{1}{2}(3)^2=\frac{9}{2}\).

Answer

\(\frac{9}{2}\)

Area of a polar region 25

The polar curve \(r=\sqrt{a+2\cos\theta}\), where \(a>2\), encloses an area of \(6\pi\) as \(\theta\) runs from \(0\) to \(2\pi\). Find \(a\).

Hints

- Use the given squared radius directly in the area formula. - Simplify the full-period trigonometric contribution. - Solve the resulting equation for the parameter.

Solution

1. Since \(r^2=a+2\cos\theta\), the enclosed area is \(A=\frac{1}{2}\int_{0}^{2\pi}(a+2\cos\theta)\,d\theta\). 2. The cosine term contributes \(0\) over a full revolution, so \(A=\pi a\). 3. Set \(\pi a=6\pi\) to obtain \(a=6\). 4. This value satisfies the condition \(a>2\).

Answer

\(a=6\)

Area of a polar region 26

For the polar curve \(r=\sqrt{1+\theta}\), find the area swept on \(0\le\theta\le1\), the area swept on \(1\le\theta\le3\), and the ratio of the second area to the first.

Hints

- The square of the radius is simpler than the radius itself. - Use a separate sector-area integral for each angle interval. - Divide the second area by the first only after finding both.

Solution

1. The first area is \(A_{1}=\frac{1}{2}\int_{0}^{1}(1+\theta)\,d\theta=\frac{3}{4}\). 2. The second area is \(A_{2}=\frac{1}{2}\int_{1}^{3}(1+\theta)\,d\theta=3\). 3. Their requested ratio is \(\frac{A_{2}}{A_{1}}=\frac{3}{3/4}=4\).

Answer

First area: \(\frac{3}{4}\). Second area: \(3\). Second-to-first ratio: \(4\).

Area of a polar region 27

The polar curve \(r^2=9\cos(2\theta)\) has two loops. Its right-hand loop is traced for \(-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}\). Find that loop's exact area.

Hints

- The equation already gives the quantity that appears in the area integrand. - Use the stated pole-to-pole interval for one loop. - Pay attention to the factor introduced when integrating a double angle.

Solution

1. Use \(r^2=9\cos(2\theta)\) directly in the polar-area formula. 2. The loop area is \(A=\frac{1}{2}\int_{-\pi/4}^{\pi/4}9\cos(2\theta)\,d\theta\). 3. Evaluating gives \(A=\frac{9}{4}\left[\sin(2\theta)\right]_{-\pi/4}^{\pi/4}=\frac{9}{2}\).

Answer

\(\frac{9}{2}\)

Area of a polar region 28

For \(r=\sin(2\theta)\) on \(0\le\theta\le\frac{\pi}{2}\), how many petals are traced? Justify your answer using the zeros and sign of \(r\) on this interval.

Hints

- Find where the radius first leaves and returns to zero. - Check whether it changes sign before reaching the second endpoint.

Solution

1. The radius is zero at both endpoints and positive between them. 2. The trace leaves the pole, draws one lobe in the first quadrant, and returns to the pole.

Answer

One petal. The radius is zero at \(\theta=0\) and \(\theta=\frac{\pi}{2}\), and it is positive between those angles. The curve leaves the pole and returns to it without starting another petal.

Area of a polar region 29

The polar radii below were measured at the midpoints of four equal angle intervals from \(0\) to \(\pi\). Use the midpoint rule to estimate the swept area. <table><tr><th>Angle</th><th>Radius</th></tr><tr><td>\(\frac{\pi}{8}\)</td><td>\(1\)</td></tr><tr><td>\(\frac{3\pi}{8}\)</td><td>\(2\)</td></tr><tr><td>\(\frac{5\pi}{8}\)</td><td>\(2\)</td></tr><tr><td>\(\frac{7\pi}{8}\)</td><td>\(1\)</td></tr></table>

Hints

- How wide is each of the four equal angular intervals? - Polar sector area uses the squared radius, not the radius alone. - Make one midpoint contribution for each interval.

Solution

1. Each angular interval has width \(\Delta\theta=\frac{\pi}{4}\). 2. Square the four midpoint radii to obtain \(1,4,4,1\). 3. The midpoint estimate of polar area is \(A\approx\frac{1}{2}\cdot\frac{\pi}{4}(1+4+4+1)=\frac{5\pi}{4}\).

Answer

\(A\approx\frac{5\pi}{4}\) square units.

Area of a polar region 30

The curve \(r=1+2\cos\theta\) has a small inner loop traced for \(\frac{2\pi}{3}\le\theta\le\frac{4\pi}{3}\). Is \(r\) positive or negative inside this interval, and why does the polar area formula still work?

Hints

- Test the radius at the middle angle \(\theta=\pi\). - Check the two zeros that close the small loop. - The area integrand is the square of the radius.

Solution

1. On the open interval, \(\cos\theta<-\frac{1}{2}\), so \(r=1+2\cos\theta<0\). 2. The radius is zero at both endpoints, closing the inner loop at the pole. 3. Its swept area remains \(\frac{1}{2}\int r^2\,d\theta\) because the square is nonnegative.

Answer

Negative between the endpoints; the area formula still uses \(r^2\).

Area of a polar region 31

The polar equation \(r=2\cos\theta\) traces its entire circle for \(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\). Find the enclosed area with a polar integral.

Hints

- The equation describes a circle traced once over this angle interval. - Use the square of the radius in the area formula. - Symmetry can simplify the trigonometric integral.

Solution

1. Use \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,d\theta\). 2. By symmetry, \(A=4\int_{0}^{\pi/2}\cos^2\theta\,d\theta=4\cdot\frac{\pi}{4}=\pi\).

Answer

\(A=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(2\cos\theta)^2\,d\theta=\pi\) square units.

Area of a polar region 32

Find the exact area of one petal of \(r=3\cos(2\theta)\). Use the tracing interval \(-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}\).

Hints

- What evidence in the given interval shows that one complete petal, rather than several petals, is traced? - Relate the petal area to the accumulation of polar sectors over that interval. - Preserve the inner angular factor when using an exact trigonometric identity.

Solution

1. The specified interval runs between consecutive zeros of \(r\) and traces one petal exactly once. 2. The area is \(A=\frac{1}{2}\int_{-\pi/4}^{\pi/4}(3\cos(2\theta))^{2}\,d\theta\). 3. Thus, \(A=\frac{9}{2}\int_{-\pi/4}^{\pi/4}\cos^{2}(2\theta)\,d\theta=\frac{9\pi}{8}\).

Answer

\(A=\frac{9\pi}{8}\)

Area of a polar region 33

Find the exact area of one petal of \(r=2\sin(3\theta)\). Use the tracing interval \(0\le\theta\le\frac{\pi}{3}\).

Hints

- Check what happens to the radius at the endpoints and between them to understand the tracing. - Model the area swept during exactly that tracing interval. - When evaluating exactly, account for how quickly the trigonometric angle changes compared with \(\theta\).

Solution

1. The radial function is zero at both endpoints and nonnegative between them, so the interval traces one petal. 2. The area is \(A=\frac{1}{2}\int_{0}^{\pi/3}(2\sin(3\theta))^{2}\,d\theta\). 3. Therefore, \(A=2\int_{0}^{\pi/3}\sin^{2}(3\theta)\,d\theta=\frac{\pi}{3}\).

Answer

\(A=\frac{\pi}{3}\)

Area of a polar region 34

Find the exact area of one petal of \(r=4\cos(3\theta)\). Use the tracing interval \(-\frac{\pi}{6}\le\theta\le\frac{\pi}{6}\).

Hints

- Decide why the symmetric interval corresponds to one petal rather than a repeated trace. - Build the area from the radius as the angle sweeps across that interval. - Symmetry can simplify the exact evaluation, but keep track of the factor inside the cosine.

Solution

1. The interval lies between consecutive zeros of the radial function and traces one petal exactly once. 2. The area is \(A=\frac{1}{2}\int_{-\pi/6}^{\pi/6}(4\cos(3\theta))^{2}\,d\theta\). 3. Thus, \(A=8\int_{-\pi/6}^{\pi/6}\cos^{2}(3\theta)\,d\theta=\frac{4\pi}{3}\).

Answer

\(A=\frac{4\pi}{3}\)

Area of a polar region 35

Find the exact area of one petal of \(r=2\sin(4\theta)\). Use the tracing interval \(0\le\theta\le\frac{\pi}{4}\).

Hints

- Use the endpoint and sign behavior of the radius to decide what the interval traces. - Think in terms of accumulated polar sectors over one complete petal. - In the exact evaluation, distinguish the outer coefficient from the angular multiplier inside the sine.

Solution

1. The radial function is zero at both endpoints and nonnegative between them, so the interval traces one petal exactly once. 2. The area is \(A=\frac{1}{2}\int_{0}^{\pi/4}(2\sin(4\theta))^{2}\,d\theta=2\int_{0}^{\pi/4}\sin^{2}(4\theta)\,d\theta\). 3. Applying a power-reduction identity gives \(A=\frac{\pi}{4}\).

Answer

\(A=\frac{\pi}{4}\)

Area of a polar region 36

Find the exact area of the inner loop of \(r=1+2\cos(\theta)\). Use the tracing interval \(\frac{2\pi}{3}\le\theta\le\frac{4\pi}{3}\).

Hints

- Use the radius at the interval endpoints and its sign inside the interval to interpret the loop being traced. - Once the tracing is understood, model the loop area by accumulating polar sectors. - Keep the exact trigonometric values at the endpoint angles when simplifying the integral.

Solution

1. The endpoints satisfy \(r=0\), and the stated interval traces the inner loop exactly once. 2. The area is \(A=\frac{1}{2}\int_{2\pi/3}^{4\pi/3}(1+2\cos(\theta))^{2}\,d\theta\). 3. Expanding and integrating gives \(A=\frac{1}{2}(2\pi-3\sqrt{3})=\pi-\frac{3\sqrt{3}}{2}\).

Answer

\(A=\pi-\frac{3\sqrt{3}}{2}\)

Area of a polar region 37

Find the exact area of the inner loop of \(r=2-3\cos(\theta)\). Use the tracing interval \(-\arccos\left(\frac{2}{3}\right)\le\theta\le\arccos\left(\frac{2}{3}\right)\).

Hints

- What do the endpoint angles tell you about where this loop begins and ends? - Look for symmetry in both the interval and the radial expression before setting up the full calculation. - Keep endpoint trigonometric values exact when simplifying the result.

Solution

1. The endpoints satisfy \(r=0\), and the symmetric interval traces the inner loop exactly once. 2. By symmetry, \(A=\int_{0}^{\arccos(2/3)}(2-3\cos(\theta))^{2}\,d\theta\). 3. Expanding and integrating gives \(A=\frac{17}{2}\arccos\left(\frac{2}{3}\right)-3\sqrt{5}\).

Answer

\(A=\frac{17}{2}\arccos\left(\frac{2}{3}\right)-3\sqrt{5}\)

Area of a polar region 38

Set up, but do not evaluate, an integral for the area of one petal centered on the positive x-axis for \(r=2\cos(3\theta)\).

Hints

- Start by identifying the angular interval over which the petal centered at \(\theta=0\) is traced once. - Use the behavior of \(r\) at the ends of that interval to check your choice. - Your final response should model the swept area but should not carry out the integration.

Solution

1. The petal centered on the positive x-axis is traced between the adjacent zeros \(\theta=-\frac{\pi}{6}\) and \(\theta=\frac{\pi}{6}\). 2. Therefore, the required setup is \(A=\frac{1}{2}\int_{-\pi/6}^{\pi/6}(2\cos(3\theta))^{2}\,d\theta\).

Answer

\(A=\frac{1}{2}\int_{-\pi/6}^{\pi/6}(2\cos(3\theta))^{2}\,d\theta\)

Area of a polar region 39

Set up, but do not evaluate, an integral for the area of the petal in the first quadrant for \(r=3\sin(2\theta)\).

Hints

- Determine which consecutive angles at the pole bound the first-quadrant petal. - Check the sign and direction of the radius between those angles to confirm the petal being traced. - Stop once you have an area integral that represents exactly one tracing of that petal.

Solution

1. The petal in the first quadrant is traced from one zero of the radial function to the next, over \(0\le\theta\le\frac{\pi}{2}\). 2. Therefore, the required setup is \(A=\frac{1}{2}\int_{0}^{\pi/2}(3\sin(2\theta))^{2}\,d\theta\).

Answer

\(A=\frac{1}{2}\int_{0}^{\pi/2}(3\sin(2\theta))^{2}\,d\theta\)

Area of a polar region 40

The polar region defined by \(r=a\) over \(0\le\theta\le\frac{\pi}{2}\) has area \(8\pi\), where \(a>0\). Find \(a\).

Hints

- Work backward from the given area by expressing how the constant radius affects sector area. - What quantity involving \(a\) is determined by the area before you solve for \(a\) itself? - Use the stated sign restriction when choosing the final value.

Solution

1. The area is \(A=\frac{1}{2}\int_{0}^{\pi/2}a^{2}\,d\theta=\frac{\pi a^{2}}{4}\). 2. Set \(\frac{\pi a^{2}}{4}=8\pi\), so \(a^{2}=32\). 3. Because \(a>0\), \(a=4\sqrt{2}\).

Answer

\(a=4\sqrt{2}\)

Area of a polar region 41

The polar region \(r=a\cos(\theta)\), \(-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\), has area \(9\pi\), where \(a>0\). Write an area equation in \(a\) and find \(a\).

Hints

- How does changing the radius by a factor affect area? - Set the polar sector-area integral equal to the given area. - Use the condition on the sign of \(a\) when choosing a solution.

Solution

1. \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}a^{2}\cos^{2}(\theta)\,d\theta=\frac{\pi a^{2}}{4}\). 2. Equating this area with \(9\pi\) gives \(a^{2}=36\). 3. The positive value is \(a=6\).

Answer

\(\frac{1}{2}\int_{-\pi/2}^{\pi/2}a^{2}\cos^{2}(\theta)\,d\theta=9\pi\) \(a=6\)

Area of a polar region 42

The polar region defined by \(r=a(1+\cos(\theta))\) over \(0\le\theta\le2\pi\) has area \(6\pi\), where \(a>0\). Find \(a\).

Hints

- Compare this family with the same cardioid when the scale factor is \(1\). - Decide how multiplying every radius by \(a\) changes area. - Use the positivity condition only when selecting the final value from the scale relation.

Solution

1. The area is \(A=\frac{1}{2}\int_{0}^{2\pi}a^{2}(1+\cos(\theta))^{2}\,d\theta=\frac{3\pi a^{2}}{2}\). 2. Set \(\frac{3\pi a^{2}}{2}=6\pi\), so \(a^{2}=4\). 3. Because \(a>0\), \(a=2\).

Answer

\(a=2\)

Area of a polar region 43

The polar region defined by \(r=a\sin(2\theta)\) over \(0\le\theta\le\frac{\pi}{2}\) has area \(4\pi\), where \(a>0\). Find \(a\).

Hints

- First decide what part of the rose the stated interval traces. - Think about how the scale factor \(a\) affects the area of that fixed-shape petal. - Preserve exact values until the positive scale factor is isolated.

Solution

1. The interval traces one petal, so \(A=\frac{1}{2}\int_{0}^{\pi/2}a^{2}\sin^{2}(2\theta)\,d\theta=\frac{\pi a^{2}}{8}\). 2. Set \(\frac{\pi a^{2}}{8}=4\pi\), so \(a^{2}=32\). 3. Because \(a>0\), \(a=4\sqrt{2}\).

Answer

\(a=4\sqrt{2}\)

Area of a polar region 44

Region A is inside \(r=2\) for \(0\leq\theta\leq\pi\). Region B is inside \(r=\sqrt{2}\) for \(0\leq\theta\leq2\pi\). Write and evaluate an area integral for each region. Which region is larger?

Hints

- Does each region use a full revolution of angles? - Square the radius in each polar-area integral. - Compare the two exact areas once each integral is evaluated.

Solution

1. For region A, \(A_A=\frac{1}{2}\int_{0}^{\pi}2^{2}\,d\theta=2\pi\). 2. For region B, \(A_B=\frac{1}{2}\int_{0}^{2\pi}(\sqrt{2})^{2}\,d\theta=2\pi\). 3. The areas are equal, despite the different radii and angle intervals.

Answer

\(A_A=\frac{1}{2}\int_{0}^{\pi}2^{2}\,d\theta=2\pi\) \(A_B=\frac{1}{2}\int_{0}^{2\pi}(\sqrt{2})^{2}\,d\theta=2\pi\) The two areas are equal.

Area of a polar region 45

Region A is inside \(r=2\cos(\theta)\) over \(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\). Region B is inside \(r=1+\cos(\theta)\) over \(0\le\theta\le2\pi\). Which region has greater area?

Hints

- Set up a separate polar-area integral for each curve and tracing interval. - Use squared-cosine and full-period identities where appropriate. - Compare the exact areas only after both integrals are evaluated.

Solution

1. Region A has area \(A_A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}(2\cos(\theta))^{2}\,d\theta=\pi\). 2. Region B has area \(A_B=\frac{1}{2}\int_{0}^{2\pi}(1+\cos(\theta))^{2}\,d\theta=\frac{3\pi}{2}\). 3. Therefore, Region B has greater area.

Answer

\(A_A=\pi\), \(A_B=\frac{3\pi}{2}\); Region B has greater area.

Area of a polar region 46

For \(r=2+\cos\theta\) over \(0\le\theta\le2\pi\), Tiago writes \(A=\int_{0}^{2\pi}(2+\cos\theta)^2\,d\theta\) for the enclosed area. Is his setup correct? If not, fix it and find the area.

Hints

- What is the area of a thin sector compared with the square of its radius? - Test the proposed rule with a circle of fixed radius. - Integrate over the full angle interval to find the total enclosed region.

Solution

1. Since \(r=2+\cos\theta\) is positive throughout the interval, the curve traces its enclosed region once. 2. The correct area formula is \(A=\frac{1}{2}\int_{0}^{2\pi}(2+\cos\theta)^{2}\,d\theta\). 3. Expanding gives \(A=\frac{1}{2}\int_{0}^{2\pi}(4+4\cos\theta+\cos^{2}\theta)\,d\theta\). 4. The integral is \(8\pi+0+\pi=9\pi\), so \(A=\frac{9\pi}{2}\).

Answer

Correct setup: \(A=\frac{1}{2}\int_{0}^{2\pi}(2+\cos\theta)^{2}\,d\theta\). Area: \(\frac{9\pi}{2}\).

Area of a polar region 47

For \(r=3\cos(2\theta)\), Seo-yeon integrates from \(0\) to \(2\pi\) to find the area of one petal. Explain the overcount and find the area of one petal.

Hints

- Count the petals formed during one complete turn. - Find the two consecutive zero-radius angles around one petal. - Use those angles, rather than a full revolution, as the area bounds.

Solution

1. The interval \(0\leq\theta\leq2\pi\) covers all four congruent petals, not one. 2. The petal centered on the positive x-axis is traced for \(-\frac{\pi}{4}\leq\theta\leq\frac{\pi}{4}\). 3. Its area is \(A=\frac{1}{2}\int_{-\pi/4}^{\pi/4}(3\cos(2\theta))^{2}\,d\theta=\frac{9\pi}{8}\).

Answer

The interval covers all four petals. One petal has area \(\frac{9\pi}{8}\).

Area of a polar region 48

Find the area inside \(r=e^{\sin(\theta)}\) from \(\theta=0\) to \(\theta=\pi\). Round to 3 decimal places.

Hints

- Square the exponential radial function before applying the area formula. - Keep the exact angular bounds in the numerical integral. - Delay rounding until the complete integral has been evaluated.

Solution

1. The area is \(A=\frac{1}{2}\int_{0}^{\pi}(e^{\sin(\theta)})^{2}\,d\theta=\frac{1}{2}\int_{0}^{\pi}e^{2\sin(\theta)}\,d\theta\). 2. Numerical evaluation gives \(A\approx6.6240780499\). 3. Rounded to 3 decimal places, \(A\approx6.624\).

Answer

\(A\approx6.624\)

Area of a polar region 49

Find the area inside \(r=2+\cos(3\theta)\) from \(\theta=0\) to \(\theta=\frac{\pi}{2}\). Round to 3 decimal places.

Hints

- Use the stated angle interval exactly as given. - Expand the square and use a power-reduction identity for the squared-cosine term. - Round only after evaluating the full expression.

Solution

1. The area is \(A=\frac{1}{2}\int_{0}^{\pi/2}(2+\cos(3\theta))^{2}\,d\theta\). 2. Evaluating gives \(A=\frac{9\pi}{8}-\frac{2}{3}\approx2.8676250686\). 3. Rounded to 3 decimal places, \(A\approx2.868\).

Answer

\(A\approx2.868\)

Area of a polar region 50

The rose \(r=4\cos(2\theta)\) has four congruent petals. a) Find the exact area of the petal traced on \(-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}\). b) Find the total area of the rose.

Hints

- Evaluate the area over the stated one-petal interval first. - Use a power-reduction identity for the squared-cosine integral. - Multiply by the number of congruent petals only after finding one petal's area.

Solution

1. One petal has area \(A_p=\frac{1}{2}\int_{-\pi/4}^{\pi/4}(4\cos(2\theta))^{2}\,d\theta=2\pi\). 2. Because the rose has four congruent petals, its total area is \(A_T=4A_p=8\pi\).

Answer

a) \(2\pi\) b) \(8\pi\)

Area of a polar region 51

The rose \(r=3\sin(3\theta)\) has three congruent petals. a) Find the exact area of the petal traced on \(0\le\theta\le\frac{\pi}{3}\). b) Find the total area of the rose.

Hints

- Evaluate the area over the stated one-petal interval first. - Account for the inner factor \(3\) when integrating the squared sine. - Multiply by the number of congruent petals only after finding one petal's area.

Solution

1. One petal has area \(A_p=\frac{1}{2}\int_{0}^{\pi/3}(3\sin(3\theta))^{2}\,d\theta=\frac{3\pi}{4}\). 2. Because the rose has three congruent petals, its total area is \(A_T=3A_p=\frac{9\pi}{4}\).

Answer

a) \(\frac{3\pi}{4}\) b) \(\frac{9\pi}{4}\)

Area of a polar region 52

The rose \(r=2\cos(4\theta)\) has eight congruent petals. a) Find the exact area of the petal traced on \(-\frac{\pi}{8}\le\theta\le\frac{\pi}{8}\). b) Find the total area of the rose.

Hints

- Use the given symmetric interval to find one petal's area. - Apply symmetry or power reduction while retaining the inner angular factor. - Scale the one-petal area by the stated number of congruent petals.

Solution

1. One petal has area \(A_p=\frac{1}{2}\int_{-\pi/8}^{\pi/8}(2\cos(4\theta))^{2}\,d\theta=\frac{\pi}{4}\). 2. Because the rose has eight congruent petals, its total area is \(A_T=8A_p=2\pi\).

Answer

a) \(\frac{\pi}{4}\) b) \(2\pi\)

Area of a polar region 53

The spiral \(r=2\theta\) sweeps an area of \(18\) from \(\theta=0\) to \(\theta=b\), where \(b>0\). Find \(b\).

Hints

- Treat the unknown endpoint as part of the area model rather than as a value to guess. - Work backward from the given area after expressing how accumulated sector area depends on \(b\). - Check the stated sign restriction when choosing the endpoint.

Solution

1. The swept area is \(A=\frac{1}{2}\int_{0}^{b}(2\theta)^2\,d\theta=\frac{2b^3}{3}\). 2. Set \(\frac{2b^3}{3}=18\). 3. Then \(b^3=27\), so \(b=3\).

Answer

\(b=3\)

Area of a polar region 54

Use Simpson’s rule to estimate the area swept from \(\theta=0\) to \(\theta=\frac{\pi}{2}\) using the radial data. <table><tr><th>\(\theta\)</th><th>\(r\)</th></tr><tr><td>\(0\)</td><td>\(2\)</td></tr><tr><td>\(\frac{\pi}{4}\)</td><td>\(3\)</td></tr><tr><td>\(\frac{\pi}{2}\)</td><td>\(2\)</td></tr></table>

Hints

- Decide which function of the tabulated radii represents the polar-area integrand before applying the numerical rule. - Check the equal angular spacing in the table and identify the Simpson weights. - Keep the geometric area factor separate from the numerical-integration factor until the setup is complete.

Solution

1. Apply Simpson’s rule to \(r^2\) with spacing \(h=\frac{\pi}{4}\). 2. The area estimate is \(\frac{1}{2}\cdot\frac{h}{3}[2^2+4(3^2)+2^2]\). 3. Substitution gives \(A\approx\frac{\pi}{24}(44)=\frac{11\pi}{6}\).

Answer

\(A\approx\frac{11\pi}{6}\)

Area of a polar region 55

Simpson’s rule is used to estimate the area swept from \(\theta=0\) to \(\theta=\frac{\pi}{2}\). The radii at \(0\), \(\frac{\pi}{4}\), and \(\frac{\pi}{2}\) are \(2\), \(k\), and \(2\), respectively, where \(k>0\). If the estimated area is \(\frac{3\pi}{2}\), find \(k\).

Hints

- Work backward from the stated numerical area estimate. - Decide what function of each radius belongs in the Simpson weighted sum for polar area. - Keep the positive condition for the final step after the equation for \(k\) has been solved.

Solution

1. Simpson’s rule applied to \(r^2\) gives \(A\approx\frac{\pi}{24}[4+4k^2+4]\). 2. Thus \(\frac{\pi}{6}(k^2+2)=\frac{3\pi}{2}\). 3. Therefore, \(k^2+2=9\), so \(k^2=7\). 4. Since \(k>0\), \(k=\sqrt{7}\).

Answer

\(k=\sqrt{7}\)

Area of a polar region 56

Émile estimates the area swept by \(r=2\theta\) on \(0\le\theta\le1\) by using the average radius \(1\) in the sector formula and obtains \(\frac{1}{2}\). Explain why this method is invalid and find the exact area.

Hints

- Compare the structure of the polar-area formula with the proposed shortcut. - Square the actual radial function inside the integral. - Evaluate over the full angular interval.

Solution

1. Polar area depends on the average of \(r^2\), not the square of the average radius. 2. The exact area is \(A=\frac{1}{2}\int_{0}^{1}(2\theta)^2\,d\theta\). 3. Evaluating gives \(A=2\int_{0}^{1}\theta^2\,d\theta=\frac{2}{3}\). 4. Thus Émile's value \(\frac{1}{2}\) is too small.

Answer

The average-radius shortcut is invalid because area uses \(r^2\) point by point. The exact area is \(\frac{2}{3}\).

Area of a polar region 57

The lemniscate \(r^2=a^2\cos(2\theta)\), where \(a>0\), encloses a total area of \(16\) in its two loops. Find \(a\).

Hints

- First find the area of one loop over its pole-to-pole interval. - Use symmetry to obtain the total area. - Apply the positive-value condition after solving the resulting equation.

Solution

1. One loop is traced for \(-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}\). 2. Its area is \(\frac{1}{2}\int_{-\pi/4}^{\pi/4}a^2\cos(2\theta)\,d\theta=\frac{a^2}{2}\). 3. The two congruent loops therefore have total area \(a^2\). 4. Set \(a^2=16\). Since \(a>0\), \(a=4\).

Answer

\(a=4\)

Area of a polar region 58

The polar curve \(r=e^{-\theta}\) spirals toward the pole for \(0\le\theta<\infty\). Find the total accumulated polar-sector area over this interval, counting overlapping sectors again as the curve turns.

Hints

- Polar-sector area adds contributions over the angular interval, including later turns. - Square the radius before integrating. - Use a limit because there is no finite final angle.

Solution

1. The accumulated area is \(A=\frac{1}{2}\int_{0}^{\infty}e^{-2\theta}\,d\theta\). 2. Evaluate the improper integral as \(A=\frac{1}{2}\lim_{b\to\infty}\left[-\frac{1}{2}e^{-2\theta}\right]_{0}^{b}\). 3. Because \(e^{-2b}\to0\), \(A=\frac{1}{4}\).

Answer

\(\frac{1}{4}\)

Area of a polar region 59

For the polar curve \(r=3e^{-\theta}\), the area swept from \(\theta=0\) to \(\theta=b\) is \(\frac{9}{8}\), where \(b>0\). Find the exact value of \(b\).

Hints

- Express the given area as an integral with the unknown endpoint. - Isolate the exponential term after evaluating the integral. - Use a logarithm only after the exponential equation is simplified.

Solution

1. Set up the area equation: \(\frac{1}{2}\int_{0}^{b}9e^{-2\theta}\,d\theta=\frac{9}{8}\). 2. Evaluating the integral gives \(\frac{9}{4}\left(1-e^{-2b}\right)=\frac{9}{8}\). 3. Thus \(e^{-2b}=\frac{1}{2}\). 4. Taking natural logarithms gives \(b=\frac{\ln 2}{2}\).

Answer

\(b=\frac{\ln 2}{2}\)

Area of a polar region 60

Find the exact area swept by \(r=1+\cos\left(\frac{\theta}{2}\right)\) for \(-\pi\le\theta\le\pi\).

Hints

- Square the entire radial expression before integrating. - A half-angle identity simplifies the squared trigonometric term. - Use the symmetry of the interval when evaluating the terms.

Solution

1. Use \(A=\frac{1}{2}\int_{-\pi}^{\pi}\left(1+\cos\left(\frac{\theta}{2}\right)\right)^2\,d\theta\). 2. Expand the square and use \(\cos^2\left(\frac{\theta}{2}\right)=\frac{1+\cos\theta}{2}\). 3. Over \([-\pi,\pi]\), the three resulting contributions are \(2\pi\), \(8\), and \(\pi\). 4. Therefore, \(A=\frac{1}{2}(3\pi+8)=\frac{3\pi+8}{2}\).

Answer

\(\frac{3\pi+8}{2}\)

Area of a polar region 61

For the polar curve \(r=2\sin\theta\), let \(A(\theta)\) be the area swept from \(0\) to \(\theta\), where \(0<\theta<\frac{\pi}{2}\). At what angle is \(\frac{dA}{d\theta}=1\)?

Hints

- Relate the rate of swept area to the current radius. - Solve the resulting trigonometric equation within the stated interval. - Use the interval to select the correct angle.

Solution

1. The instantaneous area rate is \(\frac{dA}{d\theta}=\frac{1}{2}r^2\). 2. Therefore, \(\frac{dA}{d\theta}=\frac{1}{2}(2\sin\theta)^2=2\sin^2\theta\). 3. Set \(2\sin^2\theta=1\), so \(\sin^2\theta=\frac{1}{2}\). 4. On \(0<\theta<\frac{\pi}{2}\), the solution is \(\theta=\frac{\pi}{4}\).

Answer

\(\theta=\frac{\pi}{4}\)

Area of a polar region 62

The polar curve \(r=a\theta\), where \(a>0\), sweeps an area of \(12\) as \(\theta\) runs from \(0\) to \(2\). Find \(a\).

Hints

- Write the given area as an equation involving the unknown scale factor. - Keep the scale factor outside the integral after squaring it. - Use the positivity condition when selecting the final value.

Solution

1. Set up the area equation \(\frac{1}{2}\int_{0}^{2}(a\theta)^2\,d\theta=12\). 2. Evaluating gives \(\frac{a^2}{2}\cdot\frac{8}{3}=\frac{4a^2}{3}\). 3. Thus \(\frac{4a^2}{3}=12\), so \(a^2=9\). 4. Since \(a>0\), \(a=3\).

Answer

\(a=3\)

Area of a polar region 63

Find the exact enclosed area of the loop traced by \(r=2\cos\left(\frac{\theta}{2}\right)\) for \(-\pi\le\theta\le\pi\).

Hints

- Verify that the stated interval begins and ends at the pole. - Square the half-angle radial function. - Use a power-reduction identity or symmetry.

Solution

1. The area is \(A=\frac{1}{2}\int_{-\pi}^{\pi}4\cos^2\left(\frac{\theta}{2}\right)\,d\theta\). 2. Thus \(A=2\int_{-\pi}^{\pi}\cos^2\left(\frac{\theta}{2}\right)\,d\theta\). 3. The integral equals \(\pi\), so \(A=2\pi\).

Answer

\(2\pi\)

Area of a polar region 64

The circle \(r=2\cos\theta\) is traced for \(\frac{\pi}{2}\le\theta\le\frac{3\pi}{2}\), when the radius is nonpositive. Using that interval, write the polar area integral and find the enclosed area.

Hints

- Decide whether a negative polar radius should make a small geometric area contribution negative. - Check how the stated interval traces the displayed circle before setting up the accumulation. - Keep the area model tied to the magnitude of the radius and the actual angular interval.

Solution

1. Polar area uses \(r^2\), so a negative radius does not make the area negative. 2. The area is \(A=\frac{1}{2}\int_{\pi/2}^{3\pi/2}4\cos^2\theta\,d\theta\). 3. Since \(\int_{\pi/2}^{3\pi/2}\cos^2\theta\,d\theta=\frac{\pi}{2}\), the area is \(\pi\).

Answer

\(A=\frac{1}{2}\int_{\pi/2}^{3\pi/2}(2\cos\theta)^2\,d\theta=\pi\)

Area of a polar region 65

Linnea integrates the rose \(r=2\cos(5\theta)\) from \(0\) to \(2\pi\) and reports its total area as \(2\pi\). Explain the overcount and give the correct total area.

Hints

- Determine how much angular change traces an odd-petal rose once. - Compare that interval with the proposed interval. - Recompute the area using a single tracing.

Solution

1. For an odd-petal rose, \(0\le\theta\le\pi\) traces the entire curve once. 2. The interval \([0,2\pi]\) traces the five-petal rose twice, so Linnea's \(2\pi\) doubles the area. 3. The correct area is \(\frac{1}{2}\int_{0}^{\pi}4\cos^2(5\theta)\,d\theta=\pi\).

Answer

The interval \([0,2\pi]\) traces the rose twice. The correct total area is \(\pi\).

Area of a polar region 66

The polar curve \(r=2|\cos\theta|\) traces two circles over \(0\le\theta\le2\pi\). Write one polar-area integral for this full interval and find the combined area.

Hints

- Check which part of the curve is traced between consecutive visits to the pole. - The area formula uses the square of the radius. - Does the absolute value matter after you square the radius?

Solution

1. Each circle is traced once, and the two circles touch only at the pole. 2. The polar-area formula gives \(A=\frac{1}{2}\int_{0}^{2\pi}(2|\cos\theta|)^{2}\,d\theta=2\int_{0}^{2\pi}\cos^{2}\theta\,d\theta\). 3. Since \(\int_{0}^{2\pi}\cos^{2}\theta\,d\theta=\pi\), the combined area is \(2\pi\).

Answer

\(A=\frac{1}{2}\int_{0}^{2\pi}(2|\cos\theta|)^{2}\,d\theta=2\pi\).

Area of a polar region 67

For \(r=2+\sin\theta\), compare the areas swept on \(0\le\theta\le\pi\) and \(\pi\le\theta\le2\pi\). Find both areas and their difference.

Hints

- Use separate area integrals for the upper and lower angular intervals. - The squared sine term contributes equally to both intervals. - The sign change in the sine term creates the difference.

Solution

1. On \([0,\pi]\), \(A_1=\frac{1}{2}\int_{0}^{\pi}(2+\sin\theta)^2\,d\theta=\frac{9\pi}{4}+4\). 2. On \([\pi,2\pi]\), \(A_2=\frac{1}{2}\int_{\pi}^{2\pi}(2+\sin\theta)^2\,d\theta=\frac{9\pi}{4}-4\). 3. Therefore, the first area exceeds the second by \(8\).

Answer

\(A_1=\frac{9\pi}{4}+4\), \(A_2=\frac{9\pi}{4}-4\), and \(A_1-A_2=8\).

Area of a polar region 68

For the cardioid \(r=1+\sin\theta\), find the exact area swept as \(\theta\) goes from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

Hints

- Which part of the cardioid is traced over the stated interval? - Use the square of the radius, not the radius alone, in the area formula. - A symmetric interval can simplify the integral of a sine term.

Solution

1. Use the polar-area formula: \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}(1+\sin\theta)^2\,d\theta\). 2. Expand the square: \(1+2\sin\theta+\sin^2\theta\). 3. The sine term integrates to zero over the symmetric interval. The integrals of the other two terms are \(\pi\) and \(\frac{\pi}{2}\). 4. Thus \(A=\frac{1}{2}\left(\pi+\frac{\pi}{2}\right)=\frac{3\pi}{4}\).

Answer

\(\frac{3\pi}{4}\)

Area of a polar region 69

Find the exact area swept by \(r=1-\cos(2\theta)\) for \(0\le\theta\le\pi\), from one pole crossing to the next.

Hints

- Use the pole-to-pole interval stated in the problem. - Expand the squared radius before integrating. - A power-reduction identity handles the squared cosine term.

Solution

1. Set up \(A=\frac{1}{2}\int_{0}^{\pi}(1-\cos(2\theta))^2\,d\theta\). 2. Expand the square to \(1-2\cos(2\theta)+\cos^2(2\theta)\). 3. On \([0,\pi]\), the middle term integrates to \(0\), and the other terms contribute \(\pi\) and \(\frac{\pi}{2}\). 4. Therefore, \(A=\frac{1}{2}\left(\frac{3\pi}{2}\right)=\frac{3\pi}{4}\).

Answer

\(\frac{3\pi}{4}\)

Area of a polar region 70

For the limaçon \(r=1-2\cos\theta\), identify the inner loop, find the interval that traces it once, and calculate its exact area.

Hints

- Find consecutive angles where the curve passes through the pole. - A negative radius can trace a loop; the polar-area formula still uses \(r^2\). - Use symmetry or a power-reduction identity when evaluating the integral.

Solution

1. The loop begins and ends at the pole. Solve \(1-2\cos\theta=0\): the consecutive angles around \(\theta=0\) are \(-\frac{\pi}{3}\) and \(\frac{\pi}{3}\). 2. On that interval the radius is negative and traces the small loop once. 3. The loop area is \(A=\frac{1}{2}\int_{-\pi/3}^{\pi/3}(1-2\cos\theta)^2\,d\theta\). 4. Evaluating gives \(A=\frac{1}{2}(2\pi-3\sqrt{3})=\pi-\frac{3\sqrt{3}}{2}\).

Answer

Inner-loop interval: \(\left[-\frac{\pi}{3},\frac{\pi}{3}\right]\) Area: \(\pi-\frac{3\sqrt{3}}{2}\)

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Calculus math worksheets

Area of a polar region

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