Free math worksheets

- Imagine a ray from the pole through the first quadrant. - The outer boundary is the curve reached farther from the pole along that ray. - Use the curve labels rather than color to state your answer.

1. Along every first-quadrant ray, curve \(c\) is farther from the pole than curve \(d\). 2. Therefore, \(c\) is the outer boundary and \(d\) is the inner boundary.

Outer boundary: \(c\). Inner boundary: \(d\).

- Compare the area of a narrow sector reaching the outer radius with one reaching only the inner radius. - Sector area depends on the square of the radius. - Subtract the inner sector contribution from the outer one before simplifying.

1. The area between two narrow sectors is \(dA=\frac{1}{2}(R^2-r^2)\,d\theta\). 2. Therefore, \(\frac{dA}{d\theta}=\frac{1}{2}(4^2-2^2)=\frac{1}{2}(12)=6\).

\(\frac{dA}{d\theta}=6\)

- Use the coordinates of \(P\) to determine the angle of the ray from the pole to that point. - Compare which labeled curve is farther from the pole on each side of \(P\). - A single outer-minus-inner order works only while the same curve remains outer.

1. The point \(P=(1,1)\) lies on the ray with \(\tan(\theta)=\frac{1}{1}=1\), so \(\theta=\frac{\pi}{4}\). 2. From the figure, \(c\) is farther from the pole before \(P\), while \(d\) is farther from the pole after \(P\). 3. Therefore, the outer-minus-inner order changes at \(\theta=\frac{\pi}{4}\), so one integral with a fixed order would have the wrong sign on one side. The area integral must be split at \(\frac{\pi}{4}\).

The switch occurs at \(\theta=\frac{\pi}{4}\). The integral must be split there because the outer and inner boundaries exchange roles.

- Picture one narrow sector of the region and compare the two radii that bound it. - How could the area of the outer sector and the inner sector be combined to represent only the band between them? - Use the exact first-quadrant angular width once your area model is established.

1. The outer radius is \(3\) and the inner radius is \(1\) throughout the stated interval. 2. The area is \(A=\frac{1}{2}\int_{0}^{\pi/2}(3^{2}-1^{2})\,d\theta\). 3. Therefore, \(A=\frac{1}{2}(8)\left(\frac{\pi}{2}\right)=2\pi\).

\(A=\frac12\int_0^{\pi/2}(3^2-1^2)\,d\theta=2\pi\)

- Check whether the same curve stays farther from the pole for the entire interval. - Think of the desired band as the difference between two sector areas at each angle. - Full-period trigonometric behavior can simplify the exact evaluation after the setup is established.

1. Since \(2+\cos(\theta)\ge1\), the limaçon is outer throughout the interval. 2. The area is \(A=\frac{1}{2}\int_{0}^{2\pi}((2+\cos(\theta))^{2}-1^{2})\,d\theta\). 3. Expanding and integrating gives \(A=\frac{1}{2}(6\pi+\pi)=\frac{7\pi}{2}\).

\(A=\frac{7\pi}{2}\)

- Before comparing the two radii, determine the sign of their common trigonometric factor on the interval. - Look for structure shared by the two radial functions that can simplify the area comparison. - Symmetry can help with the exact evaluation after the between-curves model is set up.

1. Because \(\cos(\theta)\ge0\) on the stated interval, \(4\cos(\theta)\) is the outer radius throughout. 2. The area is \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}((4\cos(\theta))^{2}-(2\cos(\theta))^{2})\,d\theta\). 3. Thus, \(A=6\int_{-\pi/2}^{\pi/2}\cos^{2}(\theta)\,d\theta=3\pi\).

\(A=\frac12\int_{-\pi/2}^{\pi/2}\left((4\cos\theta)^2-(2\cos\theta)^2\right)\,d\theta=3\pi\)

- Compare the two radial expressions structurally before expanding anything. - How does multiplying every radius of one curve by a constant affect the area between the curves? - Use full-period behavior only after you have simplified the geometric area relationship.

1. The outer radius is twice the inner radius, and both are nonnegative throughout the interval. 2. The area is \(A=\frac{1}{2}\int_{0}^{2\pi}((2+2\cos(\theta))^{2}-(1+\cos(\theta))^{2})\,d\theta\). 3. Factoring gives \(A=\frac{3}{2}\int_{0}^{2\pi}(1+\cos(\theta))^{2}\,d\theta=\frac{9\pi}{2}\).

\(A=\frac{9\pi}{2}\)

- Determine whether one radial function stays larger for every angle in the interval. - Represent the desired band as the difference between the areas swept to the two radii. - Use full-period trigonometric behavior to check the exact evaluation after simplifying.

1. The difference between the radii is \(2+\sin(\theta)\ge1\), so the first limaçon is outer throughout the interval. 2. The area is \(A=\frac{1}{2}\int_{0}^{2\pi}((3+2\sin(\theta))^{2}-(1+\sin(\theta))^{2})\,d\theta\). 3. Expanding and integrating gives \(A=\frac{1}{2}(16\pi+3\pi)=\frac{19\pi}{2}\).

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

- Compare the proposed expression with the geometry of subtracting a smaller sector from a larger sector at the same angle. - Ask where the radius enters the area of each individual sector before combining them. - Use the first-quadrant angular width only after the corrected area expression is established.

1. The polar area between curves uses the difference of the squared radii, not the square of their difference. 2. The correct setup is \(A=\frac12\int_0^{\pi/2}(3^2-1^2)\,\text{d}\theta\). 3. Therefore, \(A=\frac12(8)\left(\frac{\pi}{2}\right)=2\pi\).

\(A=2\pi\)

- A negative area result is a clue to inspect how the two radial boundaries were ordered. - Compare the radii over the full interval before deciding which sector area should be removed from which. - Check that the corrected integrand represents a nonnegative geometric area contribution.

1. Since \(2+\cos\theta\ge1\), the limaçon is the outer curve throughout the interval. 2. The correct setup is \(A=\frac12\int_0^{2\pi}((2+\cos\theta)^2-1^2)\,\text{d}\theta\). 3. Expanding and integrating gives \(A=\frac{7\pi}{2}\).

\(A=\frac{7\pi}{2}\)

- Compare the two radii over the entire interval. - Simplify the square of the square root before integrating. - Preserve four digits after the decimal point in the final answer.

1. The square-root radius is greater than \(1\) throughout the interval, so it is the outer radius. 2. The area is \(A=\frac{1}{2}\int_{0}^{2}((\sqrt{\theta+2})^{2}-1)\,d\theta=\frac{1}{2}\int_{0}^{2}(\theta+1)\,d\theta=2\). 3. Rounded to 4 decimal places, \(A\approx2.0000\).

\(A\approx2.0000\)

- View the band area as an accumulation function with a variable upper limit. - Differentiate before substituting the requested angle. - Use the squared outer and inner radii in the instantaneous area rate.

1. The accumulated band area is \(A(\theta)=\frac{1}{2}\int_0^{\theta}\left((3+u)^2-(1+u)^2\right)\,du\). 2. By the Fundamental Theorem of Calculus, \(\frac{dA}{d\theta}=\frac{1}{2}\left((3+\theta)^2-(1+\theta)^2\right)\). 3. At \(\theta=2\), \(\frac{dA}{d\theta}=\frac{1}{2}(25-9)=8\).

\(8\)

- Identify which radius is larger throughout the stated interval. - Subtract the squared radii rather than the radii themselves. - Simplify the difference before integrating.

1. On the interval, \(\theta+2\) is the outer radius and \(\theta\) is the inner radius. 2. The area is \(A=\frac{1}{2}\int_0^1\left((\theta+2)^2-\theta^2\right)\,d\theta\). 3. Simplify the integrand to \(4\theta+4\). 4. Therefore, \(A=\frac{1}{2}\left[2\theta^2+4\theta\right]_0^1=3\).

\(3\)

- Use the sign of the trigonometric term to decide which curve is farther from the pole on the interval. - Look for symmetry before committing to a long exact calculation. - Model only the band between the two radial boundaries, not the full area inside either curve.

1. The curves intersect at the endpoints, and \(2+\cos(\theta)\ge2\) between them. 2. The area is \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}((2+\cos(\theta))^{2}-2^{2})\,d\theta\). 3. Expanding and integrating gives \(A=4+\frac{\pi}{4}\).

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

- Verify the intersection angles by equating the two radial functions. - Use symmetry to reduce the outer-minus-inner integral. - Expand the squared cardioid term before applying exact trigonometric integrals.

1. The curves intersect when \(3\cos(\theta)=1+\cos(\theta)\), which gives the stated bounds, and the circle is outer between them. 2. By symmetry, \(A=\int_{0}^{\pi/3}(9\cos^{2}(\theta)-(1+\cos(\theta))^{2})\,d\theta\). 3. Expanding and evaluating gives \(A=\pi\).

\(A=\pi\)

- Ask whether the same curve is farther from the pole across the entire first quadrant. - Identify any angle where the two radial distances are equal, then examine the order on either side. - Your final response should represent the complete region but remain unevaluated.

1. The curves intersect at \(\theta=\frac{\pi}{4}\). The cosine cardioid is outer before this angle, and the sine cardioid is outer afterward. 2. Therefore, the required setup is \(A=\frac{1}{2}\int_{0}^{\pi/4}((1+\cos(\theta))^{2}-(1+\sin(\theta))^{2})\,d\theta+\frac{1}{2}\int_{\pi/4}^{\pi/2}((1+\sin(\theta))^{2}-(1+\cos(\theta))^{2})\,d\theta\).

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

- Use the supplied bounds to investigate what happens to the two radii at the endpoints. - Check one representative interior angle to determine the radial ordering on the interval. - Stop after constructing an integral for only the band between the curves.

1. The curves meet at both endpoints, and \(3\cos(2\theta)\ge1\) throughout the interval. 2. Therefore, the required setup is \(A=\frac{1}{2}\int_{-\frac{1}{2}\arccos(1/3)}^{\frac{1}{2}\arccos(1/3)}((3\cos(2\theta))^{2}-1^{2})\,d\theta\).

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

- Determine what the stated endpoint angles mean for the two radial boundaries. - Use an interior angle to decide which boundary is farther from the pole between the endpoints. - Your answer should be an unevaluated area model for the region between the curves.

1. The curves meet at both endpoints, and \(2\sin(\theta)\ge1\) throughout the interval. 2. Therefore, the required setup is \(A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}((2\sin(\theta))^{2}-1^{2})\,d\theta\).

\(A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}((2\sin(\theta))^{2}-1^{2})\,d\theta\)

- Determine the range of the varying radius on the stated interval before choosing an outer curve. - Think of the desired band as the difference between two sector areas at each angle. - Stop after writing a complete unevaluated model for the region.

1. Since \(\cos(\theta)\ge\frac{1}{2}\) on the stated interval, \(1+2\cos(\theta)\) is the outer radius throughout. 2. Therefore, the required setup is \(A=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}((1+2\cos(\theta))^{2}-1^{2})\,d\theta\).

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

- Do not assume in advance which constant radius is larger; express the band area in a way that works in either order. - The given area determines the magnitude of the difference between the squared radii. - After solving the resulting cases, use the condition \(a>0\) and reject any impossible value of \(a^{2}\).

1. Without assuming which circle is outer, the area is \(A=\frac{1}{2}\int_{0}^{\pi}|a^{2}-1|\,d\theta=\frac{\pi}{2}|a^{2}-1|\). 2. Setting this equal to \(3\pi\) gives \(|a^{2}-1|=6\). 3. The equations \(a^{2}-1=6\) and \(a^{2}-1=-6\) give \(a^{2}=7\) or \(a^{2}=-5\). Only \(a^{2}=7\) is possible. 4. Since \(a>0\), \(a=\sqrt{7}\).

\(\frac12\int_0^\pi|a^2-1|\,d\theta=3\pi\) \(a=\sqrt7\)

- The two curves have the same angular factor, but do not assume which scale is larger before using the given area. - Express the area so that either possible radial ordering is handled correctly. - Use the exact symmetric squared-cosine contribution, then test the algebraic cases for \(a^{2}\).

1. Since \(\cos(\theta)\ge0\) on the interval, the area can be written without assuming the scale order as \(A=\frac{1}{2}|a^{2}-1|\int_{-\pi/2}^{\pi/2}\cos^{2}(\theta)\,d\theta=\frac{\pi}{4}|a^{2}-1|\). 2. Setting this equal to \(4\pi\) gives \(|a^{2}-1|=16\). 3. The two algebraic cases give \(a^{2}=17\) or \(a^{2}=-15\), so only \(a^{2}=17\) is possible. 4. Because \(a>0\), \(a=\sqrt{17}\).

\(\frac12\int_{-\pi/2}^{\pi/2}|a^2-1|\cos^2\theta\,d\theta=4\pi\) \(a=\sqrt{17}\)

- Compare the two curves as radial scalings of the same base cardioid without deciding the scale order too early. - Represent the area using the magnitude of the squared scale difference. - After relating that magnitude to the given area, reject any algebraic case that cannot equal \(a^{2}\).

1. Without assuming whether \(a\) is greater or less than \(1\), the area is \(A=\frac{1}{2}|a^{2}-1|\int_{0}^{2\pi}(1+\cos(\theta))^{2}\,d\theta=\frac{3\pi}{2}|a^{2}-1|\). 2. Setting this equal to \(9\pi\) gives \(|a^{2}-1|=6\). 3. The two cases give \(a^{2}=7\) or \(a^{2}=-5\), so only \(a^{2}=7\) is possible. 4. Since \(a>0\), \(a=\sqrt{7}\).

\(a=\sqrt{7}\)

- Test whether the same radial function is larger near both ends of the first-quadrant interval. - If the ordering changes, determine where the two radii become equal. - Build the total area from pieces on which the radial ordering is consistent.

1. The curves intersect at \(\theta=\frac{\pi}{4}\). Before this angle, \(2\cos\theta\) is outer; afterward, \(2\sin\theta\) is outer. 2. Thus, \(A=\frac12\int_0^{\pi/4}((2\cos\theta)^2-(2\sin\theta)^2)\,\text{d}\theta+\frac12\int_{\pi/4}^{\pi/2}((2\sin\theta)^2-(2\cos\theta)^2)\,\text{d}\theta\). 3. Evaluating gives \(A=2\).

The outer curve changes at \(\theta=\frac{\pi}{4}\), so the integral must be split there. The area is \(A=2\).

- Verify the outer curve using the sign of \(\sin(\theta)\) on the interval. - Square each radius separately before subtracting. - Keep full numerical precision until the final rounding step.

1. Since \(\sin(\theta)\ge0\) on the interval, \(e^{\sin(\theta)}\ge1\), so the exponential curve is outer. 2. The area is \(A=\frac{1}{2}\int_{0}^{\pi}(e^{2\sin(\theta)}-1)\,d\theta\approx5.0532817232\). 3. Rounded to 3 decimal places, \(A\approx5.053\).

\(A\approx5.053\)

- Check the minimum possible value of the variable radius. - Expand the squared outer radius and use a power-reduction identity. - Round only after evaluating the complete definite integral.

1. Since \(2+\cos(3\theta)\ge1\), the first curve is outer throughout the interval. 2. The area is \(A=\frac{1}{2}\int_{0}^{\pi/2}((2+\cos(3\theta))^{2}-1)\,d\theta=\frac{7\pi}{8}-\frac{2}{3}\approx2.0822269052\). 3. Rounded to 3 decimal places, \(A\approx2.082\).

\(A\approx2.082\)

- Before applying the trapezoidal rule, decide what quantity each table row must contribute to an area-between-polar-curves calculation. - Check the angular spacing and identify which transformed values are endpoints and which are interior values. - Keep the geometry of the polar area model separate from the numerical quadrature weights while building the estimate.

1. Compute the polar-area integrand values \(f(\theta)=\frac{1}{2}(R^{2}-r^{2})\): \(4\), \(6\), \(\frac{5}{2}\), \(12\), and \(6\). 2. The angle spacing is \(h=\frac{1}{2}\). 3. The trapezoidal approximation is \(A\approx\frac{h}{2}\left[f_0+2(f_1+f_2+f_3)+f_4\right]=\frac{1}{4}(51)=\frac{51}{4}\).

\(A\approx\frac{51}{4}=12.75\)

- Check whether the same boundary remains farther from the pole throughout. - Simplify the outer-minus-inner expression before expanding both squares. - Use the behavior of a complete trigonometric period on the interval.

1. The first radius is larger by \(1\) throughout the interval. 2. The area is \(A=\frac{1}{2}\int_0^{\pi}\left((5+\cos(2\theta))^2-(4+\cos(2\theta))^2\right)\,d\theta\). 3. The difference of squares simplifies to \(9+2\cos(2\theta)\). 4. The cosine term integrates to \(0\), so \(A=\frac{9\pi}{2}\).

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

- Compute the reference spiral area before using the stated ratio. - Express the band as outer squared radius minus inner squared radius. - Use the positivity condition after solving the resulting equation.

1. The area enclosed by \(r=\theta\) is \(\frac{1}{2}\int_0^2\theta^2\,d\theta=\frac{4}{3}\). 2. The band area is therefore \(6\cdot\frac{4}{3}=8\). 3. Since \(a+\theta\) is outer, the band area is \(\frac{1}{2}\int_0^2\left((a+\theta)^2-\theta^2\right)\,d\theta=a^2+2a\). 4. Solve \(a^2+2a=8\) to obtain \(a=2\), using \(a>0\).

\(a=2\)

- Determine the common angular width of the two subintervals. - At each midpoint, compare the squared outer and inner radii. - Apply the numerical rule to the polar-area integrand, including its constant factor.

1. Each subinterval has width \(\Delta\theta=\frac{\pi}{4}\). 2. The midpoint approximation is \(A\approx\frac{1}{2}\Delta\theta\left((4^2-1^2)+(3^2-2^2)\right)\). 3. The squared-radius differences are \(15\) and \(5\). 4. Therefore, \(A\approx\frac{1}{2}\cdot\frac{\pi}{4}\cdot20=\frac{5\pi}{2}\).

\(A\approx\frac{5\pi}{2}\)

- Comparing the radii directly is harder than comparing their squares. - Simplify the squared-radius difference before identifying the outer curve. - Use the simplified expression as the area integrand.

1. Compare squared radii: \((\theta+1)^2-(2\theta+1)=\theta^2\ge0\). 2. Thus \(r=\theta+1\) is the outer curve, with equality only at \(\theta=0\). 3. The area is \(A=\frac{1}{2}\int_0^2\theta^2\,d\theta\). 4. Therefore, \(A=\frac{1}{2}\cdot\frac{8}{3}=\frac{4}{3}\).

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

- Use the given restriction to confirm the same outer boundary applies throughout. - Expand the outer squared radius and use full-period trigonometric integrals. - Select the geometrically meaningful solution after solving for the squared radius.

1. The condition \(a\le2\) keeps the circle inside the outer curve throughout the interval. 2. Set up \(\frac{1}{2}\int_0^{2\pi}\left((3+\cos\theta)^2-a^2\right)\,d\theta=\frac{11\pi}{2}\). 3. Evaluating gives \(\frac{\pi}{2}(19-2a^2)=\frac{11\pi}{2}\). 4. Thus \(a^2=4\), and the positive radius is \(a=2\).

\(a=2\)

- Find the intersection by comparing the squared radii. - Determine which squared radius is larger on each side of the intersection. - Split the integral where the outer curve changes.

1. The curves intersect when \(6-2\theta=2+2\theta\), so \(\theta=1\). 2. The first curve is outer on \([0,1]\), and the second is outer on \([1,2]\). 3. The area is \(A=\frac{1}{2}\int_0^1(4-4\theta)\,d\theta+\frac{1}{2}\int_1^2(4\theta-4)\,d\theta\). 4. Each integral contributes \(1\), so the total area is \(2\).

\(2\)

- Write one area expression for the band and another for the inner region. - Compare squared radii so the common trigonometric term can cancel in the band. - Equate the two areas only after evaluating each integral.

1. The band area is \(\frac{1}{2}\int_0^{\pi}(a-1)\,d\theta=\frac{\pi}{2}(a-1)\). 2. The inner area is \(\frac{1}{2}\int_0^{\pi}(1+\sin\theta)\,d\theta=\frac{\pi}{2}+1\). 3. Equate the areas: \(\frac{\pi}{2}(a-1)=\frac{\pi}{2}+1\). 4. Solving gives \(a=2+\frac{2}{\pi}\), which satisfies \(a>1\).

\(a=2+\frac{2}{\pi}\)

- Confirm that the same spiral stays farther from the pole. - Factor the common exponential after subtracting the squared radii. - Evaluate the unbounded interval as an improper integral.

1. The first spiral is the outer boundary for all \(\theta\ge0\). 2. The improper area integral is \(A=\frac{1}{2}\int_0^{\infty}(9e^{-2\theta}-e^{-2\theta})\,d\theta\). 3. Thus \(A=4\int_0^{\infty}e^{-2\theta}\,d\theta\). 4. The improper integral equals \(\frac{1}{2}\), so \(A=2\).

\(2\)

- Simplify the difference of squares before integrating. - Evaluate the exponential at the logarithmic endpoint exactly. - Keep the constant and exponential contributions separate.

1. The first radius is larger throughout the interval. 2. The area is \(A=\frac{1}{2}\int_0^{\ln 2}\left((e^{\theta}+1)^2-e^{2\theta}\right)\,d\theta\). 3. The squared-radius difference simplifies to \(2e^{\theta}+1\). 4. Evaluating gives \(A=\frac{1}{2}\left(2(e^{\ln2}-1)+\ln2\right)=1+\frac{\ln2}{2}\).

\(1+\frac{\ln2}{2}\)

- Confirm which curve is outer on the full interval. - Expand the squared radical expression carefully. - Integrate each power of the angle separately.

1. The first radius is the outer boundary throughout the interval. 2. The area is \(A=\frac{1}{2}\int_0^4\left((1+\sqrt{\theta})^2-1\right)\,d\theta\). 3. Simplify the integrand to \(\theta+2\sqrt{\theta}\). 4. Therefore, \(A=\frac{1}{2}\left[\frac{\theta^2}{2}+\frac{4}{3}\theta^{3/2}\right]_0^4=\frac{28}{3}\).

\(\frac{28}{3}\)

- Write a separate outer-minus-inner area integral for each band. - Compare the constant and oscillating parts of the squared-radius differences. - Examine the net contribution of each trigonometric term over the full interval.

1. Band A has area \(A_A=\frac{1}{2}\int_0^{\pi}(3+2\cos\theta)\,d\theta\). 2. The cosine term integrates to \(0\), so \(A_A=\frac{3\pi}{2}\). 3. Band B has area \(A_B=\frac{1}{2}\int_0^{\pi}(3+2\sin(2\theta))\,d\theta\). 4. The sine term also integrates to \(0\), so \(A_B=\frac{3\pi}{2}\). 5. Therefore, the two bands have equal area.

The bands have equal area: \(A_A=A_B=\frac{3\pi}{2}\).

- Compare the squared radii before considering the radicals. - Look for a common parameter term in the outer-minus-inner expression. - Use the stated condition to confirm both radii are real.

1. The first curve is the outer boundary because its squared radius exceeds the second by \(2\theta^2\). 2. The area is \(A=\frac{1}{2}\int_0^1\left((a+\theta^2)-(a-\theta^2)\right)\,d\theta\). 3. The parameter terms cancel, leaving \(A=\int_0^1\theta^2\,d\theta=\frac{1}{3}\). 4. Therefore, the area is independent of \(a\) because both squared radii contain the same additive parameter.

\(\frac{1}{3}\), independent of \(a\).

- Express the accumulated band area using a variable upper limit. - Differentiate before solving for the angle. - Use the interval restriction to select the correct trigonometric solution.

1. The instantaneous band-area rate is \(\frac{dA}{d\theta}=\frac{1}{2}\left(3^2-(2\cos\theta)^2\right)\). 2. Set \(\frac{1}{2}(9-4\cos^2\theta)=\frac{7}{2}\). 3. This gives \(\cos^2\theta=\frac{1}{2}\). 4. On \(0<\theta<\frac{\pi}{2}\), the solution is \(\theta=\frac{\pi}{4}\).

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

- Determine where the two radii exchange order. - Simplify the difference of their squares before integrating. - Use symmetry to evaluate the resulting absolute-value expression.

1. The curves exchange which one is outer when \(\cos\theta=0\). 2. The magnitude of the squared-radius difference is \(4a|\cos\theta|\). 3. Thus the total area is \(A=\frac{1}{2}\int_0^{2\pi}4a|\cos\theta|\,d\theta=8a\). 4. Set \(8a=24\) to obtain \(a=3\), which satisfies \(a>1\).

\(a=3\)

- Find where the two radii are equal before setting up the area. - Test one angle on each side of the intersection to identify the outer curve. - Split the area integral where the order changes.

1. Solve \(2\theta=3-\theta\) to get the intersection angle \(\theta=1\). 2. On \([0,1]\), \(3-\theta\) is the outer radius. On \([1,2]\), \(2\theta\) is the outer radius. 3. Thus \(A=\frac{1}{2}\int_0^1\left((3-\theta)^2-(2\theta)^2\right)\,d\theta+\frac{1}{2}\int_1^2\left((2\theta)^2-(3-\theta)^2\right)\,d\theta\). 4. Evaluating the two integrals gives \(A=6\).

\(6\)

- First determine where the two radii are equal. - The identity of the outer curve changes at that angle. - Symmetry can combine the two resulting integrals.

1. The curves intersect when \(e^{\theta}=e^{-\theta}\), which gives \(\theta=0\). 2. On \([-1,0]\), \(e^{-\theta}\) is the outer radius. On \([0,1]\), \(e^{\theta}\) is the outer radius. 3. By symmetry, \(A=\int_0^1\left(e^{2\theta}-e^{-2\theta}\right)\,d\theta\). 4. Evaluating gives \(A=\frac{e^2+e^{-2}}{2}-1\).

\(\frac{e^2+e^{-2}}{2}-1\)

- Use the interval that traces one corresponding pair of petals. - Factor the common trigonometric term after subtracting the squared radii. - A power-reduction identity can simplify the remaining integral.

1. Both radii are nonnegative on the stated petal interval, and the first is the outer radius. 2. The band area is \(A=\frac{1}{2}\int_{-\pi/6}^{\pi/6}\left(16\cos^2(3\theta)-4\cos^2(3\theta)\right)\,d\theta\). 3. This becomes \(A=6\int_{-\pi/6}^{\pi/6}\cos^2(3\theta)\,d\theta\). 4. The integral equals \(\frac{\pi}{6}\), so \(A=\pi\).

\(\pi\)

- Locate all angles where the radii are equal. - Determine how the outer curve changes across those angles. - Symmetry can replace several separate integrals with an absolute-value integral.

1. The curves intersect when \(\cos\theta=0\), at \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\). 2. The curve \(r=2+\cos\theta\) is outer where \(\cos\theta\ge0\), and \(r=2-\cos\theta\) is outer where \(\cos\theta\le0\). 3. The squared-radius difference has magnitude \(8|\cos\theta|\). 4. Therefore, \(A=\frac{1}{2}\int_0^{2\pi}8|\cos\theta|\,d\theta=16\).

\(16\)

- Compare the possible values of the two radii first. - Squaring removes the absolute value from the inner radius. - A power-reduction identity evaluates the remaining full-period integral.

1. Since \(0\le2|\sin\theta|\le2\), the circle is the outer boundary throughout the interval. 2. The area is \(A=\frac{1}{2}\int_0^{2\pi}\left(4-4\sin^2\theta\right)\,d\theta\). 3. This simplifies to \(A=2\int_0^{2\pi}\cos^2\theta\,d\theta\). 4. Since the integral of \(\cos^2\theta\) over \([0,2\pi]\) is \(\pi\), \(A=2\pi\).

\(A=\frac12\int_0^{2\pi}\left(4-4\sin^2\theta\right)\,d\theta=2\pi\)

- Solve for every intersection in the stated interval. - Test the ordering of the radii on each interval between intersections. - Split the area calculation where the outer curve changes.

1. The curves intersect when \(2\theta=\theta^2\), so \(\theta=0\) and \(\theta=2\). 2. On \([0,2]\), \(2\theta\) is the outer radius. On \([2,3]\), \(\theta^2\) is the outer radius. 3. The area is \(A=\frac{1}{2}\int_0^2(4\theta^2-\theta^4)\,d\theta+\frac{1}{2}\int_2^3(\theta^4-4\theta^2)\,d\theta\). 4. Evaluating gives \(A=\frac{317}{30}\).

\(\frac{317}{30}\)

- Compare the range of the inner radius with the circle’s radius. - Expand the squared inner boundary before integrating. - Use full-period trigonometric averages for the oscillating terms.

1. Since \(1\le2+\cos(2\theta)\le3\), the circle is the outer boundary throughout. 2. The area is \(A=\frac{1}{2}\int_0^{2\pi}\left(16-(2+\cos(2\theta))^2\right)\,d\theta\). 3. Expand the inner square. The linear cosine term integrates to \(0\), and \(\int_0^{2\pi}\cos^2(2\theta)\,d\theta=\pi\). 4. Therefore, \(A=\frac{1}{2}(24\pi-\pi)=\frac{23\pi}{2}\).

\(\frac{23\pi}{2}\)

- Compare the two radii using a familiar inequality for the exponential function. - Integrate the squared boundaries separately. - Combine the exact exponential and polynomial contributions with a common denominator.

1. Since \(e^{\theta}\ge1+\theta\) for \(\theta\ge0\), the exponential curve is the outer boundary. 2. The area is \(A=\frac{1}{2}\int_0^1\left(e^{2\theta}-(1+\theta)^2\right)\,d\theta\). 3. The exponential contribution is \(\frac{e^2-1}{2}\), and the polynomial contribution is \(\frac{7}{3}\). 4. Therefore, \(A=\frac{1}{2}\left(\frac{e^2-1}{2}-\frac{7}{3}\right)=\frac{3e^2-17}{12}\).

\(\frac{3e^2-17}{12}\)

- Set the two radial expressions equal and select the pair of solutions surrounding the positive x-axis. - Determine which curve is farther from the pole between those intersections. - Use symmetry and a power-reduction identity when evaluating the squared cosine.

1. On the petal centered at \(\theta=0\), solve the intersection equation: \(3\cos(2\theta)=1\), so \(\theta=\pm\frac{1}{2}\arccos\left(\frac{1}{3}\right)\). 2. The rose is farther from the pole between these angles. By symmetry, \(A=\int_{0}^{\frac{1}{2}\arccos(1/3)}\left(9\cos^{2}(2\theta)-1\right)\,d\theta\). 3. Power reduction and exact-angle evaluation give \(A=\frac{7}{4}\arccos\left(\frac{1}{3}\right)+\frac{\sqrt{2}}{2}\).

\(A=\frac{7}{4}\arccos\left(\frac{1}{3}\right)+\frac{\sqrt{2}}{2}\)

- Set the rose radius equal to the circle radius and solve within the petal centered at \(\theta=\frac{\pi}{6}\). - Check which radius is larger between the two intersections. - Use a multiple-angle substitution before applying power reduction.

1. On the petal \(0\le\theta\le\frac{\pi}{3}\), solve \(2\sin(3\theta)=1\). The bounding solutions are \(\theta=\frac{\pi}{18}\) and \(\theta=\frac{5\pi}{18}\). 2. The rose is farther from the pole between these angles, so \(A=\frac{1}{2}\int_{\pi/18}^{5\pi/18}\left(4\sin^{2}(3\theta)-1\right)\,d\theta\). 3. Using \(u=3\theta\) and power reduction gives \(A=\frac{\pi}{9}+\frac{\sqrt{3}}{6}\).

\(A=\frac{\pi}{9}+\frac{\sqrt{3}}{6}\)

- Solve for the angle at which the two radii are equal. - Test one angle on each side of the intersection to identify the outer curve. - Split the area integral and use \(\cos^{2}\theta-\sin^{2}\theta=\cos(2\theta)\).

1. Set the radii equal: \(2\cos\theta=2\sin\theta\). On the given interval, the intersection occurs at \(\theta=\frac{\pi}{4}\). 2. Testing the two sides shows that \(2\cos\theta\) is outer on \([0,\frac{\pi}{4}]\), while \(2\sin\theta\) is outer on \([\frac{\pi}{4},\frac{\pi}{2}]\). 3. Thus, \(A=\frac{1}{2}\int_{0}^{\pi/4}\left(4\cos^{2}\theta-4\sin^{2}\theta\right)\,d\theta+\frac{1}{2}\int_{\pi/4}^{\pi/2}\left(4\sin^{2}\theta-4\cos^{2}\theta\right)\,d\theta=2\).

\(A=2\)

- Set the two cardioid radii equal and solve on the first-quadrant interval. - Compare sine and cosine on each side of the intersection. - Use symmetry after writing the required split area integral.

1. Set the radii equal: \(1+\cos\theta=1+\sin\theta\). On the given interval, \(\theta=\frac{\pi}{4}\). 2. The cosine cardioid is outer before \(\frac{\pi}{4}\), and the sine cardioid is outer afterward. 3. By symmetry, the two split contributions are equal, so \(A=\int_{0}^{\pi/4}\left((1+\cos\theta)^2-(1+\sin\theta)^2\right)\,d\theta=2\sqrt{2}-\frac{3}{2}\).

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

- Set the variable radius equal to the constant radius to locate the switch. - Test the radii near \(\theta=0\) and near \(\theta=\frac{\pi}{2}\). - Use power reduction and an exact right-triangle value when evaluating the split integral.

1. Solve \(3\cos\theta=2\), giving the intersection \(\alpha=\arccos\left(\frac{2}{3}\right)\). 2. The curve \(r=3\cos\theta\) is outer on \([0,\alpha]\), and \(r=2\) is outer on \([\alpha,\frac{\pi}{2}]\). 3. Therefore, \(A=\frac{1}{2}\int_{0}^{\alpha}\left(9\cos^{2}\theta-4\right)\,d\theta+\frac{1}{2}\int_{\alpha}^{\pi/2}\left(4-9\cos^{2}\theta\right)\,d\theta=\sqrt{5}+\frac{1}{2}\arccos\left(\frac{2}{3}\right)-\frac{\pi}{8}\).

\(A=\sqrt{5}+\frac{1}{2}\arccos\left(\frac{2}{3}\right)-\frac{\pi}{8}\)

- Use the endpoint restriction to confirm which boundary is outer. - Translate the stated area into an equation with an unknown upper limit. - Check every algebraic solution against the allowed interval.

1. The circle is outer on \([0,b]\) because \(b<4\). 2. Set up \(\frac{1}{2}\int_0^b(16-\theta^2)\,d\theta=\frac{44}{3}\). 3. Evaluating gives \(8b-\frac{b^3}{6}=\frac{44}{3}\), or \(b^3-48b+88=0\). 4. Factor: \((b-2)(b^2+2b-44)=0\). 5. The only solution in \(0<b<4\) is \(b=2\).

\(b=2\)

- Compare squared radii to locate all intersection angles. - Determine which curve is outer on each resulting interval. - Use symmetry to combine the two end intervals.

1. The curves intersect when \(2+2\cos\theta=1\), so \(\theta=\pm\frac{2\pi}{3}\). 2. The radical curve is outer on \(\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]\), while the circle is outer on the two remaining subintervals. 3. Split the outer-minus-inner area integral at the two intersection angles. 4. The middle contribution is \(\sqrt{3}+\frac{2\pi}{3}\), and the two congruent outer contributions total \(\sqrt{3}-\frac{\pi}{3}\). 5. Therefore, the total area is \(2\sqrt{3}+\frac{\pi}{3}\).

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

- On each ray, the common region extends only to the smaller positive radius. - Solve for the angles where the two circle boundaries intersect. - Use symmetry, and split the integral where the smaller radius changes.

1. The circle \(r=2\cos\theta\) is traced for \(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\). The circles intersect where \(1=2\cos\theta\), so \(\theta=\pm\frac{\pi}{3}\). 2. For \(0\le\theta\le\frac{\pi}{3}\), the smaller radius is \(1\). For \(\frac{\pi}{3}\le\theta\le\frac{\pi}{2}\), the smaller radius is \(2\cos\theta\). 3. By symmetry, the common area is \(A=2\left[\frac{1}{2}\int_0^{\frac{\pi}{3}}1^2\,d\theta+\frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}(2\cos\theta)^2\,d\theta\right]\). 4. Evaluating gives \(A=\frac{\pi}{3}+\left[2\theta+\sin(2\theta)\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}=\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\).

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