Free math worksheets

- 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.

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 \(9\cdot7=63\).

\(63\)

- 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.

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

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

- 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.

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 \(A=\frac{1}{2}\int_0^{2\pi}(5-\cos(\theta))\,d\theta\).

\(A=\frac{1}{2}\int_0^{2\pi}(5-\cos(\theta))\,d\theta\)

- 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.

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

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

- First check what geometric portion the stated angle interval traces and whether any part is repeated. - Model the area by accumulating polar sectors whose radii vary with \(\theta\). - A standard squared-trigonometric identity can simplify the resulting exact integral.

1. Use \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}r^{2}\,d\theta\). 2. Substituting the radial function gives \(A=\frac{9}{2}\int_{-\pi/2}^{\pi/2}\cos^{2}(\theta)\,d\theta\). 3. Since the integral of \(\cos^{2}(\theta)\) over this interval is \(\frac{\pi}{2}\), \(A=\frac{9\pi}{4}\).

\(A=\frac12\int_{-\pi/2}^{\pi/2}(3\cos\theta)^2\,d\theta=\frac{9\pi}{4}\)

- Verify that the given interval traces the intended circle exactly once. - Relate each angle's radius to a small sector-area contribution. - Use an exact identity for the squared sine term rather than converting to decimals.

1. The polar area is \(A=\frac{1}{2}\int_{0}^{\pi}(4\sin(\theta))^{2}\,d\theta\). 2. Thus, \(A=8\int_{0}^{\pi}\sin^{2}(\theta)\,d\theta\). 3. Since \(\int_{0}^{\pi}\sin^{2}(\theta)\,d\theta=\frac{\pi}{2}\), the area is \(A=4\pi\).

\(A=\frac12\int_0^\pi(4\sin\theta)^2\,d\theta=4\pi\)

- 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.

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

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

- 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.

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 \(A=6\pi\).

\(A=6\pi\)

- 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.

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

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

- Use the given interval to determine exactly which part of the polar circle is being traced. - Think about how the radius of a narrow sector determines its area contribution. - An exact squared-trigonometric identity is useful after the area has been modeled.

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

\(A=\frac12\int_0^{\pi/2}(2\cos\theta)^2\,d\theta=\frac\pi2\)

- Interpret the stated angles geometrically before deciding how much of the circle is included. - Model the swept region using narrow polar sectors rather than starting from a memorized numerical result. - Keep the final evaluation exact; a squared-trigonometric identity can help.

1. The interval traces the right half of the circle, which is the portion 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 the squared-sine integral is \(\frac{\pi}{4}\), \(A=2\pi\).

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

- Simplify the square of the square-root radial function first. - Integrate the resulting linear expression over the stated interval. - Preserve four digits after the decimal point in the final rounded answer.

1. The area is \(A=\frac{1}{2}\int_{0}^{2}(\sqrt{\theta+1})^{2}\,d\theta\). 2. Simplifying before integration gives \(A=\frac{1}{2}\int_{0}^{2}(\theta+1)\,d\theta=2\). 3. Rounded to 4 decimal places, \(A\approx2.0000\).

\(A\approx2.0000\)

- Think of the spiral as sweeping narrow sectors while \(\theta\) changes from the first given angle to the second. - Which changing quantity determines the area contribution of each small sector? - Keep the exact bounds throughout the calculation.

1. The polar-area formula gives \(A=\frac{1}{2}\int_0^2r^2\,d\theta\). 2. Substitute \(r=\theta\): \(A=\frac{1}{2}\int_0^2\theta^2\,d\theta\). 3. Evaluating gives \(A=\frac{1}{2}\left[\frac{\theta^3}{3}\right]_0^2=\frac{4}{3}\).

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

- Compare the quantity supplied in the equation with the quantity that appears in a polar sector's area. - Consider what the oscillating term contributes over one full revolution. - Keep the calculation exact rather than introducing a square root unnecessarily.

1. Because the squared radius is given directly, \(A=\frac{1}{2}\int_0^{2\pi}(4+2\cos\theta)\,d\theta\). 2. The constant term contributes \(8\pi\), and the cosine term contributes \(0\) before the factor \(\frac{1}{2}\). 3. Therefore, \(A=4\pi\).

\(4\pi\)

- Model the area swept as the radius changes continuously with the angle. - Simplify the radial expression only after deciding what quantity enters the area model. - Use the relationship between exponential and logarithmic functions at the upper endpoint.

1. The area is \(A=\frac{1}{2}\int_0^{\ln2}e^{2\theta}\,d\theta\). 2. An antiderivative is \(\frac{1}{2}e^{2\theta}\), together with the outer factor \(\frac{1}{2}\). 3. Thus \(A=\frac{1}{4}[e^{2\theta}]_0^{\ln2}=\frac{1}{4}(4-1)=\frac{3}{4}\).

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

- 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.

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

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

- 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.

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, \(A=\frac{1}{2}\left(\tan\frac{\pi}{4}-\tan\left(-\frac{\pi}{4}\right)\right)=1\).

\(A=\frac12\int_{-\pi/4}^{\pi/4}\sec^2\theta\,d\theta=1\)

- 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.

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 \(\theta=1\), \(\frac{dA}{d\theta}=\frac{1}{2}(3)^2=\frac{9}{2}\).

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

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

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 \(a>2\).

\(a=6\)

- The square of the radial function is simpler than the radius itself. - Use separate integrals for the two angular intervals. - Compare the resulting areas only after evaluating both.

1. On \([0,1]\), \(A_1=\frac{1}{2}\int_0^1(1+\theta)\,d\theta=\frac{3}{4}\). 2. On \([1,3]\), \(A_2=\frac{1}{2}\int_1^3(1+\theta)\,d\theta=3\). 3. The ratio is \(\frac{A_2}{A_1}=\frac{3}{3/4}=4\).

\(A_1=\frac{3}{4}\), \(A_2=3\), and \(\frac{A_2}{A_1}=4\).

- 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.

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 \(A=\frac{9}{4}\left[\sin(2\theta)\right]_{-\pi/4}^{\pi/4}=\frac{9}{2}\).

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

- 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.

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, \(A=\frac{9}{2}\int_{-\pi/4}^{\pi/4}\cos^{2}(2\theta)\,d\theta=\frac{9\pi}{8}\).

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

- 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 \(\theta\).

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

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

- 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.

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, \(A=8\int_{-\pi/6}^{\pi/6}\cos^{2}(3\theta)\,d\theta=\frac{4\pi}{3}\).

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

- 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.

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

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

- 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.

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 \(A=\frac{1}{2}(2\pi-3\sqrt{3})=\pi-\frac{3\sqrt{3}}{2}\).

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

- 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.

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

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

- 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.

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

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

- 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.

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

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

- 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.

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

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

- 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.

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

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

- 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.

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 \(a>0\), \(a=4\sqrt{2}\).

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

- Think about how changing \(a\) scales every radius and therefore the enclosed area. - Keep the trigonometric contribution exact while relating the unknown scale to the stated area. - Check the sign condition only after solving the resulting scale equation.

1. The area is \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}a^{2}\cos^{2}(\theta)\,d\theta=\frac{\pi a^{2}}{4}\). 2. Set \(\frac{\pi a^{2}}{4}=9\pi\), so \(a^{2}=36\). 3. Because \(a>0\), \(a=6\).

\(\frac12\int_{-\pi/2}^{\pi/2}a^2\cos^2\theta\,d\theta=9\pi\) \(a=6\)

- 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.

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 \(a>0\), \(a=2\).

\(a=2\)

- 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.

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 \(a>0\), \(a=4\sqrt{2}\).

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

- Use each region's own angle interval rather than assuming both are full circles. - Evaluate the two polar-area integrals independently. - Compare the exact results after simplifying the squared radii.

1. Region A has area \(A_A=\frac{1}{2}\int_{0}^{\pi}2^{2}\,d\theta=2\pi\). 2. Region B has area \(A_B=\frac{1}{2}\int_{0}^{2\pi}(\sqrt{2})^{2}\,d\theta=2\pi\). 3. Therefore, the regions have equal area.

\(A_A=\frac12\int_0^\pi 2^2\,d\theta=2\pi\) \(A_B=\frac12\int_0^{2\pi}(\sqrt2)^2\,d\theta=2\pi\) The areas are equal.

- 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.

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.

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

- Compare the proposed setup with the geometric area of a narrow polar sector. - Ask what constant must appear when sector area is accumulated as the angle changes. - After correcting the setup, keep the trigonometric evaluation exact over the full tracing interval.

1. The entire region is traced once over \(0\le\theta\le2\pi\). 2. The correct area is \(A=\frac12\int_0^{2\pi}(2+\cos\theta)^2\,\text{d}\theta\). 3. Expanding and integrating gives \(A=\frac12(8\pi+\pi)=\frac{9\pi}{2}\).

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

- Determine how many petals the rose has before choosing an integration interval. - Locate the consecutive zeros surrounding one specified petal. - Compare the full-cycle integral with the one-petal integral to identify the overcount factor.

1. The interval \(0\le\theta\le2\pi\) covers all four petals, so it counts four congruent petals instead of one. 2. The petal centered on the positive x-axis is traced over \(-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}\). 3. Its area is \(A=\frac12\int_{-\pi/4}^{\pi/4}(3\cos(2\theta))^2\,\text{d}\theta=\frac{9\pi}{8}\).

The interval \(0\le\theta\le2\pi\) includes all four petals. The area of one petal is \(\frac{9\pi}{8}\).

- 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.

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, \(A\approx6.624\).

\(A\approx6.624\)

- 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.

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, \(A\approx2.868\).

\(A\approx2.868\)

- 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.

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 \(A_T=4A_p=8\pi\).

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

- 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.

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 \(A_T=3A_p=\frac{9\pi}{4}\).

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

- 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.

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 \(A_T=8A_p=2\pi\).

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

- 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.

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 \(b=3\).

\(b=3\)

- 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.

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 \(A\approx\frac{\pi}{24}(44)=\frac{11\pi}{6}\).

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

- 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.

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 \(k>0\), \(k=\sqrt{7}\).

\(k=\sqrt{7}\)

- 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.

1. Polar area depends on the average of \(r^2\), not the square of the average radius. 2. The exact area is \(A=\frac12\int_0^1(2\theta)^2\,\text{d}\theta\). 3. Evaluating gives \(A=2\int_0^1\theta^2\,\text{d}\theta=\frac23\). 4. Thus Émile's value \(\frac12\) is too small.

The average-radius shortcut is invalid because area uses \(r^2\) point by point. The exact area is \(\frac23\).

- 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.

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 \(a>0\), \(a=4\).

\(a=4\)

- Treat the unbounded angular interval with an improper integral. - Squaring the radius changes the exponential rate. - Check the limiting value of the exponential term.

1. Write the improper polar-area integral \(A=\frac{1}{2}\int_0^{\infty}e^{-2\theta}\,d\theta\). 2. Evaluate it as a limit: \(A=\frac{1}{2}\lim_{b\to\infty}\left[-\frac{1}{2}e^{-2\theta}\right]_0^b\). 3. Since \(e^{-2b}\to0\), the area is \(A=\frac{1}{4}\).

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

- 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.

1. Set up the area equation: \(\frac{1}{2}\int_0^b9e^{-2\theta}\,d\theta=\frac{9}{8}\). 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 \(b=\frac{\ln 2}{2}\).

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

- 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.

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, \(A=\frac{1}{2}(3\pi+8)=\frac{3\pi+8}{2}\).

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

- 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.

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

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

- 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.

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 \(a>0\), \(a=3\).

\(a=3\)

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

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 \(A=2\pi\).

\(2\pi\)

- 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.

1. Polar area uses \(r^2\), so the sign of the 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 \(\pi\).

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

- 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.

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 \(\frac12\int_0^\pi4\cos^2(5\theta)\,\text{d}\theta=\pi\).

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

- Use the figure to check what the absolute value changes about the tracing of the curve. - Decide whether the full interval traces each displayed loop once or repeats any part. - Build one area accumulation that remains valid across the sign changes of cosine.

1. The polar-area formula gives \(A=\frac{1}{2}\int_0^{2\pi}4\cos^2\theta\,d\theta\). 2. The absolute value disappears after squaring. 3. Since \(\int_0^{2\pi}\cos^2\theta\,d\theta=\pi\), the combined area is \(2\pi\).

\(A=\frac12\int_0^{2\pi}(2|\cos\theta|)^2\,d\theta=2\pi\)

- 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.

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 \(8\).

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

- Use the stated angular interval rather than integrating over a full revolution. - Expand the squared radius before integrating. - Look for an odd term on a symmetric interval.

1. Set up \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}(1+\sin\theta)^2\,d\theta\). 2. Expand to \(1+2\sin\theta+\sin^2\theta\). 3. The odd term integrates to \(0\), while the other two terms contribute \(\pi\) and \(\frac{\pi}{2}\). 4. Therefore, \(A=\frac{1}{2}\left(\frac{3\pi}{2}\right)=\frac{3\pi}{4}\).

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

- 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.

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. Over \([0,\pi]\), the middle term integrates to \(0\), and the other terms contribute \(\pi\) and \(\frac{\pi}{2}\). 4. Therefore, \(A=\frac{1}{2}\left(\frac{3\pi}{2}\right)=\frac{3\pi}{4}\).

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

- 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.

1. The loop begins and ends at the pole, so solve \(1-2\cos\theta=0\). The consecutive solutions surrounding \(\theta=0\) are \(\theta=-\frac{\pi}{3}\) and \(\theta=\frac{\pi}{3}\). 2. On \(-\frac{\pi}{3}\le\theta\le\frac{\pi}{3}\), the radius is nonpositive and the curve traces the inner loop once. 3. The loop area is \(A=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(1-2\cos\theta)^2\,d\theta\). 4. Expanding and evaluating gives \(A=\frac{1}{2}(2\pi-3\sqrt{3})=\pi-\frac{3\sqrt{3}}{2}\).

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