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 \(9\cdot7=63\).

\(63\)

- Use the polar-sector area formula with the stated angular bounds. - Square the radial function before integrating. - Keep the result exact in terms of \(\pi\).

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

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

- Confirm that the stated interval traces the circle exactly once. - Square the entire radial function in the polar-area formula. - Use a power-reduction identity or the known integral of \(\cos^{2}(\theta)\).

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 \(\frac{\pi}{2}\), \(A=\frac{9\pi}{4}\).

\(A=\frac{9\pi}{4}\)

- Use the interval that traces the circle once without retracing it. - Square the radial function before simplifying the constant factor. - Evaluate the squared-sine integral with a power-reduction identity.

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 \(A=4\pi\).

\(A=4\pi\)

- Apply the polar-area formula over one full tracing of the cardioid. - Expand the squared binomial before integrating. - Use full-period trigonometric integrals to simplify the terms.

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

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

- Factor out the common radial scale before expanding the square. - Integrate over the full angular interval that traces the cardioid once. - Use the full-period values of the sine and squared-sine integrals.

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 \(A=6\pi\).

\(A=6\pi\)

- Use the full-period polar-area formula because the curve is traced once on the given interval. - Expand the square and keep the three resulting terms separate. - Apply standard full-period trigonometric integrals before multiplying by \(\frac{1}{2}\).

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

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

- Confirm what part of the circle is traced by the stated angle interval. - Square the radial function in the polar-area formula. - Use a power-reduction identity for the squared-cosine integral.

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 \(\frac{\pi}{4}\), \(A=\frac{\pi}{2}\).

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

- Interpret the tracing interval before setting up the area integral. - Apply the polar-area formula to the full radial expression. - Evaluate the squared-sine integral exactly rather than approximating.

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 \(\frac{\pi}{4}\), \(A=2\pi\).

\(A=2\pi\)

- Use the square of the radial function in the area integral. - The angle bounds are already provided. - Keep the factor from the polar-area formula.

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 \(A=\frac{1}{2}\left[\frac{\theta^3}{3}\right]_0^2=\frac{4}{3}\).

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

- The area formula already uses \(r^2\), so no square root is needed. - Separate the constant and oscillating contributions. - Use the full-period integral of cosine.

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, \(A=4\pi\).

\(4\pi\)

- Square the exponential radial function. - Integrate the resulting exponential exactly. - Simplify the upper endpoint using the logarithm.

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 \(A=\frac{1}{4}[e^{2\theta}]_0^{\ln2}=\frac{1}{4}(4-1)=\frac{3}{4}\).

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

- Split the angular interval where the radial definition changes. - Use the polar-area formula on each piece. - Add the two area contributions.

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

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

- Start with the polar-area formula and square the radial function. - Look for a familiar derivative in the integrand. - Use the symmetry of the angular bounds when evaluating.

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

\(1\)

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

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

- Check that the radial function is zero at both endpoints of the given interval. - Use the polar-area formula only over the interval tracing one petal. - Apply a power-reduction identity after squaring the cosine term.

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

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

- Verify that the given interval starts and ends where the radial function is zero. - Square the entire radial function before integrating. - Account for the inner factor \(3\) when evaluating the trigonometric integral.

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

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

- Use the endpoint zeros to confirm that exactly one petal is traced. - Apply the polar-area formula over the given symmetric interval. - Use symmetry or power reduction while retaining the inner angular factor.

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

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

- Confirm that the radial function is zero at both ends of the stated interval. - Square the complete radial expression in the polar-area formula. - Keep track of the inner factor \(4\) when evaluating the trigonometric integral.

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

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

- Check that the radial function is zero at both endpoints of the loop interval. - Expand the squared binomial before integrating term by term. - Use a power-reduction identity for the squared-cosine term.

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

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

- Verify that the radial function vanishes at both endpoints of the stated interval. - Expand the square before integrating the constant, sine, and squared-sine terms. - Apply a power-reduction identity to the squared-sine term.

1. The endpoints satisfy \(r=0\), and the stated interval traces the inner loop exactly once. 2. The area is \(A=\frac{1}{2}\int_{7\pi/6}^{11\pi/6}(1+2\sin(\theta))^{2}\,d\theta\). 3. Expanding and integrating gives \(A=\frac{1}{2}(2\pi-3\sqrt{3})=\pi-\frac{3\sqrt{3}}{2}\).

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

- Use the zeros of the radial function to verify the inner-loop endpoints. - Take advantage of the symmetry of the interval before expanding the square. - Use \(\sin(\arccos(2/3))\) and a power-reduction identity during evaluation.

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

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

- Locate the consecutive zeros surrounding the petal centered at \(\theta=0\). - Use those angles as the bounds for one complete tracing of the petal. - Square the entire radial function in the polar-area formula, but do not evaluate the integral.

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

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

- Find the consecutive zeros that bound the petal lying in the first quadrant. - Check that the radial function stays nonnegative between those zeros. - Use the polar-area formula with the full radial expression squared, but do not evaluate.

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

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

- Choose angles whose terminal rays sweep the upper half-plane. - Confirm that this interval traces the upper half of the cardioid once. - Square the complete radial function in the polar-area formula without evaluating the integral.

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 coordinate-axis rays into polar-angle bounds. - Check that the radial function remains positive on the chosen interval. - Apply the polar-area formula and leave the resulting integral unevaluated.

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

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

- Express the sector area with the polar-area integral before substituting the given area. - Isolate the square of the unknown radial constant. - Use the positivity condition when selecting the square-root 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 \(a>0\), \(a=4\sqrt{2}\).

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

- Factor the constant \(a^{2}\) out of the polar-area integral. - Use the exact squared-cosine integral over the symmetric interval. - Apply the positivity condition after solving the resulting quadratic 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 \(a>0\), \(a=6\).

\(a=6\)

- Factor \(a^{2}\) out of the full-period polar-area integral. - Expand the squared cardioid factor and use standard full-period trigonometric integrals. - Apply the positivity condition after solving for \(a^{2}\).

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

\(a=2\)

- Verify that the given interval traces one complete petal. - Factor \(a^{2}\) out and evaluate the squared-sine integral exactly. - Use the positivity condition when taking the square root.

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.

\(A_A=2\pi\), \(A_B=2\pi\); the regions have equal area.

- 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.

- Start from the complete polar-area formula, including its constant factor. - Confirm that a full interval of length \(2\pi\) traces this limaçon once. - Expand the square and use full-period trigonometric integrals.

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

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

- 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 \(A=\frac{1}{2}\int_{-\pi/4}^{\pi/4}(3\cos(2\theta))^{2}\,d\theta=\frac{9\pi}{8}\).

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, \(A\approx2.868\).

\(A\approx2.868\)

- 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, \(A\approx2.0000\).

\(A\approx2.0000\)

- 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) \(2\pi\)

- Write the area integral with the unknown upper angle. - Simplify the resulting power expression before solving. - Use the positive-angle condition.

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 \(b=3\).

\(b=3\)

- The numerical rule must be applied to the squared radial values. - Include the polar-area factor in addition to the Simpson factor. - Use the angular spacing, not the radial spacing.

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

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

- Apply the numerical rule to the squared radii. - Keep the unknown midpoint radius inside the weighted sum. - Use the positivity condition after solving for its square.

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 student’s 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 \(A=\frac{1}{2}\int_0^1(2\theta)^2\,d\theta\). 3. Evaluating gives \(A=2\int_0^1\theta^2\,d\theta=\frac{2}{3}\). 4. Thus the student’s value \(\frac{1}{2}\) is too small.

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

- 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 \(A=2\pi\).

\(2\pi\)

- A negative radius changes the represented point, not the sign of the area element. - Square the radial function before integrating. - The interval traces the complete circle once.

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 \(\pi\).

\(\pi\)

- Determine how much angular change traces an odd-petal rose once. - Compare that interval with the student’s 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 the student’s \(2\pi\) doubles the area. 3. The correct area is \(\frac{1}{2}\int_0^{\pi}4\cos^2(5\theta)\,d\theta=\pi\).

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

- The two circles arise from the two signs of cosine. - Squaring the radial function removes the absolute value. - Integrate over the full stated tracing interval.

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 \(2\pi\).

\(2\pi\)

- 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 \(A=\frac{1}{2}\int_{-\pi/3}^{\pi/3}(1-2\cos\theta)^2\,d\theta\). 4. Expanding and evaluating gives \(A=\frac{1}{2}(2\pi-3\sqrt{3})=\pi-\frac{3\sqrt{3}}{2}\).

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