Area of a Polar Region - AP Calculus BC

1. A region is enclosed by $r=2\theta$ for $0\le\theta\le\pi$. Which integral represents its area?
   Answer: $\displaystyle \frac12\int_{0}^{\pi} (2\theta)^2\,d\theta$ To find the area of a polar region, we use the formula A = (1/2)∫[r(θ)]² dθ evaluated over the specified interval. The curve r = 2θ is an Archimedean spiral that starts at the origin when θ = 0 and spirals outward as θ increases to π. Applying the area formula gives A = (1/2)∫₀^π (2θ)² dθ = (1/2)∫₀^π 4θ² dθ. Choice C incorrectly uses r instead of r², which would give a meaningless result with wrong dimensions (the integral of radius gives length, not area). For polar regions, always square the radius function and include the factor of 1/2 to get the correct area.
2. A region is bounded by the polar curve $r=2+\cos\theta$ for $0\le\theta\le\pi$. Which integral gives its area?
   Answer: $\displaystyle \frac12\int_{0}^{\pi}(2+\cos\theta)^2\,d\theta$ This problem involves calculating the area of a polar region. The formula for the area enclosed by a polar curve r(θ) from θ = α to θ = β is A = (1/2) ∫_α^β $r(θ)^2$ dθ. For r = 2 + cos θ and limits from 0 to π, square the function to get (2 + cos θ)^2 inside the integral. Multiply by 1/2 and integrate over the given interval to find the area swept by the curve. A tempting distractor like choice A omits the 1/2 factor, which would double the actual area since the formula derives from summing triangular sectors. Always verify that the integration limits match the specified interval and that r is squared to account for the area element in polar coordinates.
3. Select the correct setup for the area enclosed by $r=5-5\cos\theta$ for $0\le\theta\le\pi$.
   Answer: $\displaystyle \frac12\int_{0}^{\pi} (5-5\cos\theta)^2\,d\theta$ Calculating the area of a polar region is the skill being tested here. The formula for the area enclosed by a polar curve r(θ) from θ = α to θ = β is A = (1/2) ∫_α^β $r(θ)^2$ dθ. For r = 5 - 5 cos θ and 0 to π, apply the formula using the provided interval. The integral gives the area swept from the origin over this range. A tempting distractor is option A, which forgets the 1/2 and overcalculates the area. Confirm that r is squared and the 1/2 is included for all polar area problems.
4. A polar region is inside $r=4\sin\theta$ for $0\le\theta\le\pi$. Which integral gives the area?
   Answer: $\displaystyle \frac12\int_{0}^{\pi} (4\sin\theta)^2\,d\theta$ To find the area of a polar region, we apply the formula A = (1/2)∫[r(θ)]² dθ over the given interval. The curve r = 4sin(θ) creates a circle of radius 2 centered at (0, 2) in Cartesian coordinates, and for 0 ≤ θ ≤ π, we get the entire circle. Substituting r = 4sin(θ) into the area formula yields A = (1/2)∫₀^π (4sin(θ))² dθ = (1/2)∫₀^π 16sin²(θ) dθ. Choice D incorrectly uses r instead of r², which is a fundamental error that would give the wrong units (length instead of area). Remember that polar area always requires squaring the radius function and including the factor of 1/2.
5. A region is bounded by $r=5-\sin\theta$ for $\frac{\pi}{2}\le\theta\le\frac{3\pi}{2}$. Which integral gives its area?
   Answer: $\displaystyle \frac12\int_{\pi/2}^{3\pi/2}(5-\sin\theta)^2\,d\theta$ This problem involves calculating the area of a polar region. The formula for the area enclosed by a polar curve r(θ) from θ = α to θ = β is A = (1/2) ∫_α^β $r(θ)^2$ dθ. For r = 5 - sin θ and limits from π/2 to 3π/2, square the function to get (5 - sin θ)^2 inside the integral. Multiply by 1/2 and integrate over the given interval to find the area of the specified portion. A tempting distractor like choice C uses 0 to π, which covers a different part of the curve and mismatches the limits. Always adhere to the provided limits and remember to square the radius function in polar area calculations.
6. Find the area inside $r=5-5\cos\theta$ for $0\le\theta\le2\pi$. Which is the correct setup?
   Answer: $\displaystyle \frac12\int_{0}^{2\pi} (5-5\cos\theta)^2\,d\theta$ This question asks for the area inside a limaçon using the polar area formula A = (1/2)∫[r(θ)]² dθ. The curve r = 5 - 5cos(θ) = 5(1 - cos(θ)) is a cardioid that traces a complete heart shape as θ goes from 0 to 2π, with a cusp at the origin when θ = 0. To find the total area, we integrate over the full interval: A = (1/2)∫₀^{2π} (5 - 5cos(θ))² dθ. Choice D incorrectly uses r instead of r², which is a common error that gives the wrong result with incorrect units. Remember that polar area always requires squaring the entire radius expression and multiplying by 1/2.
7. Choose the correct integral for the area enclosed by $r=1+\cos\theta$ on $-\pi\le\theta\le\pi$.
   Answer: $\displaystyle \frac12\int_{-\pi}^{\pi} (1+\cos\theta)^2\,d\theta$ Calculating the area of a polar region is the skill being tested here. The formula for the area enclosed by a polar curve r(θ) from θ = α to θ = β is A = (1/2) ∫_α^β $r(θ)^2$ dθ. For r = 1 + cos θ and -π to π, use the given limits in the formula. This interval traces the full cardioid, starting and ending at the origin. A tempting distractor is option C, which lacks the 1/2 and computes twice the area. Use equivalent intervals like -π to π or 0 to 2π for symmetric curves, but match the specified bounds.
8. Which integral represents the area inside $r=4\sin(2\theta)$ for $0\le\theta\le\frac{\pi}{2}$?
   Answer: $\displaystyle \frac12\int_{0}^{\pi/2} (4\sin(2\theta))^2\,d\theta$ This problem involves finding the area of a polar region using A = (1/2)∫[r(θ)]² dθ. For r = 4sin(2θ) on [0, π/2], we substitute to get A = (1/2)∫_0^{π/2} (4sin(2θ))² dθ = (1/2)∫_0^{π/2} 16sin²(2θ) dθ. The rose curve r = 4sin(2θ) creates a four-petaled rose, and the interval [0, π/2] traces exactly one petal since sin(2θ) goes from 0 to 1 and back to 0 as θ goes from 0 to π/2. The factor of 1/2 must be included in the polar area formula. Choice A incorrectly forgets to square the radius function, using 4sin(2θ) instead of (4sin(2θ))², which is a fundamental error in polar area calculations. For polar areas, always remember to square the radius function and include the factor 1/2 in your integral setup.
9. For $r=5-3\cos\theta$ on $0\le\theta\le2\pi$, select the correct polar area integral.
   Answer: $\displaystyle \frac12\int_{0}^{2\pi}(5-3\cos\theta)^2\,d\theta$ This question involves finding the area of a limaçon using the polar area formula. For a polar curve r = f(θ), the area is A = (1/2)∫[f(θ)]² dθ integrated over the appropriate interval. With r = 5 - 3cos(θ) on 0 ≤ θ ≤ 2π, we get A = (1/2)∫₀^(2π) (5 - 3cos(θ))² dθ. The interval [0, 2π] ensures we capture the entire limaçon, which completes one full rotation. Choice B incorrectly omits squaring the radius, using (5 - 3cos(θ)) instead of (5 - 3cos(θ))². When finding polar areas, always square the entire radius expression and include the essential factor of 1/2.
10. Choose the correct setup for the area enclosed by $r=1-\sin\theta$ on $0\le\theta\le2\pi$.
    Answer: $\displaystyle \frac12\int_{0}^{2\pi} (1-\sin\theta)^2\,d\theta$ Calculating the area of a polar region is the skill being tested here. The formula for the area enclosed by a polar curve r(θ) over a full period is A = (1/2) ∫_α^β $r(θ)^2$ dθ, where α to β covers the entire curve. For r = 1 - sin θ and 0 to 2π, use the formula with these limits to get the total area. This interval ensures the full cardioid is traced without overlap. A tempting distractor is option B, which omits the 1/2 and doubles the area value. Remember to integrate over the complete interval that traces the curve once for the total enclosed area.
11. For $r=2+2\cos\theta$ on $0\le\theta\le\pi$, which integral gives the enclosed area?
    Answer: $\displaystyle \frac12\int_{0}^{\pi}(2+2\cos\theta)^2\,d\theta$ This problem asks for the area enclosed by a polar curve, which requires the polar area formula A = (1/2)∫[r(θ)]² dθ. For r = 2 + 2cos(θ) on 0 ≤ θ ≤ π, we substitute to get A = (1/2)∫₀^π (2 + 2cos(θ))² dθ. The factor of 1/2 is essential in the polar area formula and distinguishes it from rectangular area calculations. The limits of integration [0, π] match the given interval for θ. Choice A incorrectly omits the factor of 1/2, which would give twice the actual area. When finding polar areas, always remember to include the factor of 1/2 and square the radius function r(θ).
12. A polar region is traced by $r=1-2\cos\theta$ for $0\le\theta\le\pi$. Which integral gives its area?
    Answer: $\displaystyle \frac12\int_{0}^{\pi}(1-2\cos\theta)^2\,d\theta$ This problem involves calculating the area of a polar region. The formula for the area enclosed by a polar curve r(θ) from θ = α to θ = β is A = (1/2) ∫_α^β $r(θ)^2$ dθ. For r = 1 - 2 cos θ and limits from 0 to π, which traces part of the limacon including the inner loop, square the function to get (1 - 2 cos θ)^2. Multiply by 1/2 and integrate over the given interval, noting that $r^2$ handles negative r values correctly. A tempting distractor like choice E extends to 0 to 2π, which includes the full curve and overcounts the region for the specified interval. Always match the integration limits to the problem's interval and remember to square r for the area calculation.
13. Find the correct polar-area integral for $r=4\cos\theta$ over $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$.
    Answer: $\displaystyle \frac12\int_{-\pi/2}^{\pi/2} (4\cos\theta)^2\,d\theta$ Calculating the area of a polar region is the skill being tested here. The formula for the area inside a polar curve r(θ) from θ = α to θ = β is A = (1/2) ∫_α^β $r(θ)^2$ dθ. For r = 4 cos θ and -π/2 to π/2, plug in r(θ) and the specified limits. The curve starts and ends at the origin, enclosing the full region in this interval. A tempting distractor is option C, which misses the 1/2 and computes twice the area. Ensure the limits align with where the curve begins and ends at the origin for loop areas.
14. Which integral gives the area of the region inside $r=\theta+1$ for $0\le\theta\le\pi$?
    Answer: $\displaystyle \frac12\int_{0}^{\pi}(\theta+1)^2\,d\theta$ This problem involves calculating the area of a polar region. The formula for the area enclosed by a polar curve r(θ) from θ = α to θ = β is A = (1/2) ∫_α^β $r(θ)^2$ dθ. For r = θ + 1 and limits from 0 to π, which traces a spiral segment, square the function to get $(θ + 1)^2$. Multiply by 1/2 and integrate over 0 to π for the swept area. A tempting distractor like choice C omits the 1/2, doubling the result by neglecting the area formula's factor. Always verify the interval matches the problem and square r to correctly compute polar areas.
15. Choose the correct area integral for the region bounded by $r=2-\cos\theta$ on $0\le\theta\le\pi$.
    Answer: $\displaystyle \frac12\int_{0}^{\pi} (2-\cos\theta)^2\,d\theta$ This question tests the polar area formula A = (1/2)∫[r(θ)]² dθ for a limaçon curve. The curve r = 2 - cos(θ) creates a dimpled limaçon (no inner loop since 2 > 1), and we need the area for 0 ≤ θ ≤ π, which gives the upper half. Applying the formula yields A = (1/2)∫₀^π (2 - cos(θ))² dθ, where we must square the entire expression 2 - cos(θ). Choice D incorrectly uses r instead of r², which is a fundamental error in polar area calculations that gives a result with wrong dimensions. Always remember to square the radius function and include the 1/2 factor when finding polar areas.
16. A region is bounded by $r=2\cos(2\theta)$ for $-\frac\pi4\le\theta\le\frac\pi4$. Which integral gives its area?
    Answer: $\displaystyle \frac12\int_{-\pi/4}^{\pi/4} (2\cos(2\theta))^2\,d\theta$ To find the area of this polar region, we apply the formula A = (1/2)∫[r(θ)]² dθ over the given bounds. The curve r = 2cos(2θ) is a four-petaled rose, and the interval -π/4 ≤ θ ≤ π/4 captures exactly one petal (the rightmost one). Substituting into the area formula gives A = (1/2)∫_{-π/4}^{π/4} (2cos(2θ))² dθ = (1/2)∫_{-π/4}^{π/4} 4cos²(2θ) dθ. Choice B omits the essential factor of 1/2, which would double the actual area of the petal. When working with polar areas, always include the 1/2 factor and verify that your θ-interval captures exactly the region you want.
17. Which integral gives the area enclosed by $r=2\theta$ from $0\le\theta\le1$?
    Answer: $\displaystyle \frac12\int_{0}^{1}(2\theta)^2\,d\theta$ This problem tests the application of the polar area formula to a spiral curve. The polar area formula states that A = (1/2)∫[r(θ)]² dθ, where r is expressed as a function of θ. For r = 2θ from θ = 0 to θ = 1, we substitute to get A = (1/2)∫₀¹ (2θ)² dθ. The limits of integration [0, 1] match the given range for θ, not the range of r values. Choice D incorrectly uses r instead of r², failing to square the radius function. Remember that polar area always requires squaring the radius function, regardless of its form.
18. What is the correct setup for the area inside $r=1+\sin\theta$ for $0\le\theta\le\pi$?
    Answer: $\displaystyle \frac12\int_{0}^{\pi} (1+\sin\theta)^2\,d\theta$ This problem requires finding the area inside a polar curve using the formula A = (1/2)∫[r(θ)]² dθ. For r = 1 + sin(θ) on [0, π], we substitute to get A = (1/2)∫_0^π (1 + sin(θ))² dθ. The limits [0, π] are appropriate because the cardioid r = 1 + sin(θ) traces its complete shape exactly once over this interval. Beyond π, the curve would retrace itself since sin(θ + 2π) = sin(θ). Choice D incorrectly forgets to square the radius function, using (1 + sin(θ)) instead of (1 + sin(θ))², which would give an incorrect area calculation. For polar areas, always square the entire radius expression before integrating, and verify that your limits trace the curve exactly once.
19. Find the correct setup for the area inside $r=3+\sin\theta$ for $\frac{\pi}{2}\le\theta\le\frac{3\pi}{2}$.
    Answer: $\displaystyle \frac12\int_{\pi/2}^{3\pi/2} (3+\sin\theta)^2\,d\theta$ Calculating the area of a polar region is the skill being tested here. The formula for the area in an angular sector from θ = α to θ = β is A = (1/2) ∫_α^β $r(θ)^2$ dθ. For r = 3 + sin θ and π/2 to 3π/2, substitute r(θ) and the bounds into the integral. This setup finds the area swept over that half-circle interval. A tempting distractor is option B, which excludes the 1/2 and overestimates by a factor of two. For partial regions of limaçons, integrate directly over the given angular range.
20. Which integral gives the area of the region traced by $r=3\sin\theta$ on $0\le\theta\le\pi$?
    Answer: $\displaystyle \frac12\int_{0}^{\pi} (3\sin\theta)^2\,d\theta$ This problem involves finding the area of a polar region using A = (1/2)∫[r(θ)]² dθ. For r = 3sin(θ) on [0, π], we substitute to get A = (1/2)∫_0^π (3sin(θ))² dθ = (1/2)∫_0^π 9sin²(θ) dθ. The curve r = 3sin(θ) forms a circle that is traced completely as θ goes from 0 to π (since sin(θ) goes from 0 to 1 and back to 0). The factor of 1/2 is crucial and must be included in the polar area formula. Choice C incorrectly omits the factor of 1/2, which would double the actual area—a common mistake when students forget the polar area formula differs from rectangular integration. Remember that polar area always includes the factor 1/2, and the radius function must be squared before integrating.

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