Area of a Polar Region
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AP Calculus BC › Area of a Polar Region
Find the correct polar-area integral for $r=4\cos\theta$ over $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$.
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2} (4\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2} 4\cos\theta,d\theta$
$\displaystyle \int_{-\pi/2}^{\pi/2} (4\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi} (4\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi}^{\pi} (4\cos\theta)^2,d\theta$
Explanation
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.
What integral gives the area of the region inside $r=3\sin\theta$ for $0\le\theta\le\pi$?
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2} (3\sin\theta)^2,d\theta$
$\displaystyle \int_{0}^{\pi} (3\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{2\pi} (3\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi} (3\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi} 3\sin\theta,d\theta$
Explanation
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 = 3 sin θ and the interval 0 to π, insert r(θ) squared and integrate over the given bounds. Since the curve starts and ends at the origin, this captures the full enclosed area of the loop. A tempting distractor is option B, which lacks the 1/2 and overestimates the area by a factor of two. Always square the radius function and apply the 1/2 factor to correctly compute polar areas.
Choose the correct setup for the area enclosed by $r=1-\sin\theta$ on $0\le\theta\le2\pi$.
$\displaystyle \frac12\int_{-\pi}^{\pi} (1-\sin\theta),d\theta$
$\displaystyle \frac12\int_{0}^{\pi} (1-\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{2\pi} (1-\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{2\pi} (1-\sin\theta),d\theta$
$\displaystyle \int_{0}^{2\pi} (1-\sin\theta)^2,d\theta$
Explanation
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.
What is the correct setup for the area inside $r=1+\sin\theta$ for $0\le\theta\le\pi$?
$\displaystyle \frac12\int_{0}^{\pi/2} (1+\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi} (1+\sin\theta),d\theta$
$\displaystyle \int_{0}^{\pi} (1+\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi} (1+\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi}^{\pi} (1+\sin\theta)^2,d\theta$
Explanation
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.
For $r=5\cos(3\theta)$, what integral gives the area of one petal on $0\le\theta\le\frac{\pi}{6}$?
$\displaystyle \frac12\int_{0}^{\pi/6} (5\cos(3\theta))^2,d\theta$
$\displaystyle \int_{0}^{\pi/6} (5\cos(3\theta))^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi/3} (5\cos(3\theta))^2,d\theta$
$\displaystyle \frac12\int_{-\pi/6}^{\pi/6} (5\cos(3\theta))^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi/6} 5\cos(3\theta),d\theta$
Explanation
This problem asks for the area of one petal of a rose curve using A = (1/2)∫[r(θ)]² dθ. For r = 5cos(3θ) on [0, π/6], we substitute to get A = (1/2)∫$0^{π/6}$ (5cos(3θ))² dθ = (1/2)∫$0^{π/6}$ 25cos²(3θ) dθ. The rose curve r = 5cos(3θ) has 3 petals, and one petal is traced as θ goes from 0 to π/6 (since cos(3θ) goes from 1 to 0 over this interval). The factor of 1/2 is essential in the polar area formula. Choice E incorrectly forgets to square the radius function, using 5cos(3θ) instead of (5cos(3θ))², which is a fundamental error that would not give the area. For rose curves, identify the interval for one petal and always square the radius function in the area integral.
Which integral gives the area of the region traced by $r=3\sin\theta$ on $0\le\theta\le\pi$?
$\displaystyle \int_{0}^{\pi} (3\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2} (3\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi} (3\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{2\pi} (3\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi} 3\sin\theta,d\theta$
Explanation
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.
Choose the correct setup for area of one petal of $r=2\sin(3\theta)$ on $0\le\theta\le\frac{\pi}{3}$.
$\displaystyle \int_{0}^{\pi/3}(2\sin(3\theta))^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi/3}2\sin(3\theta),d\theta$
$\displaystyle \frac12\int_{-\pi/3}^{\pi/3}(2\sin(3\theta))^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi/3}(2\sin(3\theta))^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi}(2\sin(3\theta))^2,d\theta$
Explanation
This problem asks for the area of one petal of a three-petal rose using the polar area formula. The polar area formula A = (1/2)∫[r(θ)]² dθ must be applied with the correct limits for one petal. For r = 2sin(3θ), one petal spans from θ = 0 to θ = π/3, giving A = (1/2)∫₀^(π/3) (2sin(3θ))² dθ. The interval [0, π/3] captures exactly one of the three petals since sin(3θ) completes one positive cycle. Choice E incorrectly omits squaring the radius function, which would not give an area calculation. For polar roses, identify the interval for one petal and apply the standard polar area formula with r².
Which integral gives the area enclosed by $r=2\theta$ from $0\le\theta\le1$?
$\displaystyle \int_{0}^{1}(2\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{1}(2\theta)^2,dr$
$\displaystyle \frac12\int_{0}^{2}(2\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{1}2\theta,d\theta$
$\displaystyle \frac12\int_{0}^{1}(2\theta)^2,d\theta$
Explanation
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.
Find the correct setup for area of $r=3\sin\theta$ on $0\le\theta\le\pi$.
$\displaystyle \frac12\int_{0}^{\pi}(3\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi}3\sin\theta,d\theta$
$\displaystyle \frac12\int_{0}^{\pi}(3\sin\theta)^2,dr$
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2}(3\sin\theta)^2,d\theta$
$\displaystyle \int_{0}^{\pi}(3\sin\theta)^2,d\theta$
Explanation
This question tests understanding of the polar area formula for finding the area enclosed by a polar curve. The polar area formula is A = (1/2)∫[r(θ)]² dθ, where we must square the radius function. For r = 3sin(θ) on 0 ≤ θ ≤ π, this becomes A = (1/2)∫₀^π (3sin(θ))² dθ. The limits of integration correctly span from 0 to π as specified in the problem. Choice D incorrectly uses r instead of r², which would give the arc length rather than area. Remember that polar area always requires squaring the radius function and including the factor of 1/2.
For $r=5-3\cos\theta$ on $0\le\theta\le2\pi$, select the correct polar area integral.
$\displaystyle \frac12\int_{-\pi}^{\pi}(5-3\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{2\pi}(5-3\cos\theta),d\theta$
$\displaystyle \frac12\int_{0}^{\pi}(5-3\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{2\pi}(5-3\cos\theta)^2,d\theta$
$\displaystyle \int_{0}^{2\pi}(5-3\cos\theta)^2,d\theta$
Explanation
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.