Area of a Polar Region
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AP Calculus BC › Area of a Polar Region
A region is bounded by the polar curve $r=2\theta$ for $0\le\theta\le 1$. Which integral gives the area?
$\displaystyle \int_{0}^{1}(2\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{1}2\theta,d\theta$
$\displaystyle \frac12\int_{0}^{2}(2\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{1}(2\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{1}(\theta)^2,d\theta$
Explanation
This problem involves calculating the area of a polar region. The formula for the area enclosed by a polar curve $r(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ is $A = \frac{1}{2} \int_{\alpha}^{\beta} r(\theta)^2 , d\theta$. For $r = 2\theta$ and limits from 0 to 1, which traces a short spiral segment, square the function to get $(2\theta)^2$ inside the integral. Multiply by 1/2 and integrate over 0 to 1 to find the swept area. A tempting distractor like choice B omits the 1/2, resulting in twice the correct area by ignoring the formula's derivation from sector areas. Always confirm the limits align with the curve's path and square $r$ to apply the polar area integral correctly.
A polar region is traced by $r=1-2\cos\theta$ for $0\le\theta\le\pi$. Which integral gives its area?
$\displaystyle \frac12\int_{-\pi}^{\pi}(1-2\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi}(1-2\cos\theta),d\theta$
$\displaystyle \frac12\int_{0}^{2\pi}(1-2\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi}(1-2\cos\theta)^2,d\theta$
$\displaystyle \int_{0}^{\pi}(1-2\cos\theta)^2,d\theta$
Explanation
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.
What integral computes the area inside $r=4\cos\theta$ for $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$?
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2}(4\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi}(4\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi}^{\pi}(4\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2}4\cos\theta,d\theta$
$\displaystyle \int_{-\pi/2}^{\pi/2}(4\cos\theta)^2,d\theta$
Explanation
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 = 4 cos θ and limits from -π/2 to π/2, which traces the full circle, square the function to get (4 cos $θ)^2$ inside the integral. Multiply by 1/2 and integrate over the specified symmetric interval. A tempting distractor like choice B changes limits to 0 to π, which computes the same value but doesn't match the given interval. Always use the exact limits provided and square the radius to ensure the polar area is computed accurately.
A region is bounded by the polar curve $r=2+\cos\theta$ for $0\le\theta\le\pi$. Which integral gives its area?
$\displaystyle \frac12\int_{0}^{\pi}(2+\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi}^{\pi}(2+\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi}(2+\cos\theta),d\theta$
$\displaystyle \int_{0}^{\pi}(2+\cos\theta)^2,d\theta$
$\displaystyle \int_{0}^{\pi}(2+\cos\theta),d\theta$
Explanation
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.
Find the correct polar-area integral for $r=4\cos\theta$ over $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$.
$\displaystyle \frac12\int_{0}^{\pi} (4\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi}^{\pi} (4\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2} (4\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2} 4\cos\theta,d\theta$
$\displaystyle \int_{-\pi/2}^{\pi/2} (4\cos\theta)^2,d\theta$
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.
Select the correct setup for the area enclosed by $r=5-5\cos\theta$ for $0\le\theta\le\pi$.
$\displaystyle \frac12\int_{0}^{2\pi} (5-5\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi} (5-5\cos\theta),d\theta$
$\displaystyle \int_{0}^{\pi} (5-5\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi}^{\pi} (5-5\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{\pi} (5-5\cos\theta)^2,d\theta$
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) 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.
Which integral gives the area enclosed by $r=2-\cos(3\theta)$ for $0\le\theta\le\frac{2\pi}{3}$?
$\displaystyle \frac12\int_{0}^{\pi/3} (2-\cos(3\theta))^2,d\theta$
$\displaystyle \frac12\int_{0}^{2\pi/3} (2-\cos(3\theta))^2,d\theta$
$\displaystyle \int_{0}^{2\pi/3} (2-\cos(3\theta))^2,d\theta$
$\displaystyle \frac12\int_{0}^{2\pi/3} (2-\cos(3\theta)),d\theta$
$\displaystyle \frac12\int_{0}^{2\pi} (2-\cos(3\theta))^2,d\theta$
Explanation
Calculating the area of a polar region is the skill being tested here. The formula for the area swept 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(3θ) and 0 to 2π/3, apply the formula with these limits. The interval covers one period of the oscillation, capturing a lobe. A tempting distractor is option B, which omits the 1/2 and doubles the area. Determine the period of multi-angle functions to set appropriate limits for individual regions.
Choose the correct integral for the area enclosed by $r=1+\cos\theta$ on $-\pi\le\theta\le\pi$.
$\displaystyle \int_{-\pi}^{\pi} (1+\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi}^{\pi} (1+\cos\theta),d\theta$
$\displaystyle \frac12\int_{0}^{2\pi} (1+\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi}^{\pi} (1+\cos\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2} (1+\cos\theta)^2,d\theta$
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) 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.
What integral gives the area of the region inside $r=3\sin\theta$ for $0\le\theta\le\pi$?
$\displaystyle \frac12\int_{0}^{\pi} 3\sin\theta,d\theta$
$\displaystyle \frac12\int_{0}^{\pi} (3\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{2\pi} (3\sin\theta)^2,d\theta$
$\displaystyle \int_{0}^{\pi} (3\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2} (3\sin\theta)^2,d\theta$
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 = 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 $0\le\theta\le2\pi$.
$\displaystyle \frac12\int_{-\pi}^{\pi} (1-\sin\theta),d\theta$
$\displaystyle \frac12\int_{0}^{\pi} (1-\sin\theta)^2,d\theta$
$\displaystyle \int_{0}^{2\pi} (1-\sin\theta)^2,d\theta$
$\displaystyle \frac12\int_{0}^{2\pi} (1-\sin\theta),d\theta$
$\displaystyle \frac12\int_{0}^{2\pi} (1-\sin\theta)^2,d\theta$
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.