Area Bounded By Two Polar Curves
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AP Calculus BC › Area Bounded By Two Polar Curves
What is the correct area setup for the region enclosed by $r=2\sin\theta$ and $r=2\cos\theta$ in the first quadrant?
$\dfrac{1}{2}\displaystyle\int_{\pi/4}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big], d\theta$
$\dfrac{1}{2}\displaystyle\int_{0}^{2\pi}\big[(2\cos\theta)^2-(2\sin\theta)^2\big], d\theta$
$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\big[(2\cos\theta)^2-(2\sin\theta)^2\big], d\theta$
$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\big[(2\sin\theta)^2-(2\cos\theta)^2\big], d\theta$
$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big], d\theta$
Explanation
This problem tests the skill of finding the area bounded by two polar curves. The curves $r=2\sin\theta$ and $r=2\cos\theta$ intersect at $\theta=\pi/4$ (and origin) in the first quadrant. In $\pi/4$ to $\pi/2$, $r=2\sin\theta > r=2\cos\theta$, outer is $2\sin\theta$. The area between them in this interval is $$\frac{1}{2} \int_{\pi/4}^{\pi/2} [(2\sin\theta)^2 - (2\cos\theta)^2] , d\theta$$. A tempting distractor is choice B, which uses 0 to $\pi/4$ with $\sin$ - $\cos$, negative. A transferable strategy for polar areas is to find intersection points to determine integration limits and identify which curve is outer in each interval.
What is the correct area setup for the region enclosed by $r=4\sin\theta$ and $r=2$?
$\displaystyle \frac12\int_{0}^{\pi}\big[2^2-(4\sin\theta)^2\big]d\theta$
$\displaystyle \frac12\int_{0}^{\pi}\big[(4\sin\theta)^2-2^2\big]d\theta$
$\displaystyle \frac12\int_{\pi/6}^{5\pi/6}\big[2^2-(4\sin\theta)^2\big]d\theta$
$\displaystyle \frac12\int_{0}^{\pi/2}\big[(4\sin\theta)^2-2^2\big]d\theta$
$\displaystyle \frac12\int_{\pi/6}^{5\pi/6}\big[(4\sin\theta)^2-2^2\big]d\theta$
Explanation
This problem requires finding the area between the circle r = 4sin(θ) and the horizontal line r = 2. To find intersections, set 4sin(θ) = 2, giving sin(θ) = 1/2, so θ = π/6 and θ = 5π/6. The circle r = 4sin(θ) exists only for 0 ≤ θ ≤ π (where sin(θ) ≥ 0), and it's outside r = 2 when 4sin(θ) > 2, which is true for π/6 < θ < 5π/6. The area integral is (1/2)∫_${π/6}^{5π/6}$[(4sin(θ))² - 2²]dθ. Choice A uses 0 to π, which would include regions where the circle is inside r = 2, giving an incorrect area. For polar areas between curves, carefully determine which curve is farther from the origin in your integration interval.
For the region enclosed by $r=2\sin(2\theta)$ and $r=0$, which integral gives the area of one loop?
$\displaystyle \frac12\int_{0}^{\pi/2}\big(2\sin(2\theta)\big)^2d\theta$
$\displaystyle \frac12\int_{0}^{\pi/4}\big(2\sin(2\theta)\big)^2d\theta$
$\displaystyle \frac12\int_{-\pi/4}^{\pi/4}\big(2\sin(2\theta)\big)^2d\theta$
$\displaystyle \frac12\int_{0}^{2\pi}\big(2\sin(2\theta)\big)^2d\theta$
$\displaystyle \frac12\int_{0}^{\pi}\big(2\sin(2\theta)\big)^2d\theta$
Explanation
This problem involves finding the area of one loop of the four-petaled rose r = 2sin(2θ). The curve equals zero when sin(2θ) = 0, which occurs when 2θ = 0, π, 2π, 3π, 4π... or θ = 0, π/2, π, 3π/2, 2π... One complete petal occurs between consecutive zeros, such as from θ = 0 to θ = π/2. The area of one loop is (1/2)∫_$0^{π/2}$[2sin(2θ)]²dθ. Choice B incorrectly uses 0 to π, which would include two complete petals instead of one. For rose curves r = asin(nθ) or r = acos(nθ), each petal spans an angle of π/n, so always divide the period by the coefficient of θ inside the trig function.
For the region enclosed by $r=2\cos\theta$ and $r=1$, what integral setup gives its area?
$\dfrac12\int_{-\pi/3}^{\pi/3}\left((2\cos\theta)^2-1^2\right)d\theta$
$\dfrac12\int_{-\pi/2}^{\pi/2}\left((2\cos\theta)^2-1^2\right)d\theta$
$\dfrac12\int_{0}^{\pi/3}\left((2\cos\theta)^2-1^2\right)d\theta$
$\int_{-\pi/3}^{\pi/3}\left(2\cos\theta-1\right)d\theta$
$\dfrac12\int_{-\pi/3}^{\pi/3}\left(1^2-(2\cos\theta)^2\right)d\theta$
Explanation
This problem involves calculating the area bounded by two polar curves, specifically the region enclosed by r=2cosθ and r=1. To set up the integral, first find the intersection points by solving 2cosθ = 1, which gives θ = ±π/3. Between -π/3 and π/3, r=2cosθ is greater than r=1, making it the outer curve. Thus, the area is (1/2) ∫_${-π/3}^{π/3}$ $[(2cosθ)^2$ - $1^2$] dθ. A tempting distractor is choice A, which subtracts in the wrong order, resulting in a negative integrand and an incorrect negative area. When finding areas between polar curves, always subtract the inner radius squared from the outer radius squared and confirm the integration limits encompass the entire region of interest.
What is the correct area setup for the region enclosed by $r=4\sin\theta$ and $r=2\sin\theta$?
$\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[(4\sin\theta)^2-(2\sin\theta)^2\big],d\theta$
$\displaystyle\int_{0}^{\pi}\big[(4\sin\theta)-(2\sin\theta)\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(4\sin\theta)^2\big],d\theta$
$\dfrac12\displaystyle\int_{0}^{2\pi}\big[(4\sin\theta)^2-(2\sin\theta)^2\big],d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi}\big[(4\sin\theta)^2-(2\sin\theta)^2\big],d\theta$
Explanation
This problem tests the skill of finding the area bounded by two polar curves. The curves r=4sinθ and r=2sinθ are both defined for θ in 0 to π, with the larger always outside the smaller since 4sinθ >2sinθ when sinθ >0. They touch at θ=0 and θ=π, where r=0. The area between them is (1/2) ∫_0^π $[(4sinθ)^2$ - $(2sinθ)^2$] dθ. A tempting distractor is choice D, which integrates over 0 to 2π, but this would double the area since the curves trace the same path twice. A transferable strategy for polar areas is to find intersection points to determine integration limits and identify which curve is outer in each interval.
What integral correctly sets up the area enclosed by $r=4\cos\theta$ and $r=2\cos\theta$?
$\int_{-\pi/2}^{\pi/2}\left(4\cos\theta-2\cos\theta\right)d\theta$
$\dfrac12\int_{-\pi/2}^{\pi/2}\left((4\cos\theta)^2-(2\cos\theta)^2\right)d\theta$
$\dfrac12\int_{0}^{\pi}\left((4\cos\theta)^2-(2\cos\theta)^2\right)d\theta$
$\dfrac12\int_{-\pi/2}^{\pi/2}\left((2\cos\theta)^2-(4\cos\theta)^2\right)d\theta$
$\dfrac12\int_{-\pi/4}^{\pi/4}\left((4\cos\theta)^2-(2\cos\theta)^2\right)d\theta$
Explanation
This problem involves calculating the area bounded by two polar curves, specifically the region enclosed by r=4cosθ and r=2cosθ. To set up the integral, note they meet at the pole for θ=±π/2, with r=4cosθ larger where defined. The region spans -π/2 to π/2, with r=4cosθ as outer. The area is (1/2) ∫_${-π/2}^{π/2}$ $[(4cosθ)^2$ - $(2cosθ)^2$] dθ. A tempting distractor is choice A, which reverses the order, leading to a negative area value. When finding areas between polar curves, always subtract the inner radius squared from the outer radius squared and confirm the integration limits encompass the entire region of interest.
Find the correct area integral for the region enclosed by $r=1+\cos\theta$ and $r=1$.
$\dfrac12\int_{-\pi}^{\pi}\left((1+\cos\theta)^2-1^2\right)d\theta$
$\dfrac12\int_{0}^{\pi}\left((1+\cos\theta)^2-1^2\right)d\theta$
$\dfrac12\int_{-\pi/2}^{\pi/2}\left((1+\cos\theta)^2-1^2\right)d\theta$
$\int_{-\pi/2}^{\pi/2}\left((1+\cos\theta)-1\right)d\theta$
$\dfrac12\int_{-\pi/2}^{\pi/2}\left(1^2-(1+\cos\theta)^2\right)d\theta$
Explanation
This problem involves calculating the area bounded by two polar curves, specifically the region enclosed by r=1+cosθ and r=1. To set up the integral, determine intersections at 1+cosθ = 1, so cosθ = 0 and θ = ±π/2. Between -π/2 and π/2, r=1+cosθ is greater than or equal to r=1, making it the outer curve. The area is (1/2) ∫_${-π/2}^{π/2}$ $[(1+cosθ)^2$ - $1^2$] dθ. A tempting distractor is choice A, which reverses the radii, yielding a negative integrand where the curves overlap. When finding areas between polar curves, always subtract the inner radius squared from the outer radius squared and confirm the integration limits encompass the entire region of interest.
Which integral gives the area enclosed by $r=1+\cos\theta$ and $r=1-\cos\theta$?
$\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[(1+\cos\theta)^2-(1-\cos\theta)^2\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi}\big[(1-\cos\theta)^2-(1+\cos\theta)^2\big]d\theta$
$\displaystyle\int_{-\pi/2}^{\pi/2}\big[(1+\cos\theta)-(1-\cos\theta)\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(1+\cos\theta)^2-(1-\cos\theta)^2\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{2\pi}\big[(1+\cos\theta)^2-(1-\cos\theta)^2\big]d\theta$
Explanation
This problem tests the skill of finding the area bounded by two polar curves. The curves r=1+cosθ and r=1-cosθ intersect at θ = ±π/2. Between -π/2 to π/2, r=1+cosθ is outer and r=1-cosθ is inner. The area is (1/2) ∫_${-π/2}^{π/2}$ $[(1+cosθ)^2$ - $(1-cosθ)^2$] dθ. A tempting distractor is choice C, which reverses the order, resulting in negative area. A transferable strategy for polar areas is to find intersection points to determine integration limits and identify which curve is outer in each interval.
What integral setup gives the area enclosed by $r=3-\cos\theta$ and $r=1$?
$\displaystyle \frac12\int_{0}^{2\pi}\big[1^2-(3-\cos\theta)^2\big]d\theta$
$\displaystyle \frac12\int_{0}^{\pi}\big[1^2-(3-\cos\theta)^2\big]d\theta$
$\displaystyle \frac12\int_{0}^{2\pi}\big[(3-\cos\theta)^2-1^2\big]d\theta$
$\displaystyle \frac12\int_{0}^{\pi}\big[(3-\cos\theta)^2-1^2\big]d\theta$
$\displaystyle \frac12\int_{-\pi/2}^{\pi/2}\big[(3-\cos\theta)^2-1^2\big]d\theta$
Explanation
This problem asks for the area between the limaçon r = 3 - cos(θ) and the circle r = 1. To find intersections, set 3 - cos(θ) = 1, giving cos(θ) = 2, which has no real solutions since |cos(θ)| ≤ 1. This means one curve is always outside the other. Since the minimum value of r = 3 - cos(θ) is 2 (when cos(θ) = 1), the limaçon is always outside r = 1. The area between them over a full rotation is (1/2)∫_$0^{2π}$[(3-cos(θ))² - 1²]dθ. Choice B incorrectly uses 0 to π, which would only give half the enclosed area. When curves don't intersect, integrate over the full period of the outer curve to capture the entire enclosed region.
For the region enclosed by $r=2\cos\theta$ and $r=2\sin\theta$, what is the correct area setup?
$\displaystyle \frac12\int_{-\pi/4}^{\pi/4}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$
$\displaystyle \frac12\int_{0}^{\pi/4}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]d\theta$
$\displaystyle \frac12\int_{0}^{\pi/4}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$
$\displaystyle \frac12\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]d\theta$
$\displaystyle \frac12\int_{0}^{\pi/2}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$
Explanation
This problem requires finding the area between the circles r = 2cos(θ) and r = 2sin(θ). These curves intersect when 2cos(θ) = 2sin(θ), giving tan(θ) = 1, so θ = π/4 (and θ = 5π/4, but that's outside both curves' domains). The circle r = 2cos(θ) exists for -π/2 ≤ θ ≤ π/2, while r = 2sin(θ) exists for 0 ≤ θ ≤ π. For 0 < θ < π/4, cos(θ) > sin(θ), so r = 2cos(θ) is outer. The area of the enclosed region is (1/2)∫_$0^{π/4}$[(2cos(θ))² - (2sin(θ))²]dθ. Choice A incorrectly uses 0 to π/2 as bounds, including regions where r = 2sin(θ) would be outer. For polar circles centered on axes, their intersection occurs where the angle bisects the quadrant.