Area Bounded By Two Polar Curves
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AP Calculus BC › Area Bounded By Two Polar Curves
For the region enclosed by $r=2\cos\theta$ and $r=1$, what integral setup gives its area?
$\dfrac12\int_{-\pi/2}^{\pi/2}\left((2\cos\theta)^2-1^2\right)d\theta$
$\dfrac12\int_{-\pi/3}^{\pi/3}\left(1^2-(2\cos\theta)^2\right)d\theta$
$\dfrac12\int_{-\pi/3}^{\pi/3}\left((2\cos\theta)^2-1^2\right)d\theta$
$\int_{-\pi/3}^{\pi/3}\left(2\cos\theta-1\right)d\theta$
$\dfrac12\int_{0}^{\pi/3}\left((2\cos\theta)^2-1^2\right)d\theta$
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 $r=2\sin\theta$?
$\displaystyle\int_{0}^{\pi}\big[(4\sin\theta)-(2\sin\theta)\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi}\big[(4\sin\theta)^2-(2\sin\theta)^2\big],d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(4\sin\theta)^2\big],d\theta$
$\dfrac12\displaystyle\int_{0}^{2\pi}\big[(4\sin\theta)^2-(2\sin\theta)^2\big],d\theta$
$\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[(4\sin\theta)^2-(2\sin\theta)^2\big],d\theta$
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 $r=2\cos\theta$?
$\dfrac12\int_{0}^{\pi}\left((4\cos\theta)^2-(2\cos\theta)^2\right)d\theta$
$\dfrac12\int_{-\pi/4}^{\pi/4}\left((4\cos\theta)^2-(2\cos\theta)^2\right)d\theta$
$\dfrac12\int_{-\pi/2}^{\pi/2}\left((4\cos\theta)^2-(2\cos\theta)^2\right)d\theta$
$\dfrac12\int_{-\pi/2}^{\pi/2}\left((2\cos\theta)^2-(4\cos\theta)^2\right)d\theta$
$\int_{-\pi/2}^{\pi/2}\left(4\cos\theta-2\cos\theta\right)d\theta$
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 $r=1$.
$\dfrac12\int_{-\pi}^{\pi}\left((1+\cos\theta)^2-1^2\right)d\theta$
$\int_{-\pi/2}^{\pi/2}\left((1+\cos\theta)-1\right)d\theta$
$\dfrac12\int_{-\pi/2}^{\pi/2}\left(1^2-(1+\cos\theta)^2\right)d\theta$
$\dfrac12\int_{-\pi/2}^{\pi/2}\left((1+\cos\theta)^2-1^2\right)d\theta$
$\dfrac12\int_{0}^{\pi}\left((1+\cos\theta)^2-1^2\right)d\theta$
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 $r=1-\cos\theta$?
$\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(1+\cos\theta)^2-(1-\cos\theta)^2\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{2\pi}\big[(1+\cos\theta)^2-(1-\cos\theta)^2\big]d\theta$
$\displaystyle\int_{-\pi/2}^{\pi/2}\big[(1+\cos\theta)-(1-\cos\theta)\big]d\theta$
$\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[(1+\cos\theta)^2-(1-\cos\theta)^2\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi}\big[(1-\cos\theta)^2-(1+\cos\theta)^2\big]d\theta$
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 is the correct area setup for the region enclosed by $r=2\sin\theta$ and $r=2\cos\theta$ in the first quadrant?
$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\big[(2\cos\theta)^2-(2\sin\theta)^2\big], d\theta$
$\dfrac{1}{2}\displaystyle\int_{0}^{2\pi}\big[(2\cos\theta)^2-(2\sin\theta)^2\big], d\theta$
$\dfrac{1}{2}\displaystyle\int_{\pi/4}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big], d\theta$
$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big], d\theta$
$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\big[(2\sin\theta)^2-(2\cos\theta)^2\big], d\theta$
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.
For the region enclosed by $r=2+2\cos\theta$ and $r=2$, which integral gives its area?
$\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[(2+2\cos\theta)^2-2^2\big],d\theta$
$\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\big[2^2-(2+2\cos\theta)^2\big],d\theta$
$\dfrac12\displaystyle\int_{0}^{2\pi}\big[(2+2\cos\theta)^2-2^2\big],d\theta$
$\displaystyle\int_{-\pi/2}^{\pi/2}\big[(2+2\cos\theta)-2\big],d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi}\big[(2+2\cos\theta)^2-2^2\big],d\theta$
Explanation
This problem tests the skill of finding the area bounded by two polar curves. The curves r=2+2cosθ and r=2 intersect at θ = -π/2 and θ = π/2, where r=2. Between these angles, the cardioid r=2+2cosθ is outside the circle r=2, as cosθ ≥0 in this interval, making r≥2. The area is therefore (1/2) ∫_${-π/2}^{π/2}$ $[(2+2cosθ)^2$ - $2^2$] dθ. A tempting distractor is choice A, which integrates over 0 to 2π, but this includes regions where the cardioid is inside the circle, leading to incorrect 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.
Find the correct area setup for the region enclosed by $r=2\cos\theta$ and $r=2\sin\theta$.
$\dfrac12\displaystyle\int_{0}^{\pi/4}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{2\pi}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(2\cos\theta)^2-(2\sin\theta)^2\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi/2}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]d\theta$
$\dfrac12\displaystyle\int_{0}^{\pi/4}\big[(2\sin\theta)^2-(2\cos\theta)^2\big]d\theta$
Explanation
This problem tests the skill of finding the area bounded by two polar curves. The curves r=2cosθ and r=2sinθ intersect at θ=π/4 (and origin). In 0 to π/4, r=2cosθ > r=2sinθ, so outer is 2cosθ. The area between them in this interval is (1/2) ∫_$0^{π/4}$ $[(2cosθ)^2$ - $(2sinθ)^2$] dθ. A tempting distractor is choice C, which reverses the order, giving 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.
Which integral sets up the area enclosed by $r=2\sin(2\theta)$ and $r=0$ for one petal?
$\dfrac{1}{2}\displaystyle\int_{-\pi/4}^{\pi/4}\big[2\sin(2\theta)\big]^2,d\theta$
$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\big[2\sin(2\theta)\big]^2,d\theta$
$\dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}\big[2\sin(2\theta)\big]^2,d\theta$
$\dfrac{1}{2}\displaystyle\int_{0}^{\pi}\big[2\sin(2\theta)\big]^2,d\theta$
$\displaystyle\int_{0}^{\pi/4}2\sin(2\theta),d\theta$
Explanation
This problem tests the skill of finding the area bounded by two polar curves, here the rose and the origin. The petal is between $\theta=0$ and $\theta=\pi/4$, where $r=2\sin(2\theta) \geq 0$. Since inner is $r=0$, the area is $\frac{1}{2} \int_0^{\pi/4} [2\sin(2\theta)]^2 , d\theta$. There are no bounds issues as it's a single petal. A tempting distractor is choice A, which integrates to $\pi/2$, covering two petals. A transferable strategy for polar areas is to find intersection points to determine integration limits and identify which curve is outer in each interval.
Which integral sets up the area enclosed by $r=2\cos\theta$ and $r=2\sin\theta$ in the first quadrant?
$\dfrac12\int_{0}^{\pi/4}\left((2\sin\theta)^2-(2\cos\theta)^2\right)d\theta$
$\dfrac12\int_{0}^{\pi/2}\left((2\cos\theta)^2-(2\sin\theta)^2\right)d\theta$
$\int_{0}^{\pi/4}\left(2\cos\theta-2\sin\theta\right)d\theta$
$\dfrac12\int_{\pi/4}^{\pi/2}\left((2\cos\theta)^2-(2\sin\theta)^2\right)d\theta$
$\dfrac12\int_{0}^{\pi/4}\left((2\cos\theta)^2-(2\sin\theta)^2\right)d\theta$
Explanation
This problem involves calculating the area bounded by two polar curves, specifically the region enclosed by r=2cosθ and r=2sinθ in the first quadrant. To set up the integral, intersections occur at θ=π/4, with r=2cosθ > r=2sinθ from 0 to π/4. Thus, r=2cosθ is the outer curve in this interval. The area is (1/2) ∫_$0^{π/4}$ $[(2cosθ)^2$ - $(2sinθ)^2$] dθ. A tempting distractor is choice D, which reverses the radii, producing a negative integrand. 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.