The polar curve $r( heta)= heta \cdot \sin (2 heta)$ is graphed for $0 \leq heta \leq \pi$.
Let $R$ be the region in the fourth quadrant enclosed by the curve and the $x$ -axis.

Which integral represents the area of $R$ ?
Choose 1 answer:
(A) $\int_{0}^{\frac{\pi}{2}} \frac{1}{2} \cdot heta^{2} \cdot \sin ^{2}(2 heta) d heta$
(B) $\int_{0}^{\frac{\pi}{2}} \frac{1}{4} \cdot heta^{2} \cdot \sin ^{2}(2 heta) d heta$
(C) $\int_{\frac{\pi}{2}}^{\pi} \frac{1}{4} \cdot heta^{2} \cdot \sin ^{2}(2 heta) d heta$
(ㄷ) $\int_{\frac{\pi}{2}}^{\pi} \frac{1}{2} \cdot heta^{2} \cdot \sin ^{2}(2 heta) d heta$

Q: The polar curve $r(\theta)=\theta \cdot \sin (2 \theta)$ is graphed for $0 \leq \theta \leq \pi$. Let $R$ be the region in the fourth quadrant enclosed by the curve and the $x$ -axis. Which integral represents the area of $R$ ? Choose 1 answer: (A) $\int_{0}^{\frac{\pi}{2}} \frac{1}{2} \cdot \theta^{2} \cdot \sin ^{2}(2 \theta) d \theta$ (B) $\int_{0}^{\frac{\pi}{2}} \frac{1}{4} \cdot \theta^{2} \cdot \sin ^{2}(2 \theta) d \theta$ (C) $\int_{\frac{\pi}{2}}^{\pi} \frac{1}{4} \cdot \theta^{2} \cdot \sin ^{2}(2 \theta) d \theta$ (ㄷ) $\int_{\frac{\pi}{2}}^{\pi} \frac{1}{2} \cdot \theta^{2} \cdot \sin ^{2}(2 \theta) d \theta$

∻Solution by Steps∻ ‖ step 1 ⋮ The given polar curve is $r(\theta) = \theta \cdot \sin(2\theta)$ for $0 \leq \theta \leq \pi$. We need to find the area of the region $R$ in the fourth quadrant enclosed by the curve and the $x$-axis. ‖ ‖ step 2 ⋮ The area $A$ of a region enclosed by a polar curve $r(\theta)$ from $\alpha$ to $\beta$ is given by the integral $A = \frac{1}{2} \int_{\alpha}^{\beta} r(\theta)^2 d\theta$. ‖ ‖ step 3 ⋮ For the region $R$ in the fourth quadrant, $\theta$ ranges from $\frac{\pi}{2}$ to $\pi$. Thus, we need to evaluate the integral $A = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} (\theta \cdot \sin(2\theta))^2 d\theta$. ‖ ‖ step 4 ⋮ Simplifying the integrand, we get $A = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \theta^2 \cdot \sin^2(2\theta) d\theta$. ‖ ‖ step 5 ⋮ Comparing this with the given choices, the correct integral is $D: \int_{\frac{\pi}{2}}^{\pi} \frac{1}{2} \cdot \theta^2 \cdot \sin^2(2\theta) d\theta$. ‖ ***D*** ∻Key Concept∻ ⚹Polar Area Formula⚹ ∻Explanation∻ ⚹The area of a region enclosed by a polar curve $r(\theta)$ from $\alpha$ to $\beta$ is given by $\frac{1}{2} \int_{\alpha}^{\beta} r(\theta)^2 d\theta$.⚹◊What is the relationship between the polar curve and the definite integral representing the area of the region enclosed by the curve and the x-axis⍭ Generate me a similar question◊

Question

The polar curve $r(	heta)=	heta \cdot \sin (2 	heta)$ is graphed for $0 \leq 	heta \leq \pi$.
Let $R$ be the region in the fourth quadrant enclosed by the curve and the $x$ -axis.

Which integral represents the area of $R$ ?
Choose 1 answer:
(A) $\int_{0}^{\frac{\pi}{2}} \frac{1}{2} \cdot 	heta^{2} \cdot \sin ^{2}(2 	heta) d 	heta$
(B) $\int_{0}^{\frac{\pi}{2}} \frac{1}{4} \cdot 	heta^{2} \cdot \sin ^{2}(2 	heta) d 	heta$
(C) $\int_{\frac{\pi}{2}}^{\pi} \frac{1}{4} \cdot 	heta^{2} \cdot \sin ^{2}(2 	heta) d 	heta$
(ㄷ) $\int_{\frac{\pi}{2}}^{\pi} \frac{1}{2} \cdot 	heta^{2} \cdot \sin ^{2}(2 	heta) d 	heta$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ The given polar curve is $r( heta) = heta \cdot \sin(2 heta)$ for $0 \leq heta \leq \pi$. We need to find the area of the region $R$ in the fourth quadrant enclosed by the curve and the $x$-axis. ‖ ‖ step 2 ⋮ The area $A$ of a region enclosed by a polar curve $r( heta)$ from $\alpha$ to $\beta$ is given by the integral $A = \frac{1}{2} \int_{\alpha}^{\beta} r( heta)^2 d heta$. ‖ ‖ step 3 ⋮ For the region $R$ in the fourth quadrant, $ heta$ ranges from $\frac{\pi}{2}$ to $\pi$. Thus, we need to evaluate the integral $A = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} ( heta \cdot \sin(2 heta))^2 d heta$. ‖ ‖ step 4 ⋮ Simplifying the integrand, we get $A = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} heta^2 \cdot \sin^2(2 heta) d heta$. ‖ ‖ step 5 ⋮ Comparing this with the given choices, the correct integral is $D: \int_{\frac{\pi}{2}}^{\pi} \frac{1}{2} \cdot heta^2 \cdot \sin^2(2 heta) d heta$. ‖ ***D*** ∻Key Concept∻ ⚹Polar Area Formula⚹ ∻Explanation∻ ⚹The area of a region enclosed by a polar curve $r( heta)$ from $\alpha$ to $\beta$ is given by $\frac{1}{2} \int_{\alpha}^{\beta} r( heta)^2 d heta$.⚹◊What is the relationship between the polar curve and the definite integral representing the area of the region enclosed by the curve and the x-axis⍭ Generate me a similar question◊

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