Let $R$ be the region enclosed by the polar curve $r( heta)=4 \cos (2 heta)$ where $-\frac{\pi}{8} \leq heta \leq \frac{\pi}{8}$.

Which integral represents the area of $R$ ?
Choose 1 answer:
(A) $\int_{0}^{-\frac{\pi}{4}} 4 \cos ^{2}(2 heta) d heta$
(B) $\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 8 \cos ^{2}(2 heta) d heta$
(C) $\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 \cos ^{2}(2 heta) d heta$
() $\int_{0}^{-\frac{\pi}{4}} 8 \cos ^{2}(2 heta) d heta$

Question

Let $R$ be the region enclosed by the polar curve $r(	heta)=4 \cos (2 	heta)$ where $-\frac{\pi}{8} \leq 	heta \leq \frac{\pi}{8}$.

Which integral represents the area of $R$ ?
Choose 1 answer:
(A) $\int_{0}^{-\frac{\pi}{4}} 4 \cos ^{2}(2 	heta) d 	heta$
(B) $\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 8 \cos ^{2}(2 	heta) d 	heta$
(C) $\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 \cos ^{2}(2 	heta) d 	heta$
() $\int_{0}^{-\frac{\pi}{4}} 8 \cos ^{2}(2 	heta) d 	heta$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ The integral we need to evaluate is given by the region enclosed by the polar curve $r( heta) = 4 \cos(2 heta)$ from $-\frac{\pi}{8}$ to $\frac{\pi}{8}$. ‖ ‖ step 2 ⋮ The area $A$ of a region enclosed by a polar curve $r( heta)$ from $\alpha$ to $\beta$ is given by the integral $A = \frac{1}{2} \int_{\alpha}^{\beta} r^2( heta) \, d heta$. ‖ ‖ step 3 ⋮ Substituting $r( heta) = 4 \cos(2 heta)$, we get $A = \frac{1}{2} \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} (4 \cos(2 heta))^2 \, d heta$. ‖ ‖ step 4 ⋮ Simplifying, we have $A = \frac{1}{2} \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 16 \cos^2(2 heta) \, d heta = 8 \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \cos^2(2 heta) \, d heta$. ‖ ‖ step 5 ⋮ Therefore, the integral that represents the area of $R$ is $8 \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \cos^2(2 heta) \, d heta$. ‖ ***B*** ∻Key Concept∻ ⚹Polar Area Formula⚹ ∻Explanation∻ ⚹The area enclosed by a polar curve $r( heta)$ from $\alpha$ to $\beta$ is given by $\frac{1}{2} \int_{\alpha}^{\beta} r^2( heta) \, d heta$.⚹◊What is the formula to calculate the area enclosed by a polar curve?⍭ Generate me a similar question◊

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