Let $R$ be the region in the fourth and first quadrants that is inside the polar curve $r=1+\cos ^{2}( heta)$ and outside the polar curve $r=\frac{5}{4}$, as shown in the graph. The curves intersect at $ heta=-\frac{\pi}{3}$ and $ heta=\frac{\pi}{3}$.

Which integral represents the area of $R$ ?
Choose 1 answer:
(A) $\frac{1}{2} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left(\left(1+\cos ^{2}( heta) ight)^{2}-\frac{25}{16} ight) d heta$
(B) $\frac{1}{2} \int_{-\frac{\pi}{2}}^{-\frac{\pi}{3}}\left(1+\cos ^{2}( heta) ight)^{2} d heta+\frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{25}{16} d heta$
(C) $\frac{1}{2} \int_{-\frac{\pi}{3}}^{0}\left(1+\cos ^{2}( heta) ight)^{2} d heta+\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \frac{25}{16} d heta$
(D) $\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\left(1+\cos ^{2}( heta) ight)^{2}-\frac{25}{16} ight) d heta$

Question

Let $R$ be the region in the fourth and first quadrants that is inside the polar curve $r=1+\cos ^{2}(	heta)$ and outside the polar curve $r=\frac{5}{4}$, as shown in the graph. The curves intersect at $	heta=-\frac{\pi}{3}$ and $	heta=\frac{\pi}{3}$.

Which integral represents the area of $R$ ?
Choose 1 answer:
(A) $\frac{1}{2} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left(\left(1+\cos ^{2}(	heta)ight)^{2}-\frac{25}{16}ight) d 	heta$
(B) $\frac{1}{2} \int_{-\frac{\pi}{2}}^{-\frac{\pi}{3}}\left(1+\cos ^{2}(	heta)ight)^{2} d 	heta+\frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{25}{16} d 	heta$
(C) $\frac{1}{2} \int_{-\frac{\pi}{3}}^{0}\left(1+\cos ^{2}(	heta)ight)^{2} d 	heta+\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \frac{25}{16} d 	heta$
(D) $\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\left(1+\cos ^{2}(	heta)ight)^{2}-\frac{25}{16}ight) d 	heta$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the area of the region $R$ between the curves $r=1+\cos^2( heta)$ and $r=\frac{5}{4}$, we use the formula for the area between two polar curves: $$A = \frac{1}{2} \int_{\alpha}^{\beta} \left( r_1^2 - r_2^2 ight) d heta$$ where $r_1$ is the outer curve and $r_2$ is the inner curve. ‖ ‖ step 2 ⋮ Here, $r_1 = 1 + \cos^2( heta)$ and $r_2 = \frac{5}{4}$. The curves intersect at $ heta = -\frac{\pi}{3}$ and $ heta = \frac{\pi}{3}$. ‖ ‖ step 3 ⋮ Substituting the given curves and limits into the formula, we get: $$A = \frac{1}{2} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left( (1 + \cos^2( heta))^2 - \left(\frac{5}{4} ight)^2 ight) d heta$$ ‖ ‖ step 4 ⋮ Simplifying the integrand: $$A = \frac{1}{2} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left( (1 + \cos^2( heta))^2 - \frac{25}{16} ight) d heta$$ ‖ ***A*** ∻Key Concept∻ ⚹Area between polar curves⚹ ∻Explanation∻ ⚹To find the area between two polar curves, use the formula $\frac{1}{2} \int_{\alpha}^{\beta} \left( r_1^2 - r_2^2 ight) d heta$, where $r_1$ is the outer curve and $r_2$ is the inner curve.⚹◊What is the general formula to calculate the area between two polar curves?⍭ Generate me a similar question◊

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