What is the area enclosed by $r = θ^{2} - 2 sin θ$ $r=theta^2-2sintheta$ for $θ \in [\frac{π}{4}, π]$ $theta in [pi/4,pi]$ ?

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What is the area enclosed by $r = θ^{2} - 2 sin θ$ $r=theta^2-2sintheta$ for $θ \in [\frac{π}{4}, π]$ $theta in [pi/4,pi]$ ?

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Answer 1 · 2018-06-10T23:20:39

Area $= 21.86658 square units$ $=21.86658""" """"square units"$

Explanation:

$r = θ^{2} - 2 \cdot sin θ$ $r=theta^2-2*sin theta$

Area $= \frac{1}{2} \int_{\frac{π}{4}}^{π} r^{2} d θ$ $=1/2int_(pi/4)^pi r^2 d theta$

Area $= \frac{1}{2} \int_{\frac{π}{4}}^{π} {(θ^{2} - 2 \cdot sin θ)}^{2} d θ$ $=1/2int_(pi/4)^pi(theta^2-2*sin theta)^2 d theta$

Area $= \frac{1}{2} \int_{\frac{π}{4}}^{π} (θ^{4} - 4 \cdot θ^{2} \cdot sin θ + 4 {sin}^{2} θ) d θ$ $=1/2int_(pi/4)^pi(theta^4-4*theta^2*sin theta+4sin^2theta) d theta$

Area $= \frac{1}{2} \cdot {[\frac{θ^{5}}{5} + 4 \cdot θ^{2} \cdot cos θ - 8 \cdot θ \cdot sin θ - 8 \cdot cos θ + 2 \cdot θ - sin 2 θ]}_{\frac{π}{4}}^{2}$ $=1/2*[theta^5/5+4*theta^2*cos theta-8*theta*sin theta-8*cos theta+2*theta-sin 2theta]_(pi/4)^pi$ evaluate from $\frac{π}{4}$ $pi/4$ to $π$ $pi$

Area $= 21.86658 square units$ $=21.86658""" """"square units"$

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What is the area enclosed by $r = θ^{2} - 2 sin θ$ $r=theta^2-2sintheta$ for $θ \in [\frac{π}{4}, π]$ $theta in [pi/4,pi]$ ?

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Explanation:

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What is the area enclosed by r=θ2−2sinθr=theta^2-2sintheta for θ∈[π4,π]theta in [pi/4,pi]?

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What is the area enclosed by $r = θ^{2} - 2 sin θ$ $r=theta^2-2sintheta$ for $θ \in [\frac{π}{4}, π]$ $theta in [pi/4,pi]$ ?