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(c) Find the area of the region that lies inside both curves $r = 3 - 2\sin\theta$ and $r = -3 + 2\sin\theta$ $r = 3 - 2\sin\theta$ $r = -3 + 2\sin\theta$ $3 - 2\sin\theta = -3 + 2\sin\theta$ $6 - 4\sin\theta = 0$ $2r = 3 - 2\sin\theta + 2\sin\theta - 3$ $2r = 0$

Name: c find the area of the region that lies inside both curves r 3 2sintheta and r 3 2sintheta r 3 2sintheta r 3 2sintheta 3 2sintheta 3 2sintheta 6 4sintheta 0 2r 3 2sintheta 2sintheta 3 2r 0 99818
Uploaded: 2024-03-22T03:16:18-08:00
Duration: 9 min 38 s
Channel: Bcrypt_Sha256$$2B$12$We1Wwocamog01O5I.V2Tkouxdh4Ofnmgpwkor7Leaonfpu0Ubfpua Bcrypt_Sha256$$2B$12$We1Wwocamog01O5I.V2Tkokttmmj7Lscvwvlptp4Rlhbswcdg9.Wy
Description: c find the area of the region that lies inside both curves r 3 2sintheta and r 3 2sintheta r 3 2sintheta r 3 2sintheta 3 2sintheta 3 2sintheta 6 4sintheta 0 2r 3 2sintheta 2sintheta 3 2r 0 99818

          (c) Find the area of the region that lies inside both curves $r = 3 - 2\sin\theta$ and $r = -3 + 2\sin\theta$
$r = 3 - 2\sin\theta$
$r = -3 + 2\sin\theta$
$3 - 2\sin\theta = -3 + 2\sin\theta$
$6 - 4\sin\theta = 0$
$2r = 3 - 2\sin\theta + 2\sin\theta - 3$
$2r = 0$

c find the area of the region that lies inside both curves r 3 2sintheta and r 3 2sintheta r 3 2sintheta r 3 2sintheta 3 2sintheta 3 2sintheta 6 4sintheta 0 2r 3 2sintheta 2sintheta 3 2r 0 99818

Added by Michael C.

Calculus: Early Transcendentals

James Stewart 8th Edition

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03/22/2024

Step 1

First, we need to find the intersection points of the two curves by setting their r-values equal: $3 - 2\sin\theta = -3 + 2\sin\theta$ Step 2: Solve for $\sin\theta$: $3 - 2\sin\theta = -3 + 2\sin\theta$ Add $3$ to both sides: $6 - 2\sin\theta = 2\sin\theta$ Add Show more…

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(c) Find the area of the region that lies inside both curves $r = 3 - 2\sin\theta$ and $r = -3 + 2\sin\theta$ $r = 3 - 2\sin\theta$ $r = -3 + 2\sin\theta$ $3 - 2\sin\theta = -3 + 2\sin\theta$ $6 - 4\sin\theta = 0$ $2r = 3 - 2\sin\theta + 2\sin\theta - 3$ $2r = 0$

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