Pad? 2:32 PM @ * 66% A ww3.math.msu.edu WeBWorK: mth 1 ww3.m ath.msu.edu..Desmos

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Pad? 2:32 PM @ * 66% A ww3.math.msu.edu WeBWorK: mth 1 ww3.m ath.msu.edu..Desmos | Graphing... Unit circle - Wikipe... Chegg Study | Guid... MICHIGAN STATE Logged in as bakerk40 Department of Mathematics Log Out C Kwebwork / mth_133_ss18 88080 / hw33-10.4-polar-areas-and-lengths 7 MAIN MENU Courses Hmework Hw33-10.4-Polar-areas-and-lengths: Sets Problem 7 Hw33 10.4 Polar areas and lengths Previous Problem List Next (1 point) Get help entering answers See a similar example (.PDF) Find the area inside the circle r 9 cos and outside the cardioid Probl 7 User Settings Grades Achievements Area - My Answers Submit Answers Show me another Problems You have attempted this problem 0 times OYou have 12 attempts remaining Problem1 Problem2 Problem3 Problem4 Problem5 Problem6 Problem7 Problem8 Problem9 Problem Problem Problem 12

Explanation / Answer

Let parametric curves be given by
r2(?): r = 9cos?
r1(?): r = 3(1+cos?)

area of bounded region: A(?) = 1/2?(r2²- r1²).d? between limits ? = a,b...

the limits are solution to 3cos? = 1+cos? the points of intersection of curves.
2cos? = 1 => ? = ±p/3

A(?) = 1/2?(r2²- r1²).d? = 1/2?(9cos?)² - 9(1+cos?)².d?
= 1/2?(9cos?)².d? - 1/2?9(1+cos?)².d?
= 81/8[2? + sin(2?)] - 9/8[6? + 8sin? +sin(2?)] ..
.............where I have used ?(cos?)².d?=1/4[2? + sin(2?)]
= 27?/2 +9sin(2?) - 9sin(?)

Area = A(p/3) - A(-p/3)
= 9p/2 + 9sin(2p/3) -9sin(p/3) - (-9p/2) - 9sin(-2p/3) + 9sin(-p/3)
= 9p.

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