Question #722ff

1 Answer
Dharma R.
Sep 21, 2015

#area=intintrdrd(theta)#
#int_0^(2pi)d(theta)int_0^(4cos3theta)rdr=int_0^(2pi)d(theta)[r^2/2]_0^(4cos3(theta))=int_o^(2pi)8cos^2(3theta)d(theta)#
#int_0^(2pi)4(cos6theta+1)d(theta)=4[(sin6theta)/6+theta]_0^(2pi)=8pi#

#int_(pi/2)^pi4(cos6theta+1)d(theta)=4[(sin6theta)/6+theta]_(pi/2)^pi=4[(sin6pi-sin3pi)+(pi-pi/2)]=2pi#
#2pi=1/4(8pi)=>area under pi/2 to pi =1/4(area From 0to 2pi)#

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