Question

Question #722ff

Answer 1

`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)`