Find the area of the region common to the circle r = 4 and the cardioid r = 4(1

Question

Find the area of the region common to the circle r = 4 and the cardioid r = 4(1 - cos theta). The area shared by the circle and the cardioid is (Type an exact answer, using x as needed) Find the area of the region inside the circle r= - 6 sin theta and outside the circle r= 3. The area of the region is (Type an exact answer, using x as needed) Find the area inside the lemniscate r^2 = 36 cos20 and outside the circle r= squareroot 8 The area inside the lemniscate and outside the circle is (Type an exact answer, using x as needed)

Explanation / Answer

1) We have given the circle r=4 and cardioid r=4(1-cos)

we break the area enclosed by two curves into two parts.Let's call A1 the area swept by the cardioid and A2 the area swept by the circle.

The cardioid intersects the pole at =0 and hits the r -axis when theta =Pi/2

So A1=integration of from (0 to Pi/2){(4-4*cos)^2}d

A1=integration of from (0 to Pi/2){16+16cos^2()-32cos}d

A1=integration of from (0 to Pi/2){16+16(1+cos2)/2-32cos}d

A1=integration of from (0 to Pi/2){16+8+8cos2-32cos}d

A1={16+8+4*(sin2)-32sin}from (0 to Pi/2)={8Pi+4Pi+0-32}-{0}=12Pi-32

Now A2 is simply the quarter area of a circle with radius 4,so A2=1/4(Pi*r^2)

A2=1/4(Pi*r^2)=1/4(Pi*16)=4Pi

Using symmetry we have sigma A=2A1+2A2=2(12Pi-32+4Pi)=2*(16Pi-32)=32Pi-64

The area shared by the circle and cardioid is 32Pi-64

2) The given inside circle r=-6sin and outside circle r=3

-6sin=3 which implies sin=-1/2 then =-Pi/6,7Pi/6,11Pi/6

we use the angles -Pi/6,11Pi/6 to enclose the area.

So the area is

A=integration of from(-Pi/6 to 11Pi/6)(1/2)*{(-6sin)^2-(3)^2}d

A=integration of from(-Pi/6 to 11Pi/6)(1/2)*{36((1-cos2)/2)-9}d=integration of from(-Pi/6 to 11Pi/6)(1/2)*{(18-18cos2)-9}d

A=(1/2)*{18-9sin2-9} from(-Pi/6 to 11Pi/6)=(1/2)*{33Pi-9(sin(11Pi/3))-(33Pi/2)}-{-3*Pi-9sin(-Pi/3)+9Pi/6}

A=(1/2)*{30Pi-18sin(Pi/3)-18Pi} since by simplication

A=(1/2)*{12Pi-9*sqrt(3)}=(1/2)*22.110=11.055

3)The given inside the lemniscate r^2=36cos2 and outside circle r^2=18

36cos2=18 which implies cos=Pi/6 when cos2= 1/2 therefore =-Pi/6,to Pi/6

Area A=intgration of from(-Pi/6,to Pi/6)(1/2)*{36cos2-18}d

A=(1/2)*{((36sin2)/2)-18} from(-Pi/6,to Pi/6)=9*{sin2-}from(-Pi/6,to Pi/6)

A=9*{{sin2(Pi/6)-Pi/6}-{sin2(-Pi/6)+Pi/6}}=9*{2(sin2(Pi/6)-2*Pi/6)}=9*{2(1/2)-Pi/3}=9*{(3-Pi)/3}=3{3-Pi}=9-3Pi

Therefore the area the given region is A =9-3Pi

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