What is the area enclosed by r=8sin(3theta-(2pi)/4) +4theta between theta in [pi/8,(pi)/4]?

1 Answer
May 31, 2018

color(blue)[A=int_(pi/8)^(pi/4)[8(theta)^2-32thetacos(3theta)+32cos^2(3theta)]*d(theta)=4.68224]

Explanation:

the area of the polar curve given by:

color(red)[A=1/2int_(theta_1)^(theta_2)r^2*d(theta)

The interval of the integral theta in [pi/8,pi/4]

now lets setup the integral in our interval:

A=1/2int_(pi/8)^(pi/4)(8sin(3theta-(2pi)/4) +4theta)^2*d(theta)

A=1/2int_(pi/8)^(pi/4)(4theta-8cos(3theta))^2*d(theta)

A=1/2int_(pi/8)^(pi/4)(4theta-8cos(3theta))^2*d(theta)

A=int_(pi/8)^(pi/4)[8(theta)^2-32thetacos(3theta)+32cos^2(3theta)]*d(theta)

=[(24sin(6theta)-96xsin(3theta)-32cos(3theta)+24theta^3+144theta)/9]_(pi/8)^(pi/4)

[(3*2^(21/2)pisin((3pi)/8)+2^(27/2)*cos((3pi)/8)+21*2^(5/2)pi^3+(9*2^(19/2)-6144)pi-3*2^(23/2)+2048)/(9*2^(17/2))]=4.68224

Approximation:
A=4.68224

see below the sketch of the polar curve:
jamesjames

see below the region bounded by the curve from pi/8 topi/4
jamesjames
"I hope that helpful"