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

Question

Christopher Allers · Accepted Answer

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

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=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:

see below the region bounded by the curve from #pi/8# to#pi/4#

Samantha Adkison · Answer

undefined To find the area enclosed by the polar curve $ r = 8\sin(3\theta - \frac{2\pi}{4}) + 4\theta $ between $ \theta = \frac{\pi}{8} $ and $ \theta = \frac{\pi}{4} $, we use the formula for the area enclosed by a polar curve, which is given by:

$$ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta $$

where $ \alpha $ and $ \beta $ are the starting and ending angles, respectively.

Substituting the given limits of integration and the expression for $ r $, the area $ A $ can be calculated as follows:

$$ A = \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (8\sin(3\theta - \frac{2\pi}{4}) + 4\theta)^2 \, d\theta $$

Solving this integral will give the area enclosed by the given polar curve between $ \theta = \frac{\pi}{8} $ and $ \theta = \frac{\pi}{4} $.