How do you find the area inner loop of #r=4-6sintheta#?

Question

Isabella Ackerman · Accepted Answer

#13pi-26sin^-1(2/3)-12sqrt5#

First we have to find the values of #theta# that constitute one loop. A loop will begin and end at the pole (origin), when #r=0#.

#4-6sintheta=0#

#sintheta=2/3#

Then

#theta=sin^-1(2/3)#

This is the angle in Quadrant #"I"#. The corresponding angle in Quadrant #"II"# where #sintheta=2//3# is given by #pi-sin^-1(2//3)#.

So, we want the area from #alpha=sin^-1(2//3)# to #beta=pi-sin^-1(2//3)#.

The area of a polar curve #r# from #theta_1# to #theta_2# is given by #1/2int_(theta_1)^(theta_2)r^2d theta#. The area of the curve is then:

#A=1/2int_alpha^beta(4-6sintheta)^2d theta#

Also note that #(4-6sintheta)^2=(2(2-3sintheta))^2=4(2-3sintheta)^2#.

#=2int_alpha^beta(2-3sintheta)^2d theta#

#=2int_alpha^beta(4-12sintheta+9sin^2theta)d theta#

Use the simplification #sin^2theta=1/2(1-cos2theta)#.

#=2int_alpha^beta(4-12sintheta+9/2-9/2cos2theta)d theta#

#=int_alpha^beta(13-24sintheta-9cos2theta)d theta#

Integrate term-by-term. Depending on how comfortable you are with integration, you may want to use the substitution #u=2theta# for the last term.

#=[13theta+24costheta-9/2sin2theta]_alpha^beta#

Using #sin2theta=2sinthetacostheta#:

#=[13theta+24costheta-9sinthetacostheta]_alpha^beta#

Recall that #sinalpha=sinbeta=2//3#. Thus, #cosalpha=sqrt(1-sin^2alpha)=sqrt5//3#. However, since #beta# is in Quadrant #"II"#, #cosbeta=-sqrt5//3#.

#=13beta+24cosbeta-9sinbetacosbeta-13alpha-24cosalpha+9sinalphacosalpha#

#=13(beta-alpha)+24(-sqrt5/3)-9(2/3)(-sqrt5/3)-24(5/sqrt3)+9(2/3)(sqrt5/3)#

#=13(pi-sin^-1(2/3)-sin^-1(2/3))-16sqrt5+4sqrt5#

#=13pi-26sin^-1(2/3)-12sqrt5#

Charles Anctil · Answer

undefined To find the area of the inner loop of the polar curve $ r = 4 - 6 \sin(\theta) $, you need to determine the limits of integration for $ \theta $ that enclose the inner loop. The inner loop occurs when $ r = 0 $ and $ r = 4 - 6 \sin(\theta) $ intersects, which happens when $ 4 - 6 \sin(\theta) = 0 $.

Solving $ 4 - 6 \sin(\theta) = 0 $ for $ \theta $:
$$ 4 - 6 \sin(\theta) = 0 $$
$$ \sin(\theta) = \frac{4}{6} = \frac{2}{3} $$

Since $ \sin(\theta) $ can only be in the range $[-1, 1]$, the solution $ \theta $ lies in the interval $ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $. So, the area of the inner loop can be found by integrating the function $ r = 4 - 6 \sin(\theta) $ with respect to $ \theta $ over the interval $ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $.

$$ A = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (4 - 6 \sin(\theta))^2 d\theta $$

After finding the integral, you can evaluate it to determine the area of the inner loop.