Find the area of a loop of the curve #r=a sin3theta#?

Dove #a=1#, the curve looks like:

Increasing or decreasing the value of #a# will only change the radius of the curve.

To find when the curve begins and ends, set #r=0#, since this is where the curve is at the origin.

If #asin3theta=0#, poi #sin3theta=0#. Da #sintheta=0# at #theta=0,pi,2pi...# we see that for #sin3theta#, sarà #0# at #0,pi//3,2pi//3...#

So, the curve in the first quadrant varies from #theta=0# a #theta=pi//3#.

The expression for the area of any polar equation #r# da #theta=alpha# a #theta=beta# è dato da #1/2int_alpha^betar^2d theta#.

For one loop of the given equation, the corresponding integral is then #1/2int_0^(pi//3)(asin3theta)^2d theta#.

Working this integral:

#1/2int_0^(pi//3)(asin3theta)^2d theta=1/2int_0^(pi//3)a^2(sin^2 3theta)d theta#

Usa l'identità #cos2alpha=1-2sin^2alpha# to rewrite for #sin^2alpha#, Mostrando che #sin^2alpha=1/2(1-cos2alpha)#.

We can use this to say that #sin^2 3theta=1/2(1-cos6theta)#. Then the integral reduces:

#=1/2int_0^(pi//3)a^2(1/2(1-cos6theta))d theta=a^2/4int_0^(pi//3)(1-cos6theta)d theta#

Integrating term by term:

#=a^2/4(theta-1/6sin6theta)|_0^(pi//3)#

#=a^2/4[pi/3-1/6sin2pi-(0-1/6sin0)]#

#=a^2/4(pi/3)#

#=(pia^2)/12#