Find the area of a single loop in curve #r=\sin(6\theta)#?

I am told the formula is #A=1/2\int_a^br^2d\theta#, so
#A=1/2\int_a^b(\sin(6\theta))^2d\theta#

But what are the bound values, #a# and #b#?

1 Answer

The area of 1 loop of the given polar curve is #pi/24# square units.

Explanation:

Start by drawing the polar curve. It helps to picture it.

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As you can see, each loop starts and ends when #r = 0#. Thus our bounds of integration will be consecutive values of #theta# where #r = 0#.

#sin(6theta) = 0#

#6theta = 0 or 6theta = pi#

#theta = 0 or theta = pi/6#

Thus we will be finding the value of #1/2 int_0^(pi/6) sin^2(6x) dx# to find the area.

#A = 1/2int_0^(pi/6) sin^2(6x)dx#

Recall that #cos(2x) = 1 - 2sin^2(x)#, thus #cos(12x) = 1 - 2sin^2(6x)#, and it follows that #sin^2(6x) = (cos(12x) - 1)/(-2) = 1/2 - cos(12x)/2#

#\color(maroon)(A=1/2\int_0^(\pi/6)(1/2-\cos(12x)/2)dx)#
#\color(maroon)(A=1/2{:[1/2x-1/2(1/12\sin(12x))]|:}_0^(\pi/6))#
#A = 1/2{:[1/2x - 1/24sin(12x)]|:}_0^(pi/6)#
#A = 1/2(pi/12)#
#A = pi/24#

Hopefully this helps!