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

Answer 1

#(pia^2)/12#

Where #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#, then #sin3theta=0#. Since #sintheta=0# at #theta=0,pi,2pi...# we see that for #sin3theta#, it will be #0# at #0,pi//3,2pi//3...#

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

The expression for the area of any polar equation #r# from #theta=alpha# to #theta=beta# is given by #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#

Use the identity #cos2alpha=1-2sin^2alpha# to rewrite for #sin^2alpha#, showing that #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#

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Answer 2

Adam Adkins

To find the area of the loop of the curve ( r = a \sin^3(\theta) ), where ( r ) represents the distance from the origin to a point on the curve and ( \theta ) is the angle with the positive x-axis, you can use the formula for finding the area enclosed by a polar curve:

[ A = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)]^2 d\theta ]

Where ( f(\theta) ) is the function representing the polar curve, and ( \alpha ) and ( \beta ) are the angles that define the loop. In this case, since we're dealing with a loop, ( \alpha ) and ( \beta ) would correspond to the values of ( \theta ) that define one complete loop around the origin.

First, we need to find the limits of integration, which correspond to the angles that define one complete loop around the origin. Since ( \sin(\theta) ) repeats every ( 2\pi ) radians, we can start from ( \theta = 0 ) and find the first positive ( \theta ) where ( r = 0 ), which is ( \pi ), then go to the next positive ( \theta ) where ( r = 0 ), which is ( 2\pi ). Therefore, the limits of integration are ( \alpha = 0 ) and ( \beta = 2\pi ).

Now, we can plug in ( r = a \sin^3(\theta) ) into the formula for the area:

[ A = \frac{1}{2} \int_{0}^{2\pi} [a \sin^3(\theta)]^2 d\theta ]

[ = \frac{1}{2} \int_{0}^{2\pi} a^2 \sin^6(\theta) d\theta ]

[ = \frac{a^2}{2} \int_{0}^{2\pi} \sin^6(\theta) d\theta ]

[ = \frac{a^2}{2} \left[ \frac{5}{16}\theta - \frac{1}{2}\sin(2\theta) + \frac{1}{24}\sin(4\theta) - \frac{1}{80}\sin(6\theta) \right]_{0}^{2\pi} ]

[ = \frac{5}{16}a^2\pi ]

So, the area of the loop of the curve ( r = a \sin^3(\theta) ) is ( \frac{5}{16}a^2\pi ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.