How do you find the area of one petal of #r=2cos3theta#?

Answer 1

#A=pi/3#

First, graph #r=2cos(3theta)# to get an idea of what the petals look like. It can be really helpful to draw concentric circles and radial angle lines on graph paper, so that you have a polar graph, like this:

Next, using either a graphing utility or this graph paper, plot the graph using convenient points.

#{:(theta,r),(--,--),(0,2),(+-pi/12,+-sqrt(2)~~+-1.41),(+-pi/8,+-sqrt(2-sqrt(2))~~+-0.77),(+-pi/6,0):}#

The area of a petal can be determined by an integral of the form

#A=1/2int_alpha^betar(theta)^2 d theta#

Notice the petal in Quadrant I and IV does not extend past #+-pi/6# and that it is perfectly split between the two quadrants. That implies that if we can find the are of just half a petal, then we can multiply the result by two and get the area of the entire petal.

Letting the interval of integration go from #theta=0# to #theta=pi/6# and doubling the entire integration gives

#A=2xx1/2int_0^(pi/6)4cos^2(3theta)d theta#

#A=4int_0^(pi/6)cos^2(3theta)d theta#

The half-angle identity for cosine says

#cos^2(u)=(1+cos(2u))/2#, so

#cos^2(3theta)=(1+cos(6theta))/2#.

This changes our integral to

#A=4int_0^(pi/6)(1+cos(6theta))/2d theta#

#A=2int_0^(pi/6)(1+cos(6theta))d theta#

#A=2int_0^(pi/6)d theta+2int_0^(pi/6)cos(6theta)d theta#

Using #u#-substitution, you can determine the integral of the cosine function.

#A=2[theta]_0^(pi/6)+2[1/6sin(6theta)]_0^(pi/6)#

#A=pi/3#

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Answer 2

To find the area of one petal of the polar curve ( r = 2 \cos(3\theta) ), follow these steps:

  1. Determine the limits of integration for ( \theta ) that correspond to one complete petal. This typically involves finding the values of ( \theta ) where the curve intersects itself.

  2. Set up the integral for the area using the formula for the area enclosed by a polar curve:

[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} [r(\theta)]^2 , d\theta ]

where ( r(\theta) ) is the equation of the curve and ( \theta_1 ) and ( \theta_2 ) are the limits of integration.

  1. Substitute ( r(\theta) = 2 \cos(3\theta) ) into the integral.

  2. Integrate the expression obtained in step 3 with respect to ( \theta ) from ( \theta_1 ) to ( \theta_2 ).

  3. Evaluate the integral to find the area of one petal of the curve.

By following these steps, you can find the area of one petal of the polar curve ( r = 2 \cos(3\theta) ).

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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.

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.

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.

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.

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