# What is the area enclosed between the two polar curves: #r = 4 - 2cos 3theta# and #r = 5# ?

for vivid clear illustration in his answer) can be regarded as the

cos^3A=1/4(cos 9A+3 cos 3A)#

Altogether in the four quadrants, the area is

I think this shaded area is referred to as in between area. Here,

the circle is the exterior curve.

There is a set of three other equal in-between areas, for which

the circle is the interior. curve.

make it for Sam....

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#"area" = (7pi)/9 +(17sqrt(3))/12#

There are a couple of ingredients to this:

Given two curves:

Points of intersection will satisfy:

Hence:

The area is similar to that of a triangle with vertices:

which will be:

Whereas:

OK (finally)

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To find the area enclosed between the two polar curves ( r = 4 - 2\cos(3\theta) ) and ( r = 5 ), you need to set up the integral and evaluate it.

First, find the points of intersection of the two curves by equating them:

[ 4 - 2\cos(3\theta) = 5 ]

Solve for ( \theta ) to find the limits of integration.

Once you have the limits of integration, set up the integral for the area enclosed between the two curves:

[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} [r_2^2 - r_1^2] d\theta ]

where ( r_2 ) is the outer curve and ( r_1 ) is the inner curve.

Substitute the equations of the curves and integrate within the determined limits to find the area.

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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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