# How do you find the area inside of the circle #r = 3sin(theta)# and outside the cardioid #r = 1 + sin(theta)#?

Draw both curves on the same graph paper.

Notice that the cardioid intersects with the circle at

The area of interest has been shaded above.

To find the area of a polar curve, we use

#A = 1/2 int r^2 "d"theta#

We find the area of the cardioid and the circle separately on the interval

For the circle,

#A = 1/2 int_{pi/6}^{(5pi)/6} r^2 "d"theta#

#= 1/2 int_{pi/6}^{(5pi)/6} (3sin(theta))^2 "d"theta#

#= 9/4 int_{pi/6}^{(5pi)/6} 2sin^2(theta) "d"theta#

#= 9/4 int_{pi/6}^{(5pi)/6} (1-cos(2theta)) "d"theta#

#= 9/4 [theta - sin(2theta)/2]_{pi/6}^{(5pi)/6}#

#= 9/4 ([(5pi)/6 + sqrt3/4]-[pi/6 - sqrt3/4])#

#= (3pi)/2 + (9sqrt3)/8# For the cardioid,

#A = 1/2 int_{pi/6}^{(5pi)/6} r^2 "d"theta#

#= 1/2 int_{pi/6}^{(5pi)/6} (1+sin(theta))^2 "d"theta#

#= 1/2 int_{pi/6}^{(5pi)/6} (1+2sin(theta)+sin^2(theta)) "d"theta#

#= 1/4 int_{pi/6}^{(5pi)/6} (2+4sin(theta)+1-cos(2theta)) "d"theta#

#= 1/4 [3theta - 4cos(theta) - sin(2theta)/2]_{pi/6}^{(5pi)/6}#

#= 1/4 ([(5pi)/2 + 2sqrt3 + sqrt3/4]-[pi/2 - 2sqrt3 - sqrt3/4])#

#= pi/2 + (9sqrt3)/8# The difference in area is

#((3pi)/2 + (9sqrt3)/8) - (pi/2 + (9sqrt3)/8) = pi#

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To find the area inside the circle (r = 3\sin(\theta)) and outside the cardioid (r = 1 + \sin(\theta)), we need to first determine the points of intersection between the two curves.

Setting the equations of the circle and the cardioid equal to each other: [3\sin(\theta) = 1 + \sin(\theta)]

Solving for (\theta): [2\sin(\theta) = 1] [\sin(\theta) = \frac{1}{2}] [\theta = \frac{\pi}{6}, \frac{5\pi}{6}]

These are the points of intersection.

Next, we integrate the area enclosed by the two curves from (\theta = \frac{\pi}{6}) to (\theta = \frac{5\pi}{6}).

The formula for finding the area enclosed by a polar curve is given by: [A = \frac{1}{2}\int_{\alpha}^{\beta} [r(\theta)]^2 d\theta]

Where (\alpha) and (\beta) are the angles at which the curves intersect.

Substituting the equations of the curves: [A = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [(3\sin(\theta))^2 - (1 + \sin(\theta))^2] d\theta]

[A = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [9\sin^2(\theta) - (1 + 2\sin(\theta) + \sin^2(\theta))] d\theta]

[A = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [8\sin^2(\theta) - 2\sin(\theta) - 1] d\theta]

Now, integrate with respect to (\theta) from (\frac{\pi}{6}) to (\frac{5\pi}{6}), and compute 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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