# How do you find the area of the region bounded by the polar curves #r=cos(2theta)# and #r=sin(2theta)# ?

The areas of both regions are

The graph of

Since the area element in polar coordinates is

Let us evaluate the inside integral first,

By the double-angle idenitity

Hence, the area is

For

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To find the area of the region bounded by the polar curves ( r = \cos(2\theta) ) and ( r = \sin(2\theta) ), you need to first determine the points of intersection of these two curves. Then, integrate the difference between the outer curve and the inner curve over the interval where they intersect. This can be expressed as:

[ A = \frac{1}{2} \int_{\alpha}^{\beta} (r_{\text{outer}})^2 - (r_{\text{inner}})^2 , d\theta ]

where ( r_{\text{outer}} ) is the outer curve, ( r_{\text{inner}} ) is the inner curve, and ( \alpha ) and ( \beta ) are the points of intersection.

First, find the points of intersection by setting the two curves equal to each other:

[ \cos(2\theta) = \sin(2\theta) ]

Solve this equation for ( \theta ) to find the intersection points ( \alpha ) and ( \beta ). Then, integrate the difference of the squared radii from ( \alpha ) to ( \beta ) to find the area of the region bounded by the curves.

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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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