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

Intersection:

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To find the area of the region shared by the circles ( r = 2\cos(\theta) ) and ( r = 2\sin(\theta) ) in polar coordinates, you first need to determine the points of intersection between the two curves. This occurs when ( 2\cos(\theta) = 2\sin(\theta) ). Solving this equation yields ( \theta = \frac{\pi}{4} ) and ( \theta = \frac{5\pi}{4} ).

Then, integrate the function ( \frac{1}{2}(r_2^2 - r_1^2) ) from ( \theta = \frac{\pi}{4} ) to ( \theta = \frac{5\pi}{4} ), where ( r_2 = 2\cos(\theta) ) and ( r_1 = 2\sin(\theta) ).

The integral setup would be:

[ A = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (2\cos(\theta))^2 - (2\sin(\theta))^2 , d\theta ]

[ A = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (4\cos^2(\theta) - 4\sin^2(\theta)) , d\theta ]

[ A = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (4\cos(2\theta)) , d\theta ]

[ A = \frac{1}{2} [2\sin(2\theta)]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} ]

[ A = \frac{1}{2} (2\sin(\frac{\pi}{2}) - 2\sin(\frac{\pi}{2})) ]

[ A = \frac{1}{2} (2 - (-2)) ]

[ A = \frac{1}{2} (4) ]

[ A = 2 ]

So, the area of the region shared by the circles is ( 2 ) square units.

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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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