# What is the area enclosed by #r=2sin(theta+(7pi)/4) # between #theta in [pi/8,(pi)/4]#?

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To find the area enclosed by the polar curve ( r = 2 \sin(\theta + \frac{7\pi}{4}) ) between ( \theta = \frac{\pi}{8} ) and ( \theta = \frac{\pi}{4} ), you need to evaluate the integral of ( \frac{1}{2} r^2 ) with respect to ( \theta ) over the given interval.

The integral for the area ( A ) is given by:

[ A = \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \frac{1}{2} (2\sin(\theta + \frac{7\pi}{4}))^2 d\theta ]

Simplify and solve the integral:

[ A = \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} 2\sin^2(\theta + \frac{7\pi}{4}) d\theta ] [ A = \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} 2\sin^2(\theta + \frac{7\pi}{4}) d\theta ] [ A = \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (1 - \cos(2(\theta + \frac{7\pi}{4}))) d\theta ] [ A = \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (1 - \cos(2\theta + \frac{7\pi}{2})) d\theta ]

Apply the integral properties:

[ A = \left[\theta - \frac{1}{2}\sin(2\theta + \frac{7\pi}{2})\right]_{\frac{\pi}{8}}^{\frac{\pi}{4}} ]

Evaluate the integral:

[ A = \left(\frac{\pi}{4} - \frac{1}{2}\sin(\frac{\pi}{2} + \frac{7\pi}{2})\right) - \left(\frac{\pi}{8} - \frac{1}{2}\sin(\pi + \frac{7\pi}{2})\right) ]

[ A = \left(\frac{\pi}{4} - \frac{1}{2}\right) - \left(\frac{\pi}{8} - \frac{1}{2}\sin(\frac{15\pi}{2})\right) ]

[ A = \frac{\pi}{4} - \frac{1}{2} - \frac{\pi}{8} ]

[ A = \frac{\pi}{8} - \frac{1}{2} ]

Therefore, the area enclosed by the polar curve between ( \theta = \frac{\pi}{8} ) and ( \theta = \frac{\pi}{4} ) is ( \frac{\pi}{8} - \frac{1}{2} ).

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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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