# What is the area enclosed by #r=-sin(theta+(11pi)/8) -theta/4# between #theta in [0,(pi)/2]#?

Area inside polar curves:

(If this problem allows a calculator, use a graphing calculator here)

Expand what's inside the integral:

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To find the area enclosed by the curve ( r = -\sin\left(\theta + \frac{11\pi}{8}\right) - \frac{\theta}{4} ) between ( \theta = 0 ) and ( \theta = \frac{\pi}{2} ), you need to integrate ( \frac{1}{2} r^2 ) with respect to ( \theta ) over the given interval. The integral will yield the desired 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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