# What is the area enclosed by #r=sin(5theta-(13pi)/12) # between #theta in [pi/8,(pi)/4]#?

Note: There are some nuances. In strict mathematical parlance,

Yet, here, r < 0 for part of the given interval. You can see that

5 more, for conventionally added r <=0 petals.

Interested readers can ponder over this aspect of the problem.

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To find the area enclosed by the curve ( r = \sin(5\theta - \frac{13\pi}{12}) ) between ( \theta = \frac{\pi}{8} ) and ( \theta = \frac{\pi}{4} ), you need to integrate the polar function with respect to ( \theta ) over the given interval.

[ A = \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} [r(\theta)]^2 , d\theta ]

Here, ( r(\theta) = \sin(5\theta - \frac{13\pi}{12}) ).

Evaluate this integral over the given interval to find the area enclosed by the curve.

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