# What is the area enclosed by #r=8sin(3theta-(2pi)/4) +4theta# between #theta in [pi/8,(pi)/4]#?

the area of the polar curve given by:

The interval of the integral

now lets setup the integral in our interval:

Approximation:

see below the sketch of the polar curve:

see below the region bounded by the curve from

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To find the area enclosed by the polar curve ( r = 8\sin(3\theta - \frac{2\pi}{4}) + 4\theta ) between ( \theta = \frac{\pi}{8} ) and ( \theta = \frac{\pi}{4} ), we use the formula for the area enclosed by a polar curve, which is given by:

[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 , d\theta ]

where ( \alpha ) and ( \beta ) are the starting and ending angles, respectively.

Substituting the given limits of integration and the expression for ( r ), the area ( A ) can be calculated as follows:

[ A = \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (8\sin(3\theta - \frac{2\pi}{4}) + 4\theta)^2 , d\theta ]

Solving this integral will give the area enclosed by the given polar curve between ( \theta = \frac{\pi}{8} ) and ( \theta = \frac{\pi}{4} ).

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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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