# What is the area enclosed by #r=2cos((5theta)/3-(13pi)/8)-3sin((5theta)/8+(pi)/4) # between #theta in [0,pi]#?

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To find the area enclosed by the curve (r = 2\cos\left(\frac{5\theta}{3} - \frac{13\pi}{8}\right) - 3\sin\left(\frac{5\theta}{8} + \frac{\pi}{4}\right)) between (\theta) in ([0, \pi]), you need to integrate the expression (r) with respect to (\theta) and then take half of the result due to symmetry.

The formula for the area enclosed by a polar curve is given by:

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

where (r) is the polar equation of the curve, and (\alpha) and (\beta) represent the limits of integration.

In this case, (r = 2\cos\left(\frac{5\theta}{3} - \frac{13\pi}{8}\right) - 3\sin\left(\frac{5\theta}{8} + \frac{\pi}{4}\right)) and the limits of integration are (\alpha = 0) and (\beta = \pi).

You need to compute (r^2) and then integrate it with respect to (\theta) from 0 to (\pi). After integrating, multiply the result by (1/2) to find the area enclosed by the curve within the given limits.

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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.

- What is the area enclosed by #r=-3sin(2theta-(2pi)/4) -theta# between #theta in [pi/8,(3pi)/4]#?
- What is the area under the polar curve #f(theta) = theta^2sin((5theta)/2 )-cos((2theta)/3+pi/2) # over #[pi/6,(3pi)/2]#?
- What is the Cartesian form of #( -1, (4pi)/3 ) #?
- What is the Cartesian form of #(2,(15pi)/16))#?
- What is the area under the polar curve #f(theta) = 3theta^2+thetasin(4theta-(5pi)/12 ) +cos(2theta-(pi)/3)# over #[pi/8,pi/6]#?

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