# What is the area under the polar curve #f(theta) = theta^2-thetasin(2theta-pi/4 ) +cos(3theta-(5pi)/4)# over #[pi/8,pi/2]#?

Area with polar coordinates

Integration of the definate integral by WolframAlpha

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To find the area under the polar curve ( f(\theta) = \theta^2 - \theta \sin(2\theta - \frac{\pi}{4}) + \cos(3\theta - \frac{5\pi}{4}) ) over ( [\frac{\pi}{8}, \frac{\pi}{2}] ), integrate ( \frac{1}{2} f(\theta)^2 ) with respect to ( \theta ) from ( \frac{\pi}{8} ) to ( \frac{\pi}{2} ). This gives the area enclosed by the curve.

So, the integral to compute the area is:

[ \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{\pi}{2}} ( \theta^2 - \theta \sin(2\theta - \frac{\pi}{4}) + \cos(3\theta - \frac{5\pi}{4}))^2 d\theta ]

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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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