# What is the area under the polar curve #f(theta) = theta^2sin((5theta)/2 )-cos((2theta)/3+pi/2) # over #[pi/6,(3pi)/2]#?

The answer is 142.523.

The formula for finding the area of a polar curve goes like this...

With this formula in mind, plug in your f graph and your endpoints into the above formula, like so...

Yuck. This looks like a job for a calculator.

I'm getting... 142.523...sounds good to me.

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To find the area under the polar curve ( f(\theta) = \theta^2 \sin\left(\frac{5\theta}{2}\right) - \cos\left(\frac{2\theta}{3} + \frac{\pi}{2}\right) ) over ( \left[\frac{\pi}{6}, \frac{3\pi}{2}\right] ), you need to evaluate the definite integral of the function ( f(\theta) ) with respect to ( \theta ) over the given interval.

[ A = \int_{\frac{\pi}{6}}^{\frac{3\pi}{2}} \left(\theta^2 \sin\left(\frac{5\theta}{2}\right) - \cos\left(\frac{2\theta}{3} + \frac{\pi}{2}\right)\right) d\theta ]

This integral can be evaluated numerically using calculus techniques or computational tools like integration software. Once integrated, you'll obtain the value of the area under the curve.

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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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