What is the area under the polar curve #f(theta) = thetasin((3theta)/4 )-cos^3((5theta)/12-pi/2) # over #[0,2pi]#?

Answer 1

I used WolframAlpha to evaluate:
#A = int_0^(2pi)(thetasin((3theta)/4) - cos^3((5theta)/12 - pi/2))^2d theta ~~ 45.592#

From the reference Area with polar coordinates , I obtained the equation:

#A = int_alpha^betar^2d theta#
where #r = f(theta)#
In this case, #r^2 = (thetasin((3theta)/4) - cos^3((5theta)/12 - pi/2))^2#

Substitute into the integral:

#A = int_0^(2pi)(thetasin((3theta)/4) - cos^3((5theta)/12 - pi/2))^2d theta#
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Answer 2

To find the area under the polar curve ( f(\theta) = \theta \sin\left(\frac{3\theta}{4}\right) - \cos^3\left(\frac{5\theta}{12} - \frac{\pi}{2}\right) ) over the interval ([0,2\pi]), we can use the formula for finding the area under a polar curve:

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

where (\alpha) and (\beta) are the bounds of integration.

In this case, (\alpha = 0) and (\beta = 2\pi), so the area is:

[ A = \frac{1}{2} \int_{0}^{2\pi} [\theta \sin\left(\frac{3\theta}{4}\right) - \cos^3\left(\frac{5\theta}{12} - \frac{\pi}{2}\right)]^2 , d\theta ]

Now, we need to evaluate this integral to find the area.

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Answer from HIX Tutor

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.

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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