What is the area enclosed by #r=theta^2cos(theta+pi/4)sin(2thetapi/12) # for #theta in [pi/12,pi]#?
The area is about 19.1849 by computational integration.
The area
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To find the area enclosed by the polar curve ( r = \theta^2 \cos(\theta + \frac{\pi}{4})  \sin(2\theta  \frac{\pi}{12}) ) for ( \theta ) in ( [\frac{\pi}{12}, \pi] ), follow these steps:

Calculate the area enclosed by the polar curve using the formula: [ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 , d\theta ] where ( \alpha ) and ( \beta ) are the limits of integration.

Substitute the given limits of integration: [ \alpha = \frac{\pi}{12} ] [ \beta = \pi ]

Square the given polar function: [ r^2 = (\theta^2 \cos(\theta + \frac{\pi}{4})  \sin(2\theta  \frac{\pi}{12}))^2 ]

Integrate the squared function with respect to ( \theta ) over the given interval: [ A = \frac{1}{2} \int_{\frac{\pi}{12}}^{\pi} (\theta^2 \cos(\theta + \frac{\pi}{4})  \sin(2\theta  \frac{\pi}{12}))^2 , d\theta ]

Evaluate the integral to find the area enclosed by the polar curve.
You can use numerical integration methods or software tools to compute the definite integral and find the area enclosed by the given polar curve over the specified interval.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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