[Math] area and limit of the shaded region

calculus

Consider the shaded region outside the sector of a circle of radius 12 meters and inside a right triangle.

a) write the area A= f(θ) of the region as a function of θ.

I found the area of the triangle and then subtracted the area of the sector and got

My equation: $f(θ) = 72 tan \theta – 72\theta = 72(tan \theta – \theta)$ and I know for sure this is correct

b) What is the domain of the function?

I'm not sure how to find this but I'm leaning towards $0 < \theta \leq \frac{\pi}{2}$. My options are

$0 < \theta < \frac{\pi}{2}$

$0 < \theta \leq \frac{\pi}{2}$

$0 \leq \theta \leq \frac{\pi}{2}$

$-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$

c)What does the area approach as $\theta$ approaches $\frac{\pi}{2}$?

I think its $\infty$ but my options are

$-\infty$

$100$

$50$

$\infty$

$0$

$80$

or $40$
My main struggle is with b and c. Can anyone tell me if these answers are correct.

Best Answer

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Yes area of $BCD$ is $f(\theta)=\frac12r^2\tan\theta-\frac12r^2\theta=72(\tan\theta-\theta)$

For domain:

  • $0$ is acceptable as we get a finite value $f(0)=0$.
  • All values from $0$ to $\pi/2$[exclusive] are acceptable.
  • $\pi/2$ is unacceptable as $f(\pi/2)$ is undefined.

For the limit: $$\lim_{\theta\to\pi/2}f(\theta)=72\lim_{\theta\to\pi/2}(\tan\theta-\theta)=\infty$$ As $\lim_{\theta\to\pi/2}\tan\theta=\infty$ and $\lim_{\theta\to\pi/2}\theta=\pi/2$

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