# Thread: Improper integrals/ techniques, urgent help needed.

1. ## Improper integrals/ techniques, urgent help needed.

In the figure, A is at (1,0) and the segment BD is the height of OAB. Let R be the ratio of the area of DAB to that of the shaded region formed by deleting OAB from the circular sector subtended by angle theta. Find lim R theta-> 0+

Ok so I know that this basically means find the derivative of R but I have no idea how to set up the problem and as soon as I know how to set it up I can solve it. Help please?

2. Originally Posted by wmm808

In the figure, A is at (1,0) and the segment BD is the height of OAB. Let R be the ratio of the area of DAB to that of the shaded region formed by deleting OAB from the circular sector subtended by angle theta. Find lim R theta-> 0+

Ok so I know that this basically means find the derivative of R but I have no idea how to set up the problem and as soon as I know how to set it up I can solve it. Help please?
area of the sector = $\displaystyle \frac{r^2 \theta}{2} = \frac{\theta}{2}$

area of triangle OBA = $\displaystyle \frac{\sin{\theta}}{2}$

area of triangle OBD = $\displaystyle \frac{\sin{\theta}\cos{\theta}}{2}$

$\displaystyle R = \frac{\frac{\theta}{2} - \frac{\sin{\theta}}{2}}{\frac{\sin{\theta}}{2} - \frac{\sin{\theta}\cos{\theta}}{2}}$

$\displaystyle R = \frac{\theta - \sin{\theta}}{\sin{\theta}(1 - \cos{\theta})}$

I got $\displaystyle \lim_{\theta \to 0^+} R(\theta) = \frac{1}{3}$

I had to use L'Hopital to get it ... maybe someone else can work it another way.

3. Originally Posted by skeeter
area of the sector = $\displaystyle \frac{r^2 \theta}{2} = \frac{\theta}{2}$

area of triangle OBA = $\displaystyle \frac{\sin{\theta}}{2}$

area of triangle OBD = $\displaystyle \frac{\sin{\theta}\cos{\theta}}{2}$

$\displaystyle R = \frac{\frac{\theta}{2} - \frac{\sin{\theta}}{2}}{\frac{\sin{\theta}}{2} - \frac{\sin{\theta}\cos{\theta}}{2}}$

$\displaystyle R = \frac{\theta - \sin{\theta}}{\sin{\theta}(1 - \cos{\theta})}$

I got $\displaystyle \lim_{\theta \to 0^+} R(\theta) = \frac{1}{3}$

I had to use L'Hopital to get it ... maybe someone else can work it another way.
i think you got the ratio flipped. it should be the shaded region over the triangle while you did the triangle over the shaded region. the limit therefore should be 3 instead of 1/3.

4. By Taylor's method..

$\displaystyle \lim_{\theta\rightarrow{0}}\ \frac{\theta-Sin\theta}{Sin\theta-Sin\theta\ Cos\theta}=\lim_{\theta\rightarrow{0}}\ \frac{f(\theta)}{g(\theta)}$$\displaystyle = \frac{\frac{d}{dx}f(\theta)}{\frac{d}{dx}g(\theta) }\ at\ \theta=0,\ =\frac{f'(0)}{g'(0)} \displaystyle \frac{f'(\theta)}{g'(\theta)}=\frac{1-Cos\theta}{Cos\theta(1-Cos\theta)+Sin\theta\ Sin\theta}$$\displaystyle =\frac{1-Cos\theta}{Cos\theta\ (1-Cos\theta)+1-Cos^2\theta}$

$\displaystyle =\frac{1-Cos\theta}{Cos\theta(1-Cos\theta)+(1-Cos\theta)(1+Cos\theta)}$$\displaystyle =\frac{1}{Cos\theta+(1+Cos\theta)}\ evaluated\ at\ \theta=0$

$\displaystyle giving\ \frac{1}{3}=$ the ratio of the shaded region to the triangle DBA

5. No,
skeeter did it correctly.

6. Originally Posted by Archie Meade
No,
skeeter did it correctly.
Let R be the ratio of the area of DAB to that of the shaded region
oblixps is correct, I flipped it.

oh well ... fractional dyslexia strikes again.

7. Eye strain!!
i worked out the shaded to the triangle independently
and saw you posted skeeter!
so i didnt write it upside down but still got caught out!

i thought you were being ticked off for implying the triangle
was one third of the area of the shaded piece at the limit...