Arc length formula help

1. Nov 26, 2006

sherlockjones

1 Find the area bounded by the curve $$x = t - \frac{1}{t}$$, $$y = t + \frac{1}{t}$$ and the line $$y = 2.5$$.

I know that $$A = \int_{\alpha}^{\beta} g(t)f'(t) \; dt$$

I ended up with $$\int_{1}^{2} 2.5-(t+\frac{1}{t})(1+\frac{1}{t^{2}})$$

2 Find the length of the curve: $$x = a(\cos \theta + \theta \sin \theta)$$, $$y = a(\sin \theta-\theta \cos \theta)$$, $$0\leq \theta\leq \pi$$

I obtained $$\frac{a\pi^{2}}{2}$$. Does this look correct? I used the arc length formula for parametric equations.

Is this correct?

3 Find the surface area obtained by rotating the given curve about the x-axis: $$x = 3t-t^{3}$$ $$y = 3t^{2}$$, $$0\leq t\leq 1$$.

So $$S = \int_{a}^{b} 2\pi y \sqrt{(\frac{dx}{dt}^{2})+(\frac{dy}{dt}^{2})} \; dt$$

So would I do the following: $$\int_{0}^{1} 2\pi(3t^{2})\sqrt{(3-3t^{2})+36t^{2}} \; dt$$?

Last edited: Nov 26, 2006
2. Nov 27, 2006

HallsofIvy

Staff Emeritus
Now quite. if x= f(t), y= g(t), then dx= f'(t)dt and ydx= g(t)f'(t)dt but you want (2.5- y)dx. You should have (2.5- t+ 1/t)(1+ 1/t2)dt. Also, recheck your limits of integration. When t= 1, y= 2, not 2.5.

Yes, it is. Don't you just love it when things cancel out?

No. You forgot a square: it should be
$$\int_{0}^{1} 2\pi(3t^{2})\sqrt{(3-3t^{2})^2+36t^{2}} \; dt$$
(Once again, that square root simplifies nicely. Your teacher is being nice to you!)

3. Nov 27, 2006

sherlockjones

Find the surface area generated by rotating the given curve about the y-axis:

$$x = 3t^{2}, y = 2t^{3}, 0\leq t \leq 5$$ would it be:

$$\int_{0}^{5} 2\pi(3t^{2})\sqrt{72t^{2}} \; dt$$

4. Nov 27, 2006

HallsofIvy

Staff Emeritus
No, it wouldn't. You want $\sqrt{(dx/dt)^2+ (dy/dt)^2}$. Here, dx/dt= 6t and dy/dt= 6t2. You need $\sqrt{36t^2+ 36t^4}= 6t\sqrt{1+ t^2}$.