What is the surface area when rotating a curve about the x-axis?

[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: $$\int_{0}^{1} 2\pi(3t^{2})\sqrt{(3-3t^{2})+36t^{2}} \; dt$$?

 

[HallsofIvy]

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 $$\int_{1}^{2} 2.5-(t+\frac{1}{t})(1+\frac{1}{t^{2}})$$

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.

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?

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

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: $$\int_{0}^{1} 2\pi(3t^{2})\sqrt{(3-3t^{2})+36t^{2}} \; dt$$?

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!)

 

[sherlockjones]

Instead if the question read as:

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:

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

 

[HallsofIvy]

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