- #1

sherlockjones

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**1**Find the area bounded by the curve [tex] x = t - \frac{1}{t} [/tex], [tex] y = t + \frac{1}{t} [/tex] and the line [tex] y = 2.5 [/tex].

I know that [tex] A = \int_{\alpha}^{\beta} g(t)f'(t) \; dt [/tex]I ended up with [tex] \int_{1}^{2} 2.5-(t+\frac{1}{t})(1+\frac{1}{t^{2}}) [/tex]

**2**Find the length of the curve: [tex] x = a(\cos \theta + \theta \sin \theta) [/tex], [tex] y = a(\sin \theta-\theta \cos \theta) [/tex], [tex] 0\leq \theta\leq \pi [/tex]

I obtained [tex] \frac{a\pi^{2}}{2} [/tex]. 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: [tex] x = 3t-t^{3} [/tex] [tex] y = 3t^{2} [/tex], [tex] 0\leq t\leq 1 [/tex].

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

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

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