1. Find the area of the surface generated by rotating the curve $\displaystyle y=e^x, 0\leq x \leq 1$ about the x-axis

2. Prove that the length of the curve $\displaystyle x=t-sint, y=1-cost, 0\leq t\leq 2\pi$ is 8.

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- Aug 2nd 2006, 02:42 PMc_323_hQuestions on my test I didn't know how to solve.
1. Find the area of the surface generated by rotating the curve $\displaystyle y=e^x, 0\leq x \leq 1$ about the x-axis

2. Prove that the length of the curve $\displaystyle x=t-sint, y=1-cost, 0\leq t\leq 2\pi$ is 8. - Aug 2nd 2006, 03:51 PMgalactus
#1. $\displaystyle 2{\pi}\int_{0}^{1}e^{x}\sqrt{1+e^{2x}}dx$

#2. $\displaystyle \frac{d}{dt}[t-sin(t)]=1-cos(t)$

$\displaystyle \frac{d}{dt}[1-cos(t)]=sin(t)$

$\displaystyle \int_{0}^{2{\pi}}\sqrt{(1-cos(t))^{2}+(sin(t))^{2}}dt$=$\displaystyle \int_{0}^{2{\pi}}\sqrt{2(1-cos(t))}dt$