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anonymous
 4 years ago
Check my work on this please? :)
Surface integral for the parametric equations
x = 4 + te\(^t\), y = (t\(^2\) + 1)e\(^t\), 0 ≤ t ≤ 3
Reference:
Surface Area = \(\large\int\limits_{a}^{b} 2\pi\ y\ ds \)
ds for parametric = \( \large\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\ dt \)
So...
\[S.A. = \large\int\limits_{a}^{b} 2\pi\ ((t^2+1)e^t) \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\ dt \]
\(dx = te^t\ dt\)
\(\large\frac{dx}{dt} = te^t\)
\(dy = (t^2+1)e^t+e^t(2t)\ dt \)
\(\large\frac{dy}{dt} = e^t(t^2+2t+1) \)
\(\large\frac{dy}{dt} = e^t(t+1)^2 \)
anonymous
 4 years ago
Check my work on this please? :) Surface integral for the parametric equations x = 4 + te\(^t\), y = (t\(^2\) + 1)e\(^t\), 0 ≤ t ≤ 3 Reference: Surface Area = \(\large\int\limits_{a}^{b} 2\pi\ y\ ds \) ds for parametric = \( \large\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\ dt \) So... \[S.A. = \large\int\limits_{a}^{b} 2\pi\ ((t^2+1)e^t) \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\ dt \] \(dx = te^t\ dt\) \(\large\frac{dx}{dt} = te^t\) \(dy = (t^2+1)e^t+e^t(2t)\ dt \) \(\large\frac{dy}{dt} = e^t(t^2+2t+1) \) \(\large\frac{dy}{dt} = e^t(t+1)^2 \)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\large\int\limits_{0}^{3} 2\pi\ ((t^2+1)e^t) \sqrt{(te^t)^2+(e^t(t+1)^2)^2}\ dt\] \[\large\int\limits_{0}^{3} 2\pi\ ((t^2+1)e^t) \sqrt{e^{2t}(t+1)^2+e^{2t}(t+1)^4}\ dt\] \[\large\int\limits_{0}^{3} 2\pi\ e^{2t}(t^2+1) \sqrt{t^2+2t+2}\ dt\] I'm getting 35833.252388 as the answer, Wolfram says that's ok so I'm looking for a setup error. Help? http://www.wolframalpha.com/input/?i=integrate+from+0+to+3+for+2pi%28e^%282t%29%29%28t^2%2B1%29sqrt%28t^2%2B2t%2B2%29

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll have to leave soon unfortunately, but I wanted to get this up before I left because it can take several hours for these hard types of questions to be answered typically :D Hopefully I don't have any glaring errors >_<

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well actually I hope the error can be found, so the correct answer can be found

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0revolving to xaxis yes.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0^_^ ty all in advance

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes sir that is correct, xaxis rotation. That wasn't in the question's specific directions, but it was in the section's directions. Sorry about that. :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"...the surface obtained by rotating the given curves about the xaxis."

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0your dx/dt is wrong. Look at it again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 4 goes to 0, yes?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Yes and in total, dx/dt = (t+1)e^t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1341528724406:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1341528864848:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is no close solution for integral.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So your answer isn't correct.
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