Question

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...

\[S.A. = \large\int\limits_{a}^{b} 2\pi\ ((t^2+1)e^t) \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\ dt \]


\(dx = te^t\ dt\)
\(\large\frac{dx}{dt} = te^t\)


\(dy = (t^2+1)e^t+e^t(2t)\ dt \)
\(\large\frac{dy}{dt} = e^t(t^2+2t+1) \)
\(\large\frac{dy}{dt} = e^t(t+1)^2 \)

Answer 1

\[\large\int\limits_{0}^{3} 2\pi\ ((t^2+1)e^t) \sqrt{(te^t)^2+(e^t(t+1)^2)^2}\ dt\]
\[\large\int\limits_{0}^{3} 2\pi\ ((t^2+1)e^t) \sqrt{e^{2t}(t+1)^2+e^{2t}(t+1)^4}\ dt\]
\[\large\int\limits_{0}^{3} 2\pi\ e^{2t}(t^2+1) \sqrt{t^2+2t+2}\ dt\]

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

Answer 2

I'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 >_<

Answer 3

Well actually I hope the error can be found, so the correct answer can be found

Answer 4

revolving to x-axis yes.

Answer 5

^_^ ty all in advance

Answer 6

Yes sir that is correct, x-axis rotation. That wasn't in the question's specific directions, but it was in the section's directions. Sorry about that. :-/

Answer 7

"...the surface obtained by rotating the given curves about the x-axis."

Answer 8

your dx/dt is wrong. Look at it again.

Answer 9

The 4 goes to 0, yes?

Answer 10

Yes and in total,
dx/dt = (t+1)e^t