Question

Find the constant of variation for the equation z=1/3x

Answer 1

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:
SurfaceArea=\(\large\int\limits_{a}^{b} 2\pi\ y\ ds\)
ds for parametric = \(\large\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\ dt\)

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


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


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


Therefore:
S.A.=\(\large\int\limits_{0}^{3} 2\pi\ ((t^2+1)e^t) \sqrt{(te^t)^2+(e^t(t+1)^2)^2}\ dt\)
S.A.=\(\large\int\limits_{0}^{3} 2\pi\ ((t^2+1)e^t) \sqrt{e^{2t}(t+1)^2+e^{2t}(t+1)^4}\ dt\)
S.A.=\(\large\int\limits_{0}^{3} 2\pi\ e^{2t}(t^2+1) \sqrt{t^2+2t+2}\ dt\)

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