Question

Evaluate the line integral with respect to arc length
(intergral)(3*(x^2)*y*z)ds
Where c is given by
X=t, y=t^2, z=(2/3)*(t^3), 0<=t<=1

Answer 1

\(\int_0^1 3t^2\cdot t^2\cdot \frac{2}{3}t^3\sqrt{1^2+(2t)^2+(2t^2)^2}\,dt\)
\(\int_0^12t^7\sqrt{((2t^2)+1)^2}\,dt\)
\(\int_0^1 2t^7\cdot ( (2t)^2+1)\, dt\)

Answer 2

thanks mate, do you reckon you could explain how you got from
\[\sqrt{1^{2}+2t^2+(2t^2)^2}\]
to
\[\sqrt{((2t^2)+1)^2 }\]

Answer 3

no, i didn't say that :)
I said that \(1^2+(2t)^2+(2t^2)^2=4t^4+4t^2+1=((2t^2)+1)^2\)
You can easily check that if you expand \(((2t^2)+1)^2\).