Calculus 3

May 2012
92
5
slovenia
Curve [FONT=MathJax_Caligraphic]\(\displaystyle \mathcal{K}\) [/FONT]is defined as an intersection of surfaces \(\displaystyle z^2 = 2x^2 + y^2\) and \(\displaystyle z = x + 1\). The curve's projection on plane \(\displaystyle z = 0\) is positive oriented.


  • Parametrise the curve \(\displaystyle \mathcal{K}\).
  • Find the Frenet-Serret basis of the curve \(\displaystyle \mathcal{K}\) at point \(\displaystyle A(2, 1, 3)\).
  • Find the \(\displaystyle \int_{\mathcal{K}}\vec{F}\mathrm{d}\vec{r}\) of the vector field \(\displaystyle F(x, y, z) = (z, x - 1, -2y)\) along the curve \(\displaystyle \mathcal{K}\). [FONT=MathJax_Size2]∫[/FONT]
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Since z= x+ 1, \(\displaystyle z^2= (x+ 1)^2= x^2+ 2x+ 1= 2x^2+ y^2\) so \(\displaystyle x^2- 2x- 1+ y^2= 0\). From that \(\displaystyle x^2- 2x+ 1+ y^2= 2\) and then \(\displaystyle (x- 1)^2+ y^2= 2\).

Now an obvious parameterization is \(\displaystyle x- 1= \sqrt{2}cos(\theta)\), so \(\displaystyle x= \sqrt{2}cos(\theta)+ 1\), \(\displaystyle y= \sqrt{2}sin(\theta)\), and \(\displaystyle z= x+ 1= \sqrt{2}cos(\theta)+ 2\).

At least try the rest of the problem!
 
May 2012
92
5
slovenia
Ok, I know the second part. But now, how to setup the integral and find it? @HallsofIvy