# Calculus 3

#### lebdim

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
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!

#### lebdim

Ok, I know the second part. But now, how to setup the integral and find it? @HallsofIvy