A vector equation thingy

**Dr Zoidburg** · May 19th 2010, 10:04 PM

Calculate $\int_{C}f(X).dX$ where $f(x,y)=(x^2+xy,y-x^2y)$ and C is parametrized by $x=t, y= \frac{1}{t} \; (1\leq t \leq 3)$

I'm not really sure how to go about this. This is what I've attempted:
$x=t \Rightarrow \tfrac{\delta x}{\delta t} = 1$
$y= \tfrac{1}{t} \Rightarrow \tfrac{\delta y}{\delta t} = -\tfrac{1}{t^2}$

$\int_{C}f(X).dX = \int_{C} (x^2+xy).\partial x + (y-x^2y).\partial y = \int_{1}^{3} (t^2+1-\frac{1}{t^3}+\frac{1}{t}).\partial t$

integrating that makes for a unpleasant mess: $\frac{t^3}{3}+t+\frac{1}{2t^2}+ln(t)$ which is why I'm wondering if I'm doing this right.

**HallsofIvy** · May 20th 2010, 03:26 AM

Believe it or not, that is exactly right!

**DrDank** · May 20th 2010, 05:47 AM

Also try
$\int_{C}f(X(t))\cdot X'(t) dt$

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