# Vertor Line intergral

• Nov 16th 2006, 05:26 PM
Ranger SVO
Vertor Line intergral
Let C be the portion of the curve y = 2*sqrt(x) between (1,2) and (9,6) (explination is more important than the answer)

Find
• Nov 16th 2006, 07:50 PM
ThePerfectHacker
Quote:

Originally Posted by Ranger SVO
Let C be the portion of the curve y = 2*sqrt(x) between (1,2) and (9,6) (explination is more important than the answer)

Find

When I see the $ds$ I take that to mean the line intetgral with $\sqrt{[x'(t)]^2+[y'(t)]^2}$

First we have to parametrize the curve,
If you let,
$x=t$ and $y=2\sqrt{t}$
$1\leq t\leq 9$
Thus,
$\sqrt{[x'(t)]^2+[y'(t)]^2}=\sqrt{1+(3\sqrt{t})^2}=\sqrt{1+9t^2}$
Thus,
$\int_C 2yds=\int_1^9 t\sqrt{1+9t^2}dt$
But, this integral is easy to evaluate, use substitution,
$u=1+9t^2$
• Nov 17th 2006, 12:03 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
When I see the $ds$ I take that to mean the line intetgral with $\sqrt{[x'(t)]^2+[y'(t)]^2}$

First we have to parametrize the curve,
If you let,
$x=t$ and $y=2\sqrt{t}$
$1\leq t\leq 9$
Thus,
$\sqrt{[x'(t)]^2+[y'(t)]^2}=\sqrt{1+(3\sqrt{t})^2}=\sqrt{1+9t^2}$
Thus,
$\int_C 2yds=\int_1^9 t\sqrt{1+9t^2}dt$
But, this integral is easy to evaluate, use substitution,
$u=1+9t^2$

Good explanation, but a few mistakes.
Let $x = t$ and $y = 2 \sqrt{t}$.

Then
$ds = \sqrt{x'^2 + y'^2}dt = \sqrt{1 + \frac{1}{t}}dt$

So
$\int_C3yds = \int_1^9 3 \cdot 2 \sqrt{t} \sqrt{1 + \frac{1}{t}}dt$

= $6 \int_1^9 \sqrt{t + 1}dt$

-Dan