# Homework Help: Help understanding/evaluating line integral over a curve

1. Oct 11, 2012

### Kaldanis

Evaluate the line integral, where C is the given curve.

$\int_{c} xy\:ds$, when C: $x=t^{2}, \ y=2t\ , \ 0\leq t\leq4$

To solve this I should use the formula $\int^{b}_{a} f(x(t),y(t))\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}dt$

This gives me $\int^{4}_{0} (t^{2}2t)\sqrt{(2t)^{2}+(2)^{2}}dt$

When I solve this integral I get around 860.3. I've had various friends solve it and also checked on my calculator and wolfram, all getting 860.3 each time. I've been told this answer is wrong which leads me to believe I'm using the formula incorrectly. Is my integral above correct? Is it wrong to just plain solve the integral from 0 to 4?

I appreciate any help

2. Oct 11, 2012

### Simon Bridge

Who tells you the answer is wrong?

3. Oct 11, 2012

### Kaldanis

It's for online homework that only accepts the correct answer. I can only enter answers so many times before the question is locked, so I want to be sure of the answer before I enter it again

4. Oct 11, 2012

### SammyS

Staff Emeritus

Is it possible to enter square roots, etc?

5. Oct 11, 2012

### Simon Bridge

The thing that occurs to me is to check the decimal places it wants. 860.34?
If there is nothing to tell you (context - how many dp is usually wanted?) then you should tell the person who set the exercise (check with other students).

afaict, there is nothing wrong with the working.

6. Oct 11, 2012

### tiny-tim

looks ok to me!

(btw, i assume you used an algebraic substitution, and not a trig one? )

7. Oct 11, 2012

### Kaldanis

Thank you everyone, the problem was that I was entering an approximate decimal value. It accepted $\frac{8}{15}(1+391\sqrt{17})$ as the correct answer!

8. Oct 11, 2012

### Simon Bridge

grrr... we hates computer mediated assessments oh we does!

9. Oct 12, 2012

:rofl: