Help understanding/evaluating line integral over a curve

[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

 

[Simon Bridge]

Who tells you the answer is wrong?

 

[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

 

[SammyS]

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

How about entering the exact answer.

Is it possible to enter square roots, etc?

 

[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.

 

[tiny-tim]

looks ok to me!

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

 

[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!

 

[Simon Bridge]

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

 

[micromass]

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

:rofl: