Integrate curve f ds Line Integrals

[Unart]

Homework Statement

Compute ∫f ds for f(x,y)= √(1+9xy), y=x^3 for 0≤x≤1

Homework Equations

∫f ds= ∫f(c(t))||c'(t)||

||c'(t)|| is the magnitude of ∇c'(t)

The Attempt at a Solution

So, with this equation y=x^3 ... I got the that c(t)= <t,t^3>
c'(t)=<1,3t^2>

I know that from the equation y=x^3... x=t=0 and 1... I don't know how to get the magnitude of such equation. They the lower and upper limit.

Another thing is I cannot for the life of me figure out how to take the anti-derivative of √(1+9xy)... which by the time I change to t it would be √(1+9t^4)...

Of course if I'm approaching this the wrong way please, tell me what I'm doing wrong. Please let me know if it isn't clear enough.

 

[haruspex]

||c'(t)|| is the magnitude of ∇c'(t)

Do you mean ||c'(t)|| is the magnitude of ∇c(t)?

The Attempt at a Solution

So, with this equation y=x^3 ... I got the that c(t)= <t,t^3>
c'(t)=<1,3t^2>

So ||c'(t)|| = √(1+9t⁴), yes?

Another thing is I cannot for the life of me figure out how to take the anti-derivative of √(1+9xy)... which by the time I change to t it would be √(1+9t^4)...

Yes, but after you multiply that by ||c'(t)|| it will look a lot nicer.

 

[Unart]

Yes, but after you multiply that by ||c'(t)|| it will look a lot nicer.

No joke... thanks to you. It cleared up a whole lot of dark and dreary confusion. I was plugging the number into the Magnitude equation and getting just a number for the magnitude. I didn't know that you just took the derivative as and used the magnitude equation from that.

THANKYOU!
I posted another question and the past and began to worry if my wording or something was off.