# Help evaluating a line integral!

1. Apr 9, 2010

### JHans

So, my multivariable class has just started line integrals, and I could use a little help with them. The problem I'm currently working on says:

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

$$\int\limits_C \! xy \,ds$$
$$C: x=t^2, y=2t, 0 \le t \le 1$$

I realize that, by eliminating the parameter (note really necessary, but just for the sake of understanding), it is the curve $$x=(1/4)y^2, 0 \le y \le 2$$.

I've managed to get this down to 4 times the integral, from 0 to 1, of (t^3) sqrt(t^2 + 1), but I have no idea where to go from here. Is there something I've majorly screwed up on?

2. Apr 9, 2010

### rs1n

Try substitution with $u=t^2+1$. Consider

$$t^3 \sqrt{t^2+1}\ dt = t^2 (t^2+1)^{1/2} \cdot t\ dt$$

3. Apr 9, 2010

### JHans

There must be something I'm blanking on from my days in calculus one. I feel like the deeper I go into calculus, the more prone I am to making beginner's errors.

Why can I substitute with u = (t^2) +1? If I separate (t^3) into t(t^2), what does the (t^2) term become in terms of u?

4. Apr 9, 2010

### rs1n

=) I think you might be overthinking the problem. If $u=t^2+1$, can you use this equation to find $t^2$ in terms of u?

5. Apr 9, 2010

### JHans

... Wow. That was very obvious, haha. I'm so used to setting u as a convenient expression in terms of t that I didn't even consider subtracting 1.

Is this new integral in terms of u correct?

$$\frac{1}{2}\int_1^2 \! (u-1)\sqrt{u} \,du$$

In which case I would just multiply the function of interest through and antidifferentiate?

6. Apr 9, 2010

### rs1n

Looks good to me!

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