@ralph

Evaluate the line integral where C is the given curve. We’re integrating over the curve C, y to the third ds, and C is the curve with parametric equations x = t cubed, y = t. We’re going from t = 0 to t = 2. So we’re going to integrate over that curve C of y to the third ds. We’re going to convert everything into our parameter t in terms of our parameter t. So I’m going to be integrating from t = 0 to t = 2. Those will be my limits of integration. Now y is equal to t, so I’m going to replace y with what it’s equal to in terms of t. So I’m going to be integrating the function t to the third. Now ds we’re going to write as a square root of dx dt squared + dy dt squared, squared of all that as we said dt. So we’re integrating now everything with respect to t. So this is going to be equal to the integral from 0 to 2 of t to the third times the square root of — see the derivative of x with respect to t is 3 t squared. So we have 3 t squared squared + dy dt; well, that’s just 1 squared dt. So we have the integral from 0 to 2 of t to the third times the square root of 9 t to the fourth + 1 dt. So this is a pretty straightforward integration here. We’re going to let u be equal to 9 t to the fourth + 1 then du is equal to 36 t to the third dt and so that tells me I can replace a t to the third dt with a du over 36. And so we’re going to have the integral then from — well, new limits of integration. I’m just going to put some squiggly marks there to remind myself that we switched variables. So I’m not going from t = 0 to t = 2. I’m doing things in terms of you right now. But I have a 1 over 36. I’ll put that out front, and we’re going to have the square root of u. So u to the 1/2, t to the third dt was replaced by du over 36. We got the 36 out front. And so now this is a pretty easy antiderivative in terms of u. It’s u to the 3/2 times 2/3. And again, different limits of integration. We could figure out what they are in terms of u, but I’m going to convert back into t. So we’re going to have 1 over 36 times 2/3 times u to the 3/2. Now, u is 9 t to the fourth + 1, that to the 3/2 power. And now we can go ahead and go from original limits of integration 0 to 2. So let’s see, when I put a 2 in here, we’re going to have — 1 over 36 times 2/3. That’s going to be 1 over 54, isn’t it? So we’ll have 1 over 54 times — putting a 2 in, we have 9 times 2 to the fourth. That’s 9 times 16, which is 144 + 1, is 145. So we put the 2 in there, we get 145 to the 3/2 minus, putting the 0 in, we get 9 x 0 to the fourth. That’s 0. 0 + 1 is 1. So we just get 1 to the 3/2 or 1. So let’s see, what’s the best way to write this. How about 1 over 54 — I guess we could leave it like that. We could also write 145 to the 3/2 as 145 times the square root of 145 and then minus 1. And that is that line integral of y to the third ds over the given curve C.