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

$ \displaystyle \int_C (x^2 + y^2 + z^2) \, ds $,

$ C: x = t $, $ y = \cos 2t $, $ z = \sin 2t $, $ 0 \leqslant t \leqslant 2\pi $

$\sqrt{5}\left(\frac{8 \pi^{3}}{3}+2 \pi\right)$

Vector Calculus

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Johns Hopkins University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

So if we want to evaluate this line integral here, one thing that'll be useful for us to remember is that this DS here is actually equal to the square root of ex prime squared. Plus why Prime Squared plus z prime squared. Uh, DT. And if we do this, then we could just make the substitution for X, y and Z as well as, um plugging in N ds for that. And then we could just evaluate from 0 to 2 pi. So let's go out and find out what our partials we're going to be. So let me actually move the silver. So if we take the derivative of this prospective t that just gets his ex prime is one, um, why prime is going to be? It looks like negative, too. Sign to t. And then Z is going to be, er ze prime. It is going to be to co sign of two teeth. And so now if we take these and plug them in down here, that's going to be one plus eso would be four. Sign to t for co sign to t so for signs Square two teeth plus four co sign squared two teeth DT. And now if we factor that for all we have sine squared plus coastline squared, which is just one. So overall, this just turns into Route five DT because we have one plus four. Okay, so now we go ahead and actually let me skip this over and I'll move this down here. Yeah, so now our bounds are gonna be zero The two pi x was t So this is gonna be t squared. Why was supposed to be chosen to tea, so I'll be cosine squared to t, um, sign square to t breezy squared. And then all of this was supposed to be multiplied by five DT. So we have DT. And I'll just move the route five up from here. And so again, co sign square plus sine squared is just one. So this will think there's one. So we have t's word plus one so they integrate both of those. We just use power rule. So be route five times one third take. Plus, he evaluated from 0 to 2 pi. Yeah, so we go ahead and plug into pie. So I give us Route five and that would be eight pi cubed over three plus two pi. Um and then we would do minus when we plug desirable. That just gives us zero. And so, actually, let me just erase this. And then I would just say the final answer like this. I mean, you could go ahead and get a common denominator on all that, but I think this looks fine enough.

University of North Texas

Vector Calculus