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

$$\begin{array}{l}{\int_{C}\left(x^{2}+y^{2}+z^{2}\right) d s} \\ {C : x=t, y=\cos 2 t, z=\sin 2 t, \quad 0 \leqslant t \leqslant 2 \pi}\end{array}$$

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

Vector Calculus

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All right, So this question were acid Determine the mining to grow off X squared plus y squared. Plus they squared along the curb Access, equality Why is equal to co sign Titi and see is he was assigned to duty and we're giving the tea ranges from 0 to 2 pi. So the first thing we're gonna try to do is we're gonna try toe, determine the answer. This would lose share with the integral. Sorry, we're gonna write everything in the integral in terms of tea that were able to take me. So our only issue. So So extend your industy. Why can be written in terms of ts clubs on duty and he could be written in terms of ts scientist t Our only issue it with biz with DS or the click. But what we know the formula for the park length is DS is equal to the square root of d y by D T square was easy by DT Square plus plus the dx dy DT square plus D c biking square. All of them multiplied by. All right, So do you exploit eating? Use just the derivative off T but respected team which is just one square and d y by D. T is just the dirigible coastline to t I would respect a T which is just negative to sign t again. We're going to use u substitution. Use your substitution This and then the derivative off Easy by DT is just a derivative of scientific tea With respect again, you use your substitution. You get too close I to all right now you're going to square. These one squared is just one negative to sign to t swear it is Four signs square to a team and to co sign to t squared is four co sign square to And then now what we can pull out is we can float a floor is a common factor and we're left with science square to t plus coastline scored two teeth and then we know that this is simply one. So we have the square root off one plus full which is just route by Bt. So now finally we determined or ds in terms of teas. This Route five DT. All right, so now we plug everything back into our integral and then we get everything. So now So now what we notice is that we have co sign square to t plus sine square to do well, this is simply one. So we're left with t squared plus one and we can take the square root of five because it's a constant. And then pull it on the outside. So finally are integral. Look, something like this. So, Brooke, five times integral from zero to buy 50 square plus one detail. All right, now, the integral of two squared is too cute, divided by three and the integral of honest tea. And then you're gonna take the definite So this is a definite integral. So you're going from 0 to 2 pi. So we're gonna plug in to pie, and then we're gonna plug in zero, and then we're gonna take the difference of Well, the zeros are just just give us a zero and we have a square root of five on the outside. Two pi killed this eight by cubed, divided by three plus two pi. So finally, our line integral is very little five kind a pie que divided by three plus two

Vector Calculus