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Problem 135 Medium Difficulty

[T] Find line integral $\oint_{C} \mathbf{F} \cdot d r$ of vector field $\mathbf{F}(x, y, z)=3 x^{2} z \mathbf{i}+z^{2} \mathbf{j}+\left(x^{3}+2 y z\right) \mathbf{k}$ along curve $C$ parameterized $\mathrm{by}$ $r(t)=\left(\frac{\ln t}{\ln 2}\right) \mathbf{i}+t^{3 / 2} \mathbf{j}+t \cos (\pi t), 1 \leq t \leq 4$




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Video Transcript

Okay, This question asks us to find the line integral for this function. And since this is probably going to be a conservative field, we should probably find the start and end points of our curve. So we'll start our curve At T equals one. So we get 01 negative one for our starting value. And then for our final value, we get Ellen four over Ellen too. Then a four to the three halves and then four co sign of four pi in this simplifies because Ellen four is to Ellen to. So this just becomes too eight four. Those four to the three halves is just to Cuba and co sign for pies one. So now that we have our values, let's check that are that our field is indeed conservative. So the X derivative of the why is the wide derivative of the X z derivative of the why is the y derivative of the sea and Z derivative of the X is the axe derivative of Z. So checking each of these the ex derivative of the why is zero the why derivative of the X zero. So that works then the Z derivative of the why is to Z and the wide a relative of the Z is to Z So that works. And then the Z derivative of the X is three x squared and the extra bit of of the Z is three x squared. So since these conditions are met, we know that f is a Grady int field. So we confined our potential just by integrating each component with their respective variable. So first we integrate acts with respect to acts to get X cubed, dizzy plus some function of accident urge Z and why rather then if we integrate the y component with respect to why we get y z squared plus a function of X and Z and then if we integrate rz component with respect to Z we're going to get and x cubed z plus a y z squared plus function of x and y So we see that we have an ex cubes e and a y Z squared. So that's our potential up to a constant. So again we get and X cubed, dizzy And then we got a y Z squared. So the line in a grove a Grady Int field is just the potential at the end, minus the potential at the start. And in our case, we already found these values. So the end, we said was 284 And then the start, he said, was 01 minus one. So using our potential, here we get that fi of 284 would be two cubed times for plus. Then y Z squared. So eight times four squared than minus 501 negative one. So we get zero and then minus or sorry plus y Z squared. So one. So this gives 32 plus and then 32 times for ages 1 28 then minus one, giving us 1 59 as our line integral value. And that's our answer.

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