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# Evaluate the line integral, where $C$ is the given curve.$$\begin{array}{l}{\int_{C} x y z^{2} d s} \\ {C \text { is the line segment from }(-1,5,0) \text { to }(1,6,4)}\end{array}$$

## $\frac{236 \sqrt{21}}{15}$

Vector Calculus

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

All right. So this question we were asked to determine the line into group off Becks times like times e to the power five. Yes. And we're told that we're taking the injured broke along the line segment that starts at negative 150 It goes all the way to 164 All right, so the first thing we need to do is to find vector that's in the direction are parallel to this line segment. So what we do is, um, extra minus X one. Call my wife to minus Z two. So if we do that, we're gonna get the vector 214 All right, and then out to determine expects is just x one plus 80. Next one is negative. One is too. So X is just negative. One puts to t. Why is why one plus bt? So why Oneness five and B is one. So it's five plus t. That's are why and r c is t one plus c t. Where z 10 c is four. So basically our exes Negative one plus two dear lies five plus three or is he is here. Very great. So now we have our X Y and Z in terms off t and then we need to determine DS and find es in terms where we know the formula for Diaz, I guess, is just the square root of ex derivative of extra respected to square is a derivative of why, with respect to square was derivative of Zealanders, square times did too. So the expired duty is just too delighted by DT is just one and easy by t t is just for it's just regular simple derivatives. So we get that DS is the square root of 21. This is our and then now we plug everything back into our original integral. The great thing right now is that we have everything. So if we foil the first twos of before you'll bees too Bergen again. Negative five minus two, You plus 10 t plus two square And then we're gonna pull that square root of 21 to the outside and 1/2 times 16 t square. Right? So what we can do is negative. T plus 10 t is 19. So we have 90 and then we just Mr So we have screwed up 21 on the integral from 0 to 1 of negative 80 square, 144 t cube most 32 due to the power for DT. And now this is simple because just powers taking the interval. It's pretty simple. We just add one to the exponents and divide by it. And then again, our upper are lower limit is nearer or often limit is one so in May. So we're gonna plug in 1 40 we get through. Storms were here, then we plug in zero. We just get a whole bunch of zeros now. Negative 80 divided by three was 144 to budget for plus 32 divided by five. If we create a common denominator, we're going to get 236 divided by 15. And then we have screwed up 21 on the outside. So our answer is 236 square root of 21 divided precious where you can convert it into a small or you could leave

#### Topics

Vector Calculus

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp