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Problem 15 Easy Difficulty

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

Answer

35$/ 3$

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Video Transcript

so in this question as to determine the line integral off Z squared DX plus X squared y plus y squared DZ. And we're told that C is the line segment from 100 to 412 All right, so the first thing we have to do is we have to determine the line that's parallel to the segment connecting 100 412 So, basically, what we do is extrapolated sex one comma y tu minus one comma z2. So four minus one is three on minus 01 and two minus zero. This too. So this is the value of a be sick. All right, so now what we have to do is we have to determine ex lions Ys o X is just x one plus 80 and our X one is one. So it's one plus three to why, that is why one plus Bt off. My one is zero r use one. So why is just tea and C is C one plus c t Let's see t so which is zero plus two teeth or simply to? And then since we want to write everything in the integral in terms of tea. We're gonna take the derivative of X Y and see with respect to teach. So DX is just three d t d. Y is just GT because the derivative off tees just one. So it's one DT and easy the derivative of C, which is to tea with respected t is just to be all right. Now we plug everything back into the integral and when we get this right over here and we noticed that everything is in terms of So if we pull that DT on the outside, we get 12 t squared. And then if we multiply, expand one plus three t square to get one plus 60 plus 90 square, and we have. So now we can combine like terms. So we have 12 t swear plus 90 square was to t swear that's 23 teach squared, and then we have 60 plus one. So this is what we're gonna have to take the all right now. 23 t square predicting integral of that becomes 23 T cube divided by three. The integral of 60 is three squared on the integral off. Ron DT is just and they were gonna pluck in one. So our limits are integration. The lower limit of zero. The upper limit is one. So you plug in 1 40 we get 23 divided by three, plus three plus one and we plug in zero. We get zero plus. You're what? Zero. So this whole thing is zero. Now we create a common denominator. So we multiply three by three, divided by three and we multiply nine. Uh, sorry. We multiply one by three, divided by three. We get this break. So 23 plus nine plus three is 35. So we get 35 divided by three. So that's the value of the hard line.