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

$$\begin{array}{l}{\int_{C} x y e^{y z} d y} \\ {C : x=t, y=t^{2}, z=t^{3}, 0 \leqslant t \leqslant 1}\end{array}$$

$\frac{2}{5}(e-1)$

Vector Calculus

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Missouri State University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

All right. So this question we're asked for determined the line integral of back stands. Why? Times either the power of y z along the curve giving Perry metric Lee by access, equality wisely, quantity squared been C is equal to t cube Anti goes from Syria. Well, the first thing we want to do is we're gonna We want to write everything in the integral in terms of teeth while exes, Exes in terms of TV wise in terms of But our problem is that de y is not in terms of so what we can do is we can take the derivative of eyes equal to t swear. So you wise just gonna be to tee times. DT. All right, so now we're gonna plug everything back into our starting a plug in for X t. Where weren t squared for TV for a CD, cured a t to the power of three. And for d y, we're gonna plug in to Tgt. So my plug everything into the integral we get. We can pull this to to the outside, Take it to the outside. We get two times the integral from 01 cheated the power for times e to the power of to the power five. All right, now what we can do is we have to solve this. Integral. This looks like a product, but we're not gonna use integration by parts. What we can do is we notice that there's a teacher about five on the NASA inside the exponents. So we're gonna get you equal tedx in a power. All right? Now, when we take a derivative both sides, right, I d year is equal to five. T to the power 40 t and then we can divide by five on both sides, and we finally get that t to the power or GT is equal to the U divided by five. So this and this is do you divided by five and then we have to. Then then then we have e to the power of you. So basically, we could pull this 1/5 of the outside, so you have to hit integral you and then we can change the limits off our integration so that they can be in terms of you. So we just pull against a T is equal to zero. If we pluck that and we get zero to the power five, which is just zero. And again T is equal to one of the one change that you just becomes you is equal to one to the power of five or just one. So now we change their limits of integration. And we know that the integral U to the power of you use just beat of the power of you. So we have to fifth and now we they eat it the power of one minus eight to the power of zero. But we know that either the power of zero is so eight on the power of zero is one. So our final answer the line integral is equal to two e minus.