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Problem 53

Evaluate the following line integrals using a method of your choice.

$\oint_{C} \mathbf{F} \cdot d \mathbf{r},$ where $\mathbf{F}=\left\langle 2 x y+z^{2}, x^{2}, 2 x z\right\rangle$ and $C$ is the circle

$\mathbf{r}(t)=\langle 3 \cos t, 4 \cos t, 5 \sin t\rangle,$ for $0 \leq t \leq 2 \pi$

Answer

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## Discussion

## Video Transcript

Okay, So in this video were asked to determine to grow and were given the field and, uh, the half the curve along with the curve that we're gonna determine the lining to. So our field is given by two x y plus eastward come X squared. Coming to xsi on our path is per capita rised by three co sign t come before close identical five scientific where t varies from 0 to 2 pi. Now the first thing we're gonna do is try to determine if this field is so conservative. Because if it's conservative, it helps us a lot. It makes the problem easier. So, uh, so did the little left by the lies to access digital G by D. X is two x So the first condition is met. These two partial derivatives are equal. And if we continue determining the partial derivatives would realize that the F bite easy and the age by D X are also equal. A DJ by Daisy engage by the wire article. So there this means that big gap is a conservative. All right, So since big F is a conservative field, that means that our field is the Grady in of a potential function. So let's try to determine this potential. But ingredient just basically means the partial derivatives with respect to explain. See. So now let's think about this logically. So we want a partial derivative that said, Let's start with the easy. So what? What's an expression? My cake? It's derivative. With respect. Uzi will give me to ecstasy. I think about this For a minute or two. We realized that Xz Square we take its partial derivative with respect to Z. Well, give us to Ecstasy. All right. Excellent. No. What about Let's do the one after this e X squared. What's something if I take its derivative with respect to why? Well, give me X squared. Well, that's pretty simple. That's just X squared. What? All right, now let's see. What is something? If I tickets partial derivative, it will give me two x y plus z squared again. If we think about this for a minute or two will realize the expired y plus x z squared. If I take the derivative of this with respective Axel, give me two x y plus x city square. Now we noticed something very interesting. This is equal to this. So we don't really have to write this twice. So what do I mean by that? I mean, our potential functions just x squared Y plus x c square. Now, why is this if you take the derivative of picks a X squared y plus x city square? With respect to Pax, you're just going to get to Rex. Why? Let's see square again. If you take the derivative of this potential function X, we're going prospects. These squared with respect to X. Whether sorry with respect to lying again, you'll only get X squared. Same thing for taking the deer derivative of this potential function with respect to Z again will only give you to exit so you don't need to write this twice. So this is a potential. And we didn't We could have done it, you know, like when taking into girls and store. But it's a lot easier to do it. Your answer cause these air very there these air Pretty easy expressions. So now what do we know about conservative vector fields? We know that their the greedy enough potential functions, but not just that. The line into growth is independent of the path taken. So what we have to do is determine the lining to grow at the 21 point I Sorry all that. We have to just determine the value off the potential function at the 21.7 subtracted from each other. So So, uh and but there's an extra think So are our curve is pure amateur ized, So our end point is too high. But remember, or curve is per average rise. So we need to bring floor of two pi. Same thing for zero. So are other end point is zero. But again, our curves per have address so we need to find are zero. So we're gonna plug in to fly into three Ko Sai Inti Ah calma four co scientific All five Cy Inti. If we wanted to apply for in that, we're going to get three com before zero. And if we plug in zero for T, we're going to get three co side of zero calma for co sign of zero comma five sign of zero again, we get 340 So what we get is zero. So what we get is zero. Okay, so this is the value of our line into

## Recommended Questions

Evaluate the following line integrals using a method of your choice.

$\int_{C} \nabla\left(1+x^{2} y z\right) \cdot d \mathbf{r},$ where $C$ is the helix $\mathbf{r}(t)=\langle\cos 2 t, \sin 2 t, t\rangle,$ for $0 \leq t \leq 4 \pi$

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$\mathbf{F}=\left\langle y^{2},-z^{2}, x\right\rangle ; C$ is the circle $\mathbf{r}(t)=\langle 3 \cos t, 4 \cos t, 5 \sin t\rangle$ for $0 \leq t \leq 2 \pi$.

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$$\begin{aligned}

&\mathbf{F}=\left\langle y^{2},-z^{2}, x\right\rangle ; C \text { is the circle }\\

&\mathbf{r}(t)=\langle 3 \cos t, 4 \cos t, 5 \sin t\rangle, \text { for } 0 \leq t \leq 2 \pi

\end{aligned}$$

Evaluate the following line integrals.

$\int_{C} x y d s ; C$ is the unit circle $\mathbf{r}(s)=\langle\cos s, \sin s\rangle,$ for $0 \leq s \leq 2 \pi$

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$$\int_{C}(x+y+z) d s ; C \text { is the circle } \mathbf{r}(t)=\langle 2 \cos t, 0,2 \sin t\rangle$$ for $0 \leq t \leq 2 \pi$

Scalar line integrals in $\mathbb{R}^{3}$ Convert the line integral to an ordinary integral with respect to the parameter and evaluate it.

$\int_{C}(x-y+2 z) d s ; C$ is the circle $\mathbf{r}(t)=\langle 1,3 \cos t, 3 \sin t\rangle$ for $0 \leq t \leq 2 \pi$

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$\int_{C}(x+y+z) d s ; C$ is the semicircle $\mathbf{r}(t)=\langle 2 \cos t, 0,2 \sin t\rangle$ for $0 \leq t \leq \pi$

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Evaluate the following integrals using the method of your choice. A sketch is helpful.

$$\iint_{R} \frac{d A}{4+\sqrt{x^{2}+y^{2}}} ; R=\left\{(r, \theta): 0 \leq r \leq 2, \frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\right\}$$