10. Define the two curves C1 and C2 by C1: r⃗(t) = (cos(t), 1, sin(t)), 0 ≤ t ≤ 2π, C2: r⃗(t) =

Step 1: We first need to parameterize the curves C1 and C2. For C1: r(t) = (cos(t), 1, sin(t))

10. Define the two curves C1 and C2 by C1: r⃗(t) = (cos(t),...

10. Define the two curves $C_1$ and $C_2$ by $C_1: \vec{r}(t) = (\cos(t), 1, \sin(t))$, $0 \le t \le 2\pi$, $C_2: \vec{r}(t) = (\sin(t), \cos(t), t + 1)$, $0 \le t \le 2\pi$, and define the two vector fields $\vec{F}$ and $\vec{G}$ by $\vec{F}(x, y, z) = (0, z, 0)$, $\vec{G}(x, y, z) = (e^x, 3y^2, \frac{1}{z^2 + 1})$ Evaluate the following. (Which ones can you use shortcuts for?) (a) $\oint_{C_1} \vec{F} \cdot d\vec{r}$ (b) $\oint_{C_1} \vec{G} \cdot d\vec{r}$ (c) $\oint_{C_2} \vec{F} \cdot d\vec{r}$ (d) $\oint_{C_2} \vec{G} \cdot d\vec{r}$

Transcript

00:01 In this problem, we are given a two -dimensional vector field for which we want to evaluate its line integral over curve c, where curve c is equal to the combined paths of c1 and c2.

00:14 Let's start by drawing our paths.

00:24 So c1 corresponds to a line equation where x is equal to 0 and y is equal to t, for t ranging from minus 1 to 1.

00:38 So c1 will start at minus 1 and end at 1 along the x -axis.

00:52 Next, we notice that c2 has the equation of a circle, but only a semi -circle because the angle is constrained between pi over 2 and 3 pi over 2, and the circle has a radius of 1.

01:07 So the circle starting at pi over 2 will connect to the start and end of c1, like so.

01:19 So this is the path for our integral.

01:27 So we want to evaluate f dot dr over this entire path, which will equal to f dot dr over c1 plus f dot dr over c2.

01:50 Now, how do we evaluate f dot dr? this can be written as the integral of f dot r prime of t dt, and in this case, t for c1 ranges from minus 1 to 1.

02:18 And we'll do the same thing for our second part of our line integral.

02:24 This will be equal to the integral of f dot r prime t dt, where in this case, t ranges between pi over 2 to 3 pi over 2.

02:43 So now, let's calculate r prime of t for both of our paths.

02:51 R1 of t is equal to the position vector 0t, such that ry prime of t is equal to 0, 1.

03:08 R2 of t for our second path is equal to the position vector cos of t, sin of t, such that r prime 2 of t will give us minus sin t and cos t.

03:28 So now, we're going to tackle the dot product of these two derivatives with our vector field f.

03:39 So for our first path integral, f dot r prime will give us only the y component of f, which is 7y.

03:50 For our second path, f dot dr will give us minus 5x plus 6y times sin of t plus 7y cos t...

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