Let F = (langle z - y, x, -x angle). Evaluate each of the...

Let F = (langle z - y, x, -x angle). Evaluate each of the following, and notice that there may be more than one choice of parametrization available to you, so you may want to try doing some of these with more than one method. (Directly evaluate means compute the given integral as is; do not apply any theorems to change the integral). (a) Let C be the unit circle in the xy-plane, oriented CCW. Directly evaluate (oint_C mathbf{F} cdot dmathbf{r}). (b) Let $S_1$ be the upper unit hemisphere, with upward facing normal. Directly evaluate (iint_{S_1} ( abla imes mathbf{F}) cdot mathbf{n} , dS). (c) Let $S_2$ be the paraboloid $z = x^2 + y^2 - 1$, with upward facing normal, for $-1 le z le 0$. Directly evaluate (iint_{S_2} ( abla imes mathbf{F}) cdot mathbf{n} , dS). (d) Let $S_3$ be the unit disk in the xy-plane, with upward facing normal. Directly evaluate (iint_{S_3} ( abla imes mathbf{F}) cdot mathbf{n} , dS). (e) Sketch a picture of C, $S_1$, $S_2$, $S_3$ (all on the same picture). How do they relate? (f) What relationship did you get between your answers above? Explain the relationship between your answers above using Stokes' Theorem.

Transcript

00:01 Hello, in the question we have given that this function is given so this function we can write it in this way f is equal to z minus y i cap plus x j cap minus x k cap.

00:14 So now we have to find out the following quantities.

00:16 So we have to calculate the integral over f dot dr.

00:20 So we have to so they are saying that let c be the unit circle in the xy plane oriented counterclockwise.

00:29 So this will be the equation.

00:32 So f dot dr.

00:33 So this is my f and this is the dr because it is only in the xy plane.

00:37 So this will be dr.

00:39 So now here we will use the parameter parameterization.

00:43 So i will be going into the polar coordinates and solve this.

00:46 So in polar coordinates i know x is equals to r cos theta.

00:50 So dx will be minus r sin theta d theta and y is we know it is r sin theta.

00:57 So dy will be equal to so differentiation of sin theta is cos theta.

01:01 So it will be r cos theta d theta.

01:04 So now if we plug the values over here, so z i dot i is 1.

01:09 So this will be z minus 1 z is 0 because it is in the xy plane.

01:14 So minus y so minus this is the value of y which we have plugged over here and dx.

01:19 So this is the value of dx which we have plugged over here and then we have j dot j j dot j is 1.

01:27 So this is the value of x and this is the value of dx which we have plugged over here.

01:31 Now if i simplify this so i know that r into r is r square and you are also r square.

01:37 So it is we can take it common and then over here inside the bracket it is sin square and cos square and this d theta i can take common.

01:47 So it's become it becomes sin square theta plus cos square theta and it is 1.

01:51 So the integration limits will be from 0 to 2 pi because it is a circle.

01:57 So now if i integrate this and solve i get f bar dot dr bar is equals to 2 pi.

02:04 So this is my equation 1.

02:06 Now in the second part they are saying that let s1 be the upper unit of hemisphere with upward facing normal.

02:15 So upward facing normal means this normal if i draw it should be upward.

02:20 So now we have to find out this integral del cross f n cap ds.

02:29 So here del cross f first we will find out.

02:32 So del cross f can be found out by using solving this determinant.

02:37 So this i daba by daba x and the x component of this function.

02:44 So that is z minus y and j daba by daba y and the j component of this function and similarly this k.

02:57 Then if we solve this i get del cross f is equals to if i solve this determinant i get this del cross f as this.

03:04 Now what we are doing over here is so they have just given this hemisphere.

03:09 So this bottom part we are considering a circle.

03:12 So now whatever the flux passing through this circle should be come out from this one.

03:17 So if i consider this surface this surface to be s1 and this surface to be s2 then the net flux passing through this should be equal to 0.

03:29 So that is what we have done over here.

03:31 So del cross f dot n cap ds will be equal to the surface integral over 1 and the surface integral over s and that should be equal to 0.

03:41 So if i rearrange i get this equation...

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