Evaluate the surface integral ∬s 𝐅 ·d 𝐒 for the given vector field 𝐅 and the oriented surface S

step 1 So we are asked to find the flux of the given surface. And what we can notice from the problem i

Evaluate the surface integral ∬s 𝐅 ·d 𝐒 for the given vector...

Key Concepts

Surface Orientation

Surface orientation involves the choice of a normal vector direction at each point on a surface, which is essential in defining the flux integral. For closed surfaces, the outward normal is typically used to ensure consistency in applying the divergence theorem, representing the outward flow of the field.

Divergence of a Vector Field

The divergence is a scalar quantity that represents the rate at which the vector field spreads out from a point. Calculating the divergence is a critical step when applying the divergence theorem, as it converts a surface integral into a volume integral over the divergence of the field.

Surface Integral (Flux Integral)

This concept refers to the integration of a vector field over a surface to measure the total flow (or flux) passing through that surface. It involves taking the dot product of the vector field with the surface's normal vector at each point and integrating over the entire surface.

Divergence Theorem

The divergence theorem is a key result that relates the flux of a vector field through a closed surface to the volume integral of the divergence of the field over the region enclosed by the surface. It is used to simplify the computation of surface integrals, especially when the direct evaluation over complex surfaces is challenging.

Volume Integration (Triple Integral)

Volume integration, often performed as a triple integral, is used to compute quantities over a three-dimensional region. In the context of the divergence theorem, the triple integral of the divergence over the volume enclosed by the surface yields the total flux across the surface.

Transcript

00:01 So we are asked to find the flux of the given surface.

00:05 And what we can notice from the problem is that we're going to have to break the surface into four parts as it's a half cylinder.

00:14 So we're going to need s1, and s1 is going to be equal to the lower disk, aka this will be s1 is when x is equal to 0.

00:26 We'll have s2, which is at x is equal to 2, and then we will have s3, which is going to be the rectangular plane when z is equal to 0, and s4, which is going to be the curved part of the cylinder, when this curved part of cylinder, when y squared plus z squared is equal to 1, and that's going to be in between the three planes, x equals 0, x equals 2.

00:59 And z equals zero.

01:03 And so to find the flux, we're going to have to add these four things together and then add them together separately.

01:10 So let's start with s1.

01:12 We'll do s1 in red.

01:14 I'll do a different color for all four of the surfaces.

01:18 So for s1, we know that ds is equal to nds.

01:26 And we know that that since the surface is pointing downwards, that the normal vector is negative i.

01:36 So ds equals negative i ds.

01:42 And so now, if we go to plug this in, we know that the surface of the flux, of flux, sorry, that's not where the f goes.

01:53 We know that s1, the flux ds, is equal to the integral of s1 times x squared, i plus y squared j plus z squared k multiplied by negative i d s and since we're looking at the x portion of this this is going to be equal to the integral of negative x squared ds just again because we only care about the i component and we know that on the entirety of the surface for s one we said that s one is at x equals zero so this tells us that this is going to be equal to zero squared, which is equal to zero.

02:44 So s1 equals zero.

02:50 And now we're going to move on to s2.

02:54 S2, we're going to have the same thing where ds is equal to nds.

02:59 However, s2 points upwards because x is positive 2.

03:03 So that tells us that ds is equal to ids.

03:09 And we're going to have the same thing as the previous problem.

03:11 I'll rewrite it out anyways, though.

03:13 Where the integral of s2 of the flux ds is equal to the integral of s2 of x squared i plus y squared j plus z squared k multiplied by ids.

03:32 And since we're only looking at the i direction, well, this is going to be equal to the surface integral of s2 of x squared ds.

03:42 And we know that on all points to the surface, that x is equal to 2.

03:48 So we can rewrite this as this is equal to 4.

03:53 I'm just going to pull the 4 out from 2 squared of s2 of ds.

04:01 And this is going to be equal to 4 times the area, because again, when you integrate an object, you get area.

04:09 And we're talking about the area of a cylinder.

04:11 So that is going to be pi r squared over two.

04:16 So four times pi, the radius would be one because the diameter is two.

04:21 So pi 1 over 2, and that's equal to 2 pi.

04:27 So we have s1 is equal to 0, and s2 is equal to 2 pi.

04:35 And now we're going to move on to s3.

04:38 So we have the same thing, where s3 is equal to s3, ds is equal to n ds.

04:46 And in this case, we're talking about when z is at zero.

04:51 In other words, this is going to be pointing in the negative direction.

04:55 And so we write that as negative k.

04:57 So ds is equal to negative k ds.

05:01 So now when we go to do our surface integral, we'll have the surface, the third surface of the flux is equal to the surface integral of x squared i, plus y squared j plus z squared k multiplied by negative k d s so yet again we're going to be looking at just the z coordinate and specifically negative z squared i'm going to move this up here so this is going to be the integral of the surface of negative z squared because again that negative k is going to go to here so negative z squared d s well the z for the entire surface is zero.

05:53 So we have the same thing as with s1, where this is equal to the surface integral of zero squared, which is going to be zero.

06:05 So a lot of the times of these problems, setting up your balance correctly can make the problems very easy.

06:09 In this case, we're able to make two of the surfaces zero, which is very nice for us.

06:14 So we have s1 is equal to zero, s2 is equal to 2 pi, and s3 is equal to 0.

06:23 So now we have to find s4.

06:25 We can add the 4 together, and that will be the answer for a flux for the surface.

06:32 Now, we are not able to fully define s4 as one specific plane, which is meaning we're going to have to do a little more math for this part.

06:43 So we know that the surface of the flux is equal to f times and ds.

06:55 And in this case, ds is going to be equal to the magnitude of the cross product multiplied by da, as seen in previous problems.

07:06 And we can rewrite this as ds is then equal to ru crossed with rv da.

07:15 Once we incorporate the n, pardon, so nds, da.

07:23 So we have to find the cross products.

07:25 Of the two partial derivatives with respect to you and v.

07:29 So since s4 is the cylinder y squared plus z squared equals 1, we can write the parametric representation of this as 1x, y squared, and z squared, well, y squared and z squared equaling a radius we can write as v sine u and cosine u.

07:57 Now from here we have to think about our bounds...

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