Evaluate the surface integral. ∫S x z d S S is the boundary of the region enclosed by

Evaluate the surface integral. ∫S x z d S S is the...

Key Concepts

Surface Integrals

Surface integrals extend the idea of integration to functions defined on two?dimensional surfaces in space. They involve integrating a function multiplied by the differential element of area (dS) over a given surface, and are used to compute quantities like mass, flux, or other accumulative measures over curved surfaces.

Parameterization of Surfaces

Parameterization involves expressing a surface in terms of two parameters, which simplifies the computation of surface integrals. By mapping the surface to a parameter domain, one can derive expressions for the differential area element dS, making the integration tractable even on complex geometries.

Cylindrical Coordinates

Cylindrical coordinates are particularly useful when dealing with regions or surfaces that exhibit circular symmetry, such as those involving cylinders. Transforming to cylindrical coordinates often simplifies both the description of the domain and the computation of integrals over it.

Symmetry in Integration

Symmetry is a powerful tool in evaluating integrals. When a surface or the integrand possesses symmetric properties (for example, the integrand might be odd in one of the variables over a symmetric region), one can often argue that certain contributions cancel, leading to simplifications or even a zero value for the integral.

Decomposition of Closed Surfaces

When the surface of integration is closed and composed of several distinct parts, it is often beneficial to decompose it into simpler pieces. This allows each piece to be parameterized and integrated individually using the most suitable coordinate system or method before summing the contributions to obtain the overall integral.

Transcript

00:01 So for this problem, we're given three surfaces, s1 being equal to x squared plus y squared, which is equal to 9.

00:13 S2 being that x is equal to 0, and s3 being that x plus 5, x plus y is equal to 5, or that x is equal to 5 minus y.

00:29 So now we have to set up a parameterization for this, as we need something in terms of x that we can work.

00:35 With and so if we swap the integral into polar it would be easier to work with the x.

00:44 So in this case from the x squared plus or sorry y squared plus z squared plus z this is a z so in the case of z squared plus y squared is equal to nine this would give a parametization of x is equal to x of that y is equal to three cosine theta and that z is equal to three sine theta.

01:11 And then because we're in polar, our bounds will be from zero to theta to two pi.

01:19 And we're given the other one from our third surface, which is that zero is less than equal to x, which then is equal to five minus y.

01:29 And we can rewrite five minus y as being five minus three cosine y because we know that our y parametization is y equals three cosine y.

01:44 5 minus three cosine.

01:50 So now when we want to cross r theta with r of x, because they're just the two variables that we have right now.

02:01 And this will give 1 -0.

02:06 As for x, we only have x.

02:08 And for theta, this will give 0 -negative 3 -sine theta, and 3.

02:14 Cosine theta for i j and k now when we solve this cross product we will get out negative three cosine theta of j minus three sine theta of k and when we take the magnitude of the cross product that is r of x cross r of x cross r of theta that will be equal to the square roots of negative three cosine sine theta squared plus negative three, sine theta squared.

02:58 And we're going to have a sine squared and a cosine squared.

03:01 So we can pull out, we'll have the square root of nine times sine squared theta plus cosine squared theta.

03:16 And so the sine squared plus cosine squared will become equal to one.

03:19 So one times the square root of nine gives us three.

03:30 Now, when we are evaluating the surface integral of f of x, y, z, we know that this is equal to the integral of, in this case, x, of 3, sine theta, as z is equal to 3 sine theta, multiplied by the magnitude of the cross products, which in this case will be 3, and da will be equal to dx, d theta.

04:13 So we are trying now to evaluate the integral from 0 to 2 pi and the integral from 0 to 5 minus 3 cosine theta of 3x times 3 sine theta dx d theta.

04:35 And i'm moving on to actually solving the integral.

04:38 From here, there's no crazy things we have to do with this integral.

04:44 We can solve this one as normal.

04:48 And doing so we'll give the first integral to be from 0 to 2 pi of 9 times 5 minus 3 cosine theta squared sine theta over 2 d theta that's just taken by integrating the 9x squared sine theta over 2 and then plugging in the bounds and so from here we can pull out a 3 -halfs and we will be left with three halves times the integral from 0 to 2 pi of 5 minus 3 cosine theta squared times 3 sine theta d theta and now this one we will have to use a u substitution for as we have both the sine and cosine value however if we set u equal to 5 minus 3 cosine theta this will both solve this parentheses here and when we do the u substitution it will get rid of the sine theta here so if we set u equal to three cosine theta you'll have that d u is equal to three sine theta d theta and then to get our new limits i'm just going to plug zero and two pi in for um our u function right here and doing so will give new limits of 0 to 3 for 0 to 3.

06:58 And now once we plug this in, from 0 to 3 for our sides, this will give out that, one second, let me just check my notes really quick.

07:19 Okay.

07:22 So when you do this u -sub, we'll be left with that this will be left with that this is equal to 3 -house from 2 to 2.

07:37 Of u squared to u and if the bounds are the same for the integral, we will be left with an answer of zero.

07:50 And that will be the answer for our first surface that we need to calculate the area of.

07:56 And for our second surface we came up with, which is that x is equal to zero, when we integrate x z d z, so when we integrate the second surface s2 of x z d s, since x is equal to zero, this is the same as integrating s2 of 0, ds, which is equal to 0.

08:20 So for s1 and s2, s1 is equal to s2, which is equal to 0...

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