Evaluate the surface integral. ∬5 x z d S, S is the part of the plane 2 x+2 y+z=4 th

step 1 So we are given that 4 is equal to 2x plus 2y plus Z. And we can rewrite this in terms of Z to g

Evaluate the surface integral. ∬5 x z d S, S is...

Key Concepts

Surface Integral of a Scalar Function

A surface integral of a scalar function extends the idea of definite integrals to functions defined on surfaces. It involves summing the function’s values weighted by an infinitesimal surface area element over a given surface. This integral is used to compute quantities like mass, charge, or heat distribution when the density is defined over a surface.

Parameterization of Surfaces

Parameterization is a method of representing a surface using two parameters, typically transforming a complex surface into a domain in the plane. This method involves expressing the surface as a vector function of two variables, which facilitates the computation of the surface integral by converting it into a double integral over a parameter domain.

Computation of the Surface Area Element (dS)

The surface area element dS quantifies an infinitesimal piece of area on a surface and is crucial in setting up surface integrals. When a surface is given explicitly as a function z = f(x, y), the element is computed as dS = ?(1 + (?f/?x)² + (?f/?y)²) dx dy. This factor accounts for the stretching of the surface relative to the projection in the coordinate plane.

Projection onto a Coordinate Plane

When evaluating surface integrals, it is often useful to project the surface onto a coordinate plane to simplify the integration limits. This involves identifying the region in the plane over which the parameters vary, and it allows the use of standard double integration techniques once the limits of the resulting domain are determined.

Transcript

00:01 So we are given that 4 is equal to 2x plus 2y plus z.

00:11 And we can rewrite this in terms of z to guess that c is equal to 4 minus 2x minus 2y.

00:19 Since we have z in terms of x and y, we can use the formula that the surface integral of f x, y, d s is equal to the integral of f of x y g of x y f f of x y f multiplied the square root of both of the partial derivatives squared plus one d a for our integral since we are asked to integrate x z d s this will be equal to the integral of x times 4 minus 2x minus 2y multiplied by the square root of the x partial derivative which is negative 2 squared so 4 plus the y partial which is also 4 1 squared plus 1 d a and so we can pull this would be equal to 9 so we can pull that out and we'll be left with this is equal to 3 times the integral of 4x minus 2x squared minus 2xyda.

01:59 Now from here we have to come up with the bounds for the problem, and especially since we're only looking for the first octant.

02:08 So if we think about this in intercept form, we can rewrite z as x over 2 plus y over 2 plus z over 4 is equal to 1.

02:23 And we can tell that the x and y intercept, the x intercept and the y intercept, are both equal to 2.

02:34 And so this would give us a triangle with vertices at 0 ,0, 0 ,0, and 0.

02:50 And we'll have a line between 2 and 02, like so.

02:56 So we can rewrite this as x over 2 plus y over 2, which is equal to 1.

03:03 And we can rewrite that as y is equal to 2 minus x.

03:08 This would give us our bounds that we have 0 to x to 2, and 0 to y to 2 minus x.

03:24 Like so.

03:25 Now that we have our bounds, we can move on with solving this integral...

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