Question

Evaluate the surface integral. S xz ds, s is the part of the plane 2x 2y z = 4 that lies in the first octant

Answer 1

The value of the given surface integral is 4.

The given plane intercepts the coordinate axes at (2, 0, 0), (0, 2, 0), and (4, 0, 0). These point are the coordinates of a triangular region that we can parameterize using.

$S(u,v)=(1-u)((1-u)(2,0,0)+u(0,2,0)+u(4,0,0)\\S(u,v)=(2(1-u)(1-v),2u(1-v),4v)$

What is the surface integral?

A surface integral is a generalization of multiple integrals to integration over surfaces. It can be thought of as the double integral analog of the line integral. Given a surface, one may integrate a scalar field over the surface or a vector field.

with 0≤u≤1 and  0≤v≤1. Then the surface element ds  is equivalent to

$ds=\||\:\left(s_u\times \:\:S_v\right)||dudv=12\left(1-v\right)dudv$

The surface integral is then

$\int \:\int \:_Sxzds=12\int _0^1\:\int _0^1\:\left(2\left(1-u\right)\left(1-v\right),2u\left(1-v\right),4v\right)dvdu=4$

Therefore the value of the given surface integral is 4.

To learn more about the integral visit:

brainly.com/question/14295614

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