Question
Asked Nov 9, 2019
51 views

Evaluate the surface integral.

The double integral over S (xz) dS, S is the part of the plane 2x+2y+z=4 that lies in the first octant.

check_circle

Expert Answer

Step 1

Evaluate  the surface integral for the function

2x+2y+z=4

Step 2

Rewrite the given equation 2x+2y+z=4 as x/2+y/2+z/4=1
Here S is the triangular region with vertices (2,0,0),(0,2,0),(0,0,4)
Choose x,y as perimeter and the domain for the perimeter in the triangular region with vertices (2,0,0),(0,2,0),(0,0,4)is

help_outline

Image Transcriptionclose

D {x, y) R "0<x<2-x} The plane is z = 4-2x-2y ax -2 -2 Oy

fullscreen
Step 3

Calculate the surfa...

help_outline

Image Transcriptionclose

D {x,y)R2"0 <x <2-x} The plane is z = 4-2x- 2y -2 дх -2 xzds x(4-2x-2y)|-2) +(-2)^ +ldS 2 2-x 2x 0 0 2ку — х? 2 dc (2-x) x(2-x)-x (2-x)-x. dx 2 G(2-2- 2 (2 dx

fullscreen

Want to see the full answer?

See Solution

Check out a sample Q&A here.

Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in

Math

Calculus