Please don't provide handwritten solution... ]]Evaluate the surface integral 1 dS where is the portion of the plane z = 4 − 6x – 1y in the first octant. Then-=f(x, y, z) dS!!sqrt(38)Odu du

To find where is the portion of the plane in the first octant

Answered: Evaluate the surface integral 1 d.S…

Evaluate the surface integral 1 d.S where o is the portion of the plane z = 4 - 6x – 1y in the first octant. Then // f(x, y, z) dS ·!! 0 0 = sqrt(38) du du

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Please don't provide handwritten solution...

]]
Evaluate the surface integral 1 dS where is the portion of the plane z = 4 − 6x – 1y in the first octant. Then
-
=
f(x, y, z) dS
!!
sqrt(38)
O
du du

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