Evaluate the surface integral / - 2a dS. Where S is the triangular region with vertices (5,0,0), (0, 3,0) and (0,0, – 1). An equation of the plane in which the triangular region lies is given by: z = -1+x/5+y/3 Σ Therefore 02 - 2a dS -2xsqrt(1+(1/5)^2+(1/3)^2) Z dy da where = 16 2 = -5/3x+25/3 Σ 5 Σ Evaluate -2a dS -44.704 Σ M M M M
Evaluate the surface integral / - 2a dS. Where S is the triangular region with vertices (5,0,0), (0, 3,0) and (0,0, – 1). An equation of the plane in which the triangular region lies is given by: z = -1+x/5+y/3 Σ Therefore 02 - 2a dS -2xsqrt(1+(1/5)^2+(1/3)^2) Z dy da where = 16 2 = -5/3x+25/3 Σ 5 Σ Evaluate -2a dS -44.704 Σ M M M M
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.6: Quadratic Functions
Problem 35E
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