Evaluate the surface integral I. -4x dS. S Where S is the triangular region with vertices (4, 0, 0), (0, 2, 0) and (0, 0, – 1). An equation of the plane in which the triangular region lies is given by: = Z Σ Therefore • Y2 -4x dS Σ dy dr where Y1 = Y2 = -4 Σ X2 = Σ Evaluate S. -4x dS = M M M M

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