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4. Suppose S is the surface z= x² + 4y? Tying beneath the plane z=1. Orient S by taking the inner normal n to pointing in the
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Solution ZEL Regions: Z = x2 + 4y2 limit of integral, x²4442<Z st -1-442 x as √ 1-412 - 2 / 2 = 12 h and Vedas field V = {1,The flux = SSS div (v) LV (By divergence theorem) S 2 11 - 442 2xz dzdady V7-4y2 Jox²+442 . 2 W1-412 Z=1 x22 Ja 1 - - 1 - 4 4

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