(1 point) This problem will illustrate the divergence theorem by computing the outward flux of the vector field F(x, y, z) = 3xi + 3yj + 2zk across the boundary of the right rectangular prism: -2≤ x ≤5,-2≤ y ≤3, -2 ≤ z < 5 oriented outwards using a surface integral and a triple integral over the solid bounded by rectangular prism. Note: The vectors in this field point outwards from the origin, so we would expect the flux across each face of the prism to be positive. Part 1 - Using a Surface Integral First we parameterize the six faces using 0 < s < 1 and 0 <t≤1: The face with z = -2: 0₁ = (x₁(s), y₁ (t), z₁ (s, t)) x₁(s) = y₁ (t) = Z₁ (s, t) = -2 The face with z=5: 0₂ = (x2(s), y(t), Z2 (s, t)) X2(s) = y2(t) = Z2(s, t) = 5 The face with x = -2: 03 = x3(s, t), y3 (S), Z3 (t)) X3(s, t)=-2 y3(s) = Z3 (t) = The face with x = 5: 04 = (x4(s, t), y4(s), Z4 (t)) x4(s, t) = 5 y4(s) = Z4 (t) = The face with y=-2: 05 = = (xs(s), ys (s, t), Z5 (t)) X5(s) = ys (s, t)=-2 Z5 (t) = The face with y=3:06 = (x6(s), y6 (s, t), 26(t)) X6(s) = y6(s, t) = 3 The face with y = 3: 06 = (x6(s), y6 (s, t), 26(t)) X6(s) = y6(s, t)=3 Z6 (t) = Then (mind the orientation) F.nds = + k k = S'S" " F(04). дол ds × дол 004) d = Τσι F(01). × dt ds 301) ds dt + do₂ F(02). ds at 202) ds dt + d03 F(03). X ds dt Ot ds dos dos ds dt + F(05). × ds ³) ds dt + '" доб F(06). X at ds 006) ds dt Part 2 - Using the Divergence Theorem F.ndS= [[ F-n =/ div F dV = s = √ √ √ √ di dx dydz