2. (25) Let F(x, y, z): = < x + arctan(yz), 3 ln(x + 2) + 3y - V√z, cos(x) + 3ey + 2z>, and let S be the tetrahedron wit

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2 25 Let F X Y Z X Arctan Yz 3 Ln X 2 3y V Z Cos X 3ey 2z And Let S Be The Tetrahedron Wit 1 (146.89 KiB) Viewed 59 times

2. (25) Let F(x, y, z): = < x + arctan(yz), 3 ln(x + 2) + 3y - V√z, cos(x) + 3ey + 2z>, and let S be the tetrahedron with its corners at (0, 0, 0), (2, 0, 0), (0, 2, 0), and (0, 0, 3). We want to calculate the flux of F out of S. The trouble is that this would require doing 4 surface integrals, which would not be fun. So you need to apply the divergence theorem (and be happy this isn't "verify"). a) Find the equation of the plane that is the “top” of the tetrahedron, in the form z = f(x, y). b) Sketch and label the "base" of the tetrahedron (in the xy-plane). You must label the line that bounds the triangle (in the form y = mx + b). c) Use (a) and (b) to write down the inequalities in x, y, and z which determine S. (The only variable that has 2 constant limits is x. y and z have "tops" that depend on other variables.) d) Compute div(F). (This should be a super easy (non zero) function, leading to a very nice integral.) e) Compute the correct volume integral, so that you know (using the divergence theorem) the desired flux. If your integral limits are wrong, you won't earn points here. The answer is an integer.