Surface Integrals of Vector Fields When dealing with surface integrals of vector fields we encounter the surface orientation issue. Our basic method of integration is: √ √ F(x, y, z) ·ñd$ = ± √ √ F(x, y, z)· Gu× F;)dA Σ Where the + indicates that we need to decide which it is. Ideally we'd like to parametrize the surface so that the cross-product generated normal vectors point in the direction of desired orientation but this is quite difficult. Instead we take the approach from class - we look at the cross product г, X, and examine whether it agrees with or disagrees with Σ's orientation. If it disagrees we negate. For example suppose we wish to evaluate(x²+y+)ds where Σ is the part of the parabolic sheet y = x² with -2≤x≤2 and 0≤2 ≤ 3, oriented Σ outwards, meaning with positive y-component. We can parameterize Σ with 7(x, z) = x + (4-x)+zk with -2≤x≤2 and 0≤3. We can find Fxxy with Matlab: syms x z; rbar -[x,4-x^2,z]; cross(diff(rbar,x),diff(rbar,z)) This yields output: ans = [-2*x, -1, 0] These vectors have negative y-component and hence they disagree with Σ's orientation. The final integral then needs to be negated. Here is the whole thing: syms x y z; rbar = [x,4-x^2,2]; F = [x^2,y,z]; -int(int(simplify(dot(subs(F,[x,y,z],rbar), cross(diff(rbar,x),diff(rbar,z)))),x,-2,2),z,0,3) Evaluate cx+zds where C is the semicircle x² + y² = 9 with z = 5 and x ≥ 0. Assign the result to q9. Evaluate/x+zdS where Σ is the part of the cylinder x² + 2² = 9 between y = 1 and y = 3. Assign the result to q10. y Evaluate ƒƒ‹yî +xî +zk) · ñdS where Σ is the part of the paraboloid z = 9 - x² - 1² inside the cylinder r = 2 cos 0 with downwards orientation. Assign the result to q11.