Math 21a Handout on Surface AreaThe text argued that after parameterizing a surface (or part of one) by a function X(u, v),with u and v coordinates on a region R in R2, then the integral of a function f on the surface is thesame as the integralfdSS∫∫≡ fuvXXdudvuvR((,))| |X ×∫∫(1)The purpose of this handout is to provide three examples of the preceding formula, but allwith the same surface and same function f. Rather, each example will use a different parameterization of the surface, and give evidence for the (not obvious) fact that the integral in (1)is truly independent of your chosen way of parameterizing the surface. In particular, the surface inquestion is the top half of the unit sphere in R3, that is where• x2 + y2 + z2 = 1,• z ≥ 0.(2)Meanwhile, the function f is just the z-coordinate in these examples.Example 1 : This example computes the integral in 1 using the parameterization whereX(u, v) = (u, v, (1 - u2 - v2)1/2) .(3)where (u, v) are restricted to the disk D where u2 + v2 ≤ 1. In this case,• Xu = (1, 0, -u/(1 - u2 - v2)1/2) ,• Xv = (0, 1, -v/(1 - u2 - v2)1/2) ,• Xu × Xv = (u/(1 - u2 - v2)1/2, v/(1 - u2 - v2)1/2, 1)• |Xu × Xv| = 1/(1 - u2 - v2)1/2 .(4)Meanwhile, z = (1 - u2 - v2)1/2, so the integrand z |Xu × Xv| in (1) is just the constant 1. Thus, theintegral in 1 is simply the area of the unit radius disk which is π.Example 2 : In this example, the top half of sphere is parameterized asX(u, v) = ((1 - u2)1/2 cos(v), (1 - u2)1/2 sin(v), u) ,(5)where 0 ≤ u ≤ 1 and 0 ≤ v ≤ 2π. (You should convince yourself that the expression in (5) reallyparameterizes the top half of the sphere.) In this case,• Xu = (- u (1 - u2)-1/2 cos(v), -u (1 - u2)-1/2 sin(v), 1),• Xv = (-(1 - u2)1/2 sin(v), (1 - u2)1/2 cos(v), 0) ,• Xu × Xv = (-(1 - u2)1/2 cos(v), -(-1 - u2)1/2 sin(v), -u)• |Xu × Xv| = 1.(6)Thus, in this parameterization, the integrand in (1) is z |Xu × Xv| = u, and the integral in (1) isududv0102∫∫π ,(7)which is equal to π also.Example 3 : In this example, the top half of the sphere is parameterized using sphericalcoordinates, soX(u, v) = (cos(v) sin(u), sin(v) sin(u), cos(u)) ,(8)where 0 ≤ u ≤ π/2 and 0 ≤ v ≤ 2π. Here,• Xu = (cos(v) cos(u), sin(v) cos(u), - sin(u)) .• Xv = (-sin(v) sin(u), cos(v) sin(u), 0) ,• Xu × Xv = sin(u) (cos(v) sin(u), sin(v) sin(u), cos(u)) ,• |Xu × Xv| = sin(u) .(9)Thus, the integrand in (1) is z |Xu × Xv| = cos(u) sin(u) = 2−1 sin(2u) and the integral in (1) is2−1 sin( )/20202ududvππ∫∫.(10)Here, the u integral gives a factor of 2−1 (cos(2u)0 - cos(2u)π/2), which is 1, while the v integralgives a factor of 2π and then the factor of 2−1 out in front makes the whole expression in (10) equalto π as it should
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