By Mejlbro L.

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**Example text**

We therefore conclude that O is a subset of an hyperboloid of revolution with two nets and centrum (0, 0, −a). We get the equation of the hyperboloid of revolution by replacing from 1), 2 by x2 +y 2 in the expression (z + a)2 − x2 − y 2 = a2 , or in its standard form, z+a a 2 − x a 2 − y a 2 = 1. The surface O it that subset which lies between the planes z = a and z = 2a. 15 A surface of revolution O is given in semi polar coordinates ( , ϕ, z) by 0 ≤ ϕ ≤ 2π, a≤ ≤ 2a, z = a ln , a where a is some positive constant.

By assuming that g(x, y, z) is odd, it follows by the symmetry of the sphere that the surfaces can be paired in such a way that the sum of the surface integrals over each pair is zero. (The details are left to the reader). Since x f (x, y, z), y f (x, y, z) and z f (x, y, z) all are homogeneous of degree 3, we conclude that p = 0. Remark. We shall for obvious reasons skip the traditional variants which give a lot of tedious computations. The reason for including this example is of course to demonstrate that one in some cases may beneﬁt from the symmetry.

6 2 1. Show that the line element ds is given by 2 − sin2 t dt. sin t ds = A cylinder surface C with L as its leading curve is given in the following way: x = cos t, y = −2 ln sin t, z ∈ [0, sin t], 2. Compute the surface integral C t∈ π π , . 6 2 xz dS. A Curve element and surface integral. D Follow the guidelines; apply the formula of the surface integral over a cylinder surface. 8 Figure 27: The leading curve L. I 1) From dx = − sin t dt and dy cos t = −2 , dt sin t follows that dx dt 2 + dy dt 2 = sin2 t + 4 cos2 t 1 = 2 sin t sin2 t 2 − sin2 t sin t = (sin2 t)2 − 4 sin2 t + 4 2 , hence ds = dx dt 2 + dy dt 2 dt = 2 − sin2 t 2 − sin2 t dt = dt, sin t sin t t∈ π π , .