Task 2 (7p) Prerequisites: Parametric surfaces, surface elements. The surface of a bath ring, bicycle tire, or donut for that matter, is mathematically given by a torus. It can be defined as the surface that results from a circle in the zz-plane being rotated once around the z-axis. See Figure la. In this task, we want to study a slightly more general surface - namely one that results from rotating a general closed curve in the zz-plane around the z-axis. One can for instance imagine a corrugated bicycle tire (see Figure 1b). The goal will be to find a formula for the area of the surface. D.5 .0.5 (a) A torus (grey) is formed by rotating a circle (b) The more general type of surface we want (red) around the z-axis. to consider. (a) Let y : [0, 1] - R3 be a parametrication of the curve. Argue that a parametrization of the surface is given by D : [0, 1] x [0, 27] - RS, (t, 0) - R(0) 7(t), where (2p) cos(0) - sin(0) R(0) = sin(0) cos(0) o(b) Calculate the partial derivatives of the parametrization, and show that they are orthogonal in each point. Tip: Remember that y lies in the rz-plane, so 72 = 0. (2p) 2 (c) Calculate the surface area element, and derive a formula for the surface area. Tip: Remember that the length of u x v is given by [u| |v| sin(a), where a is the angle between the vectors. (2p) (d) Apply the formula to the special case of the torus. In this case, the curve is a circle of radius a centered at (b, 0,0), with b > a. (1p)