Consider the sphere x^2+y^2+z^2 = 4a^2 of radius 2a and center at the origin. Us
ID: 3193975 • Letter: C
Question
Consider the sphere x^2+y^2+z^2 = 4a^2 of radius 2a and center at the origin. Use a surface integral to find a formula for the area of the cap, S, of the sphere above the disk x^2 + y^2 = a^2 in the xy-plane. (Hint: Use polar coordinates.)Explanation / Answer
1) Assuming you mean F(x,y,z) = e^x i + e^(x-y) j + e^y k: Break this integral into two portions: (i) C1: (0, 0, 0) to (0, 0, 1) with parameterization r(t) = (0, 0, t) for t in [0, 1]: Since r'(t) = (0, 0, 1), ?c1 F · dr = ?(t = 0 to 1) (e^0, e^0, e^0) · (0, 0, 1) dt = 1. (ii) C2: (0, 0, 1) to (0, 1, 1) with parameterization r(t) = (0, t, 1) for t in [0, 1]: Since r'(t) = (0, 1, 0), ?c2 F · dr = ?(t = 0 to 1) (e^0, e^(-t), e^t) · (0, 1, 0) dt = -e^(-t) {for t = 0 to 1} = 1 - 1/e. -- By (i) and (ii), ?c F · dr = 1 + (1 - 1/e) = 2 - 1/e. --------------------- 2) Solving for z, we have z = v(4a^2 - x^2 - y^2). So, the surface area equals (with cartesian coordinates) ?? v(1 + (z_x)^2 + (z_y)^2) dA = ?? v(1 + (-x/v(4a^2 - x^2 - y^2))^2 + (-y/v(4a^2 - x^2 - y^2))^2) dA = ?? v[1 + (x^2 + y^2)/(4a^2 - x^2 - y^2)] dA = ?? v[4a^2 / (4a^2 - x^2 - y^2)] dA = 2a ?? dA / v(4a^2 - x^2 - y^2). Now, convert to polar coordinates (as the region of integration is in x^2 + y^2 = a^2): ==> 2a ?(? = 0 to 2p) ?(r = 0 to a) (r dr d?) / v(4a^2 - r^2) = 2pa ?(r = 0 to a) 2r (4a^2 - r^2)^(-1/2) dr = 2pa * [-v(4a^2 - r^2) {for r = 0 to a}] = 2pa * (2a - av3) = 2pa² (2 - v3). ----------------------- 3) We know that ?c ?L · dr = 0 on a closed curve C, by the Fundamental Theorem of Line Integrals, assuming that L is continuous inside C. So, ?c ?L · dr = ?B ?L · dr + ?T ?L · dr = 0 ==> ?B ?L · dr + 1 = 0 ==> ?B ?L · dr = -1. ----------------------- 4) Parameterize the cylinder by r(u, v) = (3 cos u, 3 sin u, v) for u in [0, p/2], v in [0, 4]. Since r_u x r_v = (3 cos u, 3 sin u, 0), the flow rate equals ??s v(x,y,z) · dS = ?? (2 * 3 cos u, 3 sin u, 3 cos u * 3 sin u) · (3 cos u, 3 sin u, 0) dA = ?(u = 0 to p/2) ?(v = 0 to 4) (18 cos^2(u) + 9 sin^2(u)) dy du = ?(u = 0 to p/2) 4 (18 cos^2(u) + 9 sin^2(u)) du = ?(u = 0 to p/2) 2 [18 (1 + cos(2u)) + 9 (1 - cos(2u))] du = ?(u = 0 to p/2) [54 + 18 cos(2u)] du = 27p. --------------------- 5) By Green's Theorem, note that A = -?c y dx So, the area equals (remembering an extra negative sign due to the clockwise orientation) ?(t = 0 to 2p) (1 - cos t) * (1 - cos t) dt = ?(t = 0 to 2p) (1 - 2 cos t + cos^2(t)) dt = ?(t = 0 to 2p) (1 - 2 cos t + (1/2)(1 + cos(2t))) dt = (1/2) ?(t = 0 to 2p) (3 - 4 cos t + cos(2t)) dt = 3p.