# 3 a we can parametrize the sphere by φ uv cos u sin

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Unformatted text preview: 3. (a) We can parametrize the sphere by Φ ( u,v ) = (cos u sin v, sin u sin v, cos v ). To determine the domain for the piece of the sphere cut out by the cone, we note that the cone is z = radicalbig x 2 + y 2 , giving z = cos v = √ cos 2 u sin 2 v + sin 2 u sin 2 v = √ sin 2 v = | sin v | . In the first quadrant of the real plane we have cos v = sin v = ⇒ tan v = 1 = ⇒ v = π 4 . Hence the required do- main is braceleftBig ( u, v ) | ≤ u ≤ 2 π, ≤ v ≤ π 4 bracerightBig . Now φ u = ( − sin u sin v, cos u sin v, 0), φ v = (cos u cos v, sin u cos v, − sin v ) and-1 1 x-1 1 y-1 1 z-1 x-1 φ u × φ v = ( − sin v ) Φ ( u,v ) so bardbl φ u × φ v bardbl = | sin v | (1) = sin v , since v ≤ π . The surface area is integraldisplay Φ bardbl φ u × φ v bardbl dA = integraldisplay 2 π integraldisplay π 4 sin v dv du = (2 π ) ( − cos v ) vextendsingle vextendsingle vextendsingle vextendsingle π 4 = (2 − √ 2) π . (b) We can parametrize the cone by Φ ( u,v ) = ( v cos u, v sin u, v ). In spherical polars this cone is ϕ = π 4 , so the z –component is no larger than cos( π 4 ) = 1 √ 2 . Thus the domain of Φ is braceleftbigg ( u,v ) | ≤ u ≤ 2 π, ≤ v ≤ 1 √ 2 bracerightbigg . Now φ u = ( − v sin u, v cos u, 0), φ v = (cos u, sin u, 1) and φ u × φ v = ( v cos u, u sin u, − v ) so bardbl φ u × φ v bardbl = √ v 2 cos 2 u + v 2 sin 2 u + v 2 = √ v 2 + v 2 = √ 2 | v | = √ 2 v since v ≥ 0. The surface area is integraldisplay Φ bardbl φ u × φ v bardbl dA = integraldisplay 2 π integraldisplay 1 √ 2 √ 2 v dv du = (2 π ) ( √ 2) parenleftbigg v 2 2 parenrightbigg vextendsingle vextendsingle vextendsingle vextendsingle 1 √ 2 = π √ 2 . 4. (a) We can parametrize by Φ ( r, θ ) = parenleftbigg r cos θ, r sin θ, 1 r parenrightbigg , < r ≤ 1, 0 ≤ θ ≤ 2 π . (b) The image on the right was produced using the Mathematica command ParametricPlot3D . (c) Volume = integraldisplay region 1 dV cylindrical = polars x y z x MATB42H Solutions # 7 page 3 integraldisplay 2 π integraldisplay 1 integraldisplay 1 r 1 r dz dr dθ = 2 π integraldisplay 1 (1 − r ) dr = 2 π bracketleftbigg r − r 2 2 bracketrightbigg 1 = π cubic units. (d) Now φ r = parenleftbigg cos θ, sin θ, − 1 r 2 parenrightbigg , φ θ = ( − r sin θ, r cos θ, ) and φ r × φ θ = parenleftbigg 1 r cos θ, 1 r sin θ, r parenrightbigg so bardbl φ r × φ θ bardbl = radicalbigg 1 r 2 + r 2 = √ 1 + r 4 r . The surface area is integraldisplay Φ dS = integraldisplay 1 integraldisplay 2 π √ 1 + r 4 r dθ dr = 2 π integraldisplay 1 √ 1 + r 4 r dr ≥ 2 π integraldisplay 1 1 r dr = lim t → + bracketleftbigg 2 π ln r bracketrightbigg 1 t = ∞ ....
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• Winter '10
• EricMoore

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