# MATH23 Homework 12 Solution - Homework#12 Solutions Math 23...

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Homework #12 SolutionsMath 23, Spring 20151.LetSbe the part of the plane containing the points(1,0,0),(0,1,0), and(0,0,1)thatlies in the first octant (that is, withx0,y0, andz0), with upward-pointingnormal.(a)Letf(x, y, z) =x2+z. ComputeRRSf dS.
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2.Find the surface area of the part of the spherex2+y2+z2= 4which is betweenthe planesz= 1andz=-1.This surface can be parametrized by~r(ϕ, θ) = 2 cosθsinϕ~ı+ 2 sinθsinϕ~+ 2 cosϕ~k.(Thiscomes from spherical coordinates, settingρ= 2. You can useuandvif you’d like, but it seemslike usingϕandθmakes it easier to keep track of.) Whenz= 1, we see that cosϕ= 1/2 (thesphere has radius 2, so we get a triangle where the hypotenuse has length 2 and the adjacentleg has length 1), and similarly, whenz=-1, we see that cosϕ=-1/2. Thus the parameterspace is given byϕ[π/3,2π/3] and (of course)θ[0,2π).In order to compute the area, we first compute~rϕ×~rθ=~ı~~k2 cosθcosϕ2 sinθcosϕ-2 sinϕ-2 sinθsinϕ2 cosθsinϕ0=(4 cosθsin2ϕ)~ı-(-4 sinθsin2ϕ)~+(4 cos2θsinϕcosϕ+ 4 sin2θsinϕcosϕ)~k= 4cosθsin2ϕ~ı+ sinθsin2ϕ~+ sinϕcosϕ~k,and then|~rϕ×~rθ|= 4qcos2θsin4ϕ+ sin2θsin4ϕ+ sin2ϕcos2ϕ= 4qsin4ϕ+ sin2ϕcos2ϕ= 4 sinϕqsin2ϕ+ cos2ϕ= 4 sinϕ.If we letSdenote the surface under discussion (the part of the sphere described in the question),we’re now in a position to computeArea (S) =ZZSdS=Z2π0Z2π/3π/34 sinϕ dϕ dθ= 4Z2π0Z2π/3π/3sinϕ dϕ!= 4·2π[-cosϕ]2π/3π/3= 8π12+12= 8π.One thing to notice is that the whole sphere (of radius 2) has surface area 16π. SoShas halfof this area, which seems roughly plausible.3.LetSbe the potion of the cylinderx2+y2= 4between the planesz=-2andz= 2, with “inward-pointing” normal (that is,~npoints toward thez-axis), and let~F(x, y, z) =y2z~ı+~+xy~k. ComputeRRS~F·d~S.We see thatSis part of a cylinder of radius 2, so we can parametrizeSby~r(u, v) = 2 cosu~ı+
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2 sinu~+v~kwhereu[0,2π) andv[-2,2]. We have~ru×~rv=~ı~~k-2 sinu2 cosu0001=~ı(2 cosu-0)-~(-2 sinu-0) +~k(0-0)= 2 (cosu~ı+ sinu~).
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