Thursday, April 17 ∗∗ Surface Parameterpalooza1. Let S be the portion of the plane x +y +z =1 which lies in the positive octant.(a) Draw a picture of S.Solution. The picture is shown below.0.00.51.00.00.51.00.00.51.0(b) Find a parametrization r: D →S, being sure to clearly indicate the domain D. Check youranswer with the instructor.Solution. One can use the parametrization r(u,v) = (u, v, 1 −u − v) with the domainD given by D ={(u, v) : 0 ≤u ≤1, 0 ≤v ≤1 −u}.(c) Use your answer in (b) to compute the area of S via an integral over D.Solution. Using the parametrization in (b), one getsru=(1,0,−1), rv=(0,1,−1),so ru×rv=(1,1,1), and kru×rvk=p3. Hence the area of S isÏDdS =Z10Z1−u0kru×rvkdvdu =p32.(d) Check your answer in (c) using only things you learned in the first few weeks of this class.Solution. The picture of S is a triangle with vertices A =(1,0,0), B =(0,1,0) and C =(0,0,1).Thus−→AB =(−1,1,0) and−→AC =(−1,0,1), and the area is12k−→AB ×−→ACk=p32.2. Consider the surface S which is the part of z +x2+y2=1 where z ≥0.(a) Draw a picture of S.Solution. The picture is shown below.(b) Find a parametrization r: D →S. Check your answer with the instructor.Solution. One can use the parametrization r(r,θ) = (r cos θ, r sinθ, 1 −r2) with the do-main 0 ≤r ≤1, 0 ≤θ ≤2π.3. Let S be the surface given by the following parametrization. Let D =[−1,1] ×[0, 2π] and definer(u, v) =(u cosv,u sinv,v).(a) Consider the vertical line segment L ={u =0} in D. Describe geometrically the image of Lunder r.Solution. The image of u =0 under r is a line segment (0, 0, v) where 0 ≤ v ≤2π.(b) Repeat for the vertical segments where u =−1 and u =1.Solution. When u =1, the image r(1, v) =(cos v, sinv, v) is a helix with 0 ≤v ≤ 2π, and sois u =−1. Thus the images of u =1 and u =−1 form the double helix.(c) Use your answers in (a) and (b) to make a sketch of S.Solution. The picture is shown below.4. Consider the ellipsoid E given byx29+y24+z2=1.(a) Draw a picture of E.Solution. The picture is shown below.(b) Find a parametrization of E. Hint: Find a transformation T : R3→R3which takes the unitsphere S to E, and combine that with our existing parametrization of the plain sphere S.Solution. One can use the following parametrizationr(θ,φ) =(3 sinθ cosφ, 2 sinθ sinφ, cos θ)with the domain 0 ≤θ ≤π, 0 ≤φ
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