# Solutions11 - Math 601 Solutions to Homework 11 1. Consider...

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Math 601 Solutions to Homework 111. Consider the curve given by the following parametric equations:x= cos(t)andy= sin(2t)LetDbe the region inside this curve and to the right of they-axis:(a) Use Green’s Theorem to computeZZDx2dA.(b) Use Green’s Theorem to find the area of the regionD.Answer:(a) We need to find a vector fieldFsuch that rot(F) =x2.Thevector fieldF=-yx2iwill work (as would13x3jor lots of otherpossibilities). Then, by Green’s Theorem:ZZDx2dA=I∂D-yx2dxFrom the parametrization of the curve, we have:x=costy=sin(2t)dx=-sint dtdy=2 cos(2t)dtThe bounds fortare-π2tπ2(we can see this by notingthatxandyare both 0 whentis a multiple ofπ/2, and thenby plotting a few points of the parametric curve to make sure wechoose the right-side of the lemniscate).We can now compute the integral (recall that sin(2t) = 2 sintcost):ZZDx2dA=I∂D-yx2dx=Zπ/2-π/2-sin(2t) cos2t(-sint)dt=Zπ/2-π/22 sin2tcos3t dt=Zπ/2-π/22 sin2t(1-sin2t) cost dt
We use the substitutionu= sint,du= cost dt:ZZDx2dA=Z1-12u2(1-u2)du=23u3-25u51-1=815Note:If we had used a different vector field, we would haveobtained the same answer at the end.(b) To compute the area we need to evaluate the integralZDdA, sowe need to find a vector fieldFsuch that rot(F) = 1. The vectorfieldF=-yiwill work (as wouldxjor lots of other possibilities).Then, by Green’s Theorem:ZZDdA=I∂D-y dxAs in part (a), we have:x=costy=sin(2t)dx=-sint dtdy=2 cos(2t)dtwith-π2tπ2.We can now compute the integral (recall that sin(2t) = 2 sintcost):ZZDdA=I∂D-y dx=Zπ/2-π/2-sin(2t)(-sint)dt=Zπ/2-π/22 sin2tcost dtWe use the substitutionu= sint,du= cost dt:ZZDdA=Z1-12u2du=23u31-1=43
2. Consider the surface integralZZSz dAwhereSis the upper half (z0) of the spherex2+y2+z2= 1.(a) Set up this integral using a parametrization wheret=xandu=y.(b) Set up this integral using a parametrization wheret=randu=θ.(c) Set up this integral using a parametrization wheret=θandu=φ.Answer:(a) Witht=xandu=y, we have the parametric equations:x=ty=uz=1-t2-u2-1-u2t1-u2-1u1Then, the tangent vectors to the surface areTt=1,0,-t1-t2-u2Tu=0,1,-u1-t2-u2And, the normal vector to the surface isTt×Tu=flflflflflflflflijk10-t1-t2-u201-u1-t2-u2flflflflflflflfl=t1-t2-u2i+u1-t2-u2j+kThus:dA=kTt×Tukdt du=st21-t2-u2+u21-t2-u2+ 1dt du=r11-t2-u2dt duThus:ZZSz dA=Z1-1Z1-u2-1-u21-t2-u2kTt×Tukdt du=Z1-1Z1-u2-1-u21-t2-u2r11-t2-u2dt du=Z1-1Z1-u2-1-u2dt du
(b) Witht=randu=θ, we have the parametric equations:x=tcosuy=tsinuz=1-t20t10u2πThen, the tangent vectors to the surface areTt=cosu,sinu,-t1-t2Tu= (-tsinu, tcosu,0)And, the normal vector to the surface isTt×Tu=flflflflflflflflijkcosusinu-t1-t2-tsinutcosu0flflflflflflflfl=t2cosu1-t2i+t2sinu1-t2j+tkThus:dA=kTt×Tukdt du=st4cos2u1-t2+t4sin2u1-t2+t2dt du=st21-t2dt duThus:ZZSz dA=Z2π0Z101-t2kTt×Tukdt du=Z2π0Z101-t2st21-t2dt du=Z2π0Z10t dt du(c) Recall thatx=ρsinφcosθ,y=ρsinφsinθandz=ρcosφ. Onthe sphere, we haveρ= 1. Thus, if we use the parameterst=θandu=φ, we have the parametric equations:x=sinucosty=sinusintz=cosu0t2π0uπ2
Then, the tangent vectors to the surface areTt= (-sinusint,sinucost,0)Tu= (cosucost,cosusint,-sinu)And, the normal vector to the surface isTt×Tu=flflflflflflflflijk-sinusintsinucost0cosucostcosusint-sinuflflflflflflflfl=-sin2ucosti-sin2usintj-sinucosukThus:dA=kTt×Tukdt du=psin4ucos2t+ sin4usin2t+ sin2ucos2u dt du=psin4u+ sin2ucosudt du=psin2u dt du= sinu dt duThus:ZZSz dA=Zπ/20Z2π0cosukTt×Tukdt du=Zπ/20Z2π0cosusinu dt du=Zπ/20Z2π0cosusinu dt du

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