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

• Homework Help
• 10

This preview shows page 1 - 6 out of 10 pages.

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

Course Hero member to access this document

Course Hero member to access this document

End of preview. Want to read all 10 pages?

Course Hero member to access this document

Term
Spring
Professor
alndy
Tags
Stokes theorem
• • • 