let F = ( exp(y+z)-2y^2 , x*exp(y+z) + y , exp(y+z) )
for the surface S:={(x,y,z) : z=exp - (x^2+4y^2), x^2+4y^2<=4}
find a parametrization for S so that the normal is pointing upward in the direction of the positive z-axis
find double integral of (curl of F) . dS when the normal os S is as above
my parametrization found (is it correct?)
x= 2 p cos t
y = p sin t
z = exp(-4p^2)
0<=p<=1 0<=t<= 2 Pi
vector calculus
2011-05-16 11:17 am
回答 (1)
2011-05-17 7:19 am
✔ 最佳答案
x= 2 p cos ty = p sin t
z = exp(-4p^2)
0<=p<=1, 0<=t<= 2π
is correct.
∫∫_S curl(F)∙dS = ∫_C F∙ds (By Stokes thm.)
where C is the curve x^2+4y^2=4 counterclockwise, thus
(x,y,z)=(2cost, sint, 0), 0<= t <= 2π
∫_C F∙ds=∫[0~2π] {2cost*[e^(sint)-2(sint)^2]+sint[2cost e^(sint)+ sint]} dt
= 2e^(sint) - (4/3)(sint)^3+2(sint-1)e^(sint)+(1/2)[t - 0.5sin(2t)] sub. t=0~2π
= π
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