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Integrate the given function over the accompanying surface in Exercises 27–34.

f(x,z)=e-(x2+z2), where S is the unit disk centered at the point (0, 2, 0)and in the plane y = 2.

Short Answer

Expert verified

The integration of the function f(x,z)=e-(x2+z2)is π-πe.

Step by step solution

01

Step 1. Given Information.

The function:

f(x,z)=e-(x2+z2)

02

Step 2. Formula for finding the integral.

The formula for finding the integral over the smooth surface is given by,

∫SdS=∫∫Df(x(u,v),y(u,v),z(u,v))ru×rvdudv

03

Step 3. Find ru,rv

The surface is parametrized as follows:

r(u,v)=(ucosv,2,usinv)

ru=∂r∂u=(cosv,0,sinv)rv=∂r∂v=(-usinv,0,ucosv)

04

Step 4. Find ru×rv.

ru×rv=(cosv,0,sinv)×(-usinv,0,ucosv)=-uj=0i-uj+0kru×rv=02+(-u)2+02=u

05

Step 5. Find f(x(u,v),z(u,v)).

f(x,z)=e-(x2+z2)f(x(u,v),z(u,v))=e-((ucosv)2+(usinv)2)=e-u2(cos2v+sin2v)=e-u2

06

Step 6. Find the region of integration and integrate.

The region of integration is given by,

D={(u,v)0≤u≤1,0≤v≤2π}∫SdS=∫∫Df(x(u,v),z(u,v))ru×rvdudv=∫02π[∫01e-u2udu]dv=∫02π[-e-u22]01dv=∫02π[-12e+12]dv=12-12e∫02πdv=12-12e[v]02π=12-12e(2π)=π-πe

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