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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Evaluate the surface integral.

19. ∫∫s xz dS, S is the boundary of the region enclosed by the cylinder y2 + z2 = 9 and the planes x = 0 and x + y = 5

To determine

To find: The value of SxzdS .

Explanation

Given data:

y2+z2=9 and planes are x=0 and x+y=5 .

Formula used:

Sf(x,y,z)dS=Df(r(u,v))|ru×rv|dA (1)

ru=xui+yuj+zuk (2)

rv=xvi+yvj+zvk (3)

As S consists of three surfaces: S1 be the lateral surface of the cylinder, S2 be the front formed by the plane x+y=5 and S3 be the back in the plane x=0 .

On S1 surface:

Let x=u , y=3cosv , z=3sinv and limits 0x5y and 0v2π .

To find limits of u .

Substitute 3cosv for y in x limits,

0u53cosv

r(u,v)=ui+3cosvj+3sinvk

Find ru .

Substitute u for x , 3cosv for y and 3sinv for z in equation (2),

ru=(u)ui+(3cosv)uj+(3sinv)uk=1i+0j+0k=1i

Find rv .

Substitute u for x , 3cosv for y and 3sinv for z in equation (3),

rv=(u)vi+(3cosv)vj+(3sinv)vk=0i+3(sinv)j+3cosvk=3sinvj+3cosvk

Find ru×rv .

ru×rv=(1i)×(3sinvj+3cosvk)=|ijk10003sinv3cosv|=(00)i(3cosv0)j+(3sinv0)k=3cosvj3sinvk

Find |ru×rv| .

|ru×rv|=|3cosvj3sinvk|=(3cosv)2+(3sinv)2=9cos2v+9sin2v=9(cos2v+sin2v)

Simplify the equation.

|ru×rv|=9 {cos2θ+sin2θ=1}=3

Find S1xzdS .

Modify equation (1) as follows.

S1xzdS=Dxz(|ru×rv|)dA

Apply limits and u for x , 3sinv for z and 3 for |ru×rv| ,

S1xzdS=02π053cosvu(3sinv)(3)dudv=902π053cosvu(sinv)dudv=902π[12u2]053cosvsinvdv=9202π[(53cosv)20]sinvdv

Simplify the equation.

S1xzdS=9202π(53cosv)2sinvdv (4)

Apply substitution method.

Consider 53cosv=t and dt=3sinvdv .

Find new limits.

tupper=53cos(2π)=53(1)=2tlower=53cos(0)=53(1)=2

Apply new limits and substitute 53cosv for t and 3sinvdv for dt in equation (4),

S1xzdS=9222(t)2dt3=922213(t)2dt=918[13t33]22=918[19(2323)]

Simplify the equation.

S1xzdS=918(0)=0

On S2 surface:

Consider x+y=5 plane.

Rearrange the equation.

x=5y

Then r(y,z)=(5y)i+yj+zk .

Find ry .

Modify equation (2) as follows.

ry=xyi+yyj+zyk

Substitute (5y) for x , y for y and z for z ,

ry=(5y)yi+(y)yj+(z)yk=1i+1j+0k=1i+1j

Find rz

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Chapter 16 Solutions

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