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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Evaluate the surface integral.

14. ∫∫s y2z2 dS, S is the part of the cone y = x 2 + z 2 given by 0 ≤ y ≤ 5

To determine

To find: The value of Sy2z2dS .

Explanation

Given data:

y=x2+z2 and 0y5 .

Formula used:

Sf(x,y,z)dS=Df(r(u,v))|ru×rv|dA (1)

rx=xxi+yxj+zxk (2)

rz=xzi+yzj+zzk (3)

Consider the general form of r(x,z) .

r(x,z)=xi+yj+zk

Substitute x2+z2 for y ,

r(x,z)=xi+x2+z2j+zk

Find rx .

Substitute x2+z2 for y in equation (2),

rx=xxi+x(x2+z2)j+zxk=i+(12(x2+z2)12(2x))j+(0)k=i+(xx2+z2)j

Find rz .

Substitute x2+z2 for y in equation (3),

rz=xzi+z(x2+z2)j+zzk=(0)i+(12(x2+z2)12(2z))j+(1)k=(zx2+z2)j+k

Find rx×rz .

rx×rz=[i+(xx2+z2)j]×[(zx2+z2)j+k]=|ijk1(xx2+z2)00(zx2+z2)1|=(xx2+z20)i(10)j+(zx2+z20)k=(xx2+z2)ij+(zx2+z2)k

Find |rx×rz| .

|rx×rz|=|(xx2+z2)ij+(zx2+z2)k|=(xx2+z2)2+(1)2+(zx2+z2)2=x2x2+z2+1+z2x2+z2=x2+z2x2+z2+1

Simplify the equation

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Chapter 16 Solutions

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