   Chapter 16.7, Problem 15E

Chapter
Section
Textbook Problem

Evaluate the surface integral.15. ∫∫s x dS, S is the surface y = x2 + 4z, 0 ≤ x ≤ 1, 0 ≤ z ≤ 1

To determine

To find: The value of SxdS .

Explanation

Given data:

y=x2+4z , 0x1 and 0z1 .

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+4z for y ,

r(x,z)=xi+(x2+4z)j+zk

Find rx .

Substitute x2+4z for y in equation (2),

rx=xxi+x(x2+4z)j+zxk=i+(2x+0)j+(0)k=i+(2x)j

Find rz .

Substitute x2+4z for y in equation (3),

rz=xzi+z(x2+4z)j+zzk=(0)i+(0+4)j+(1)k=4j+k

Find rx×rz .

rx×rz=[i+(2x)j]×[4j+k]=|ijk12x0041|=(2x0)i(10)j+(40)k=(2x)ij+4k

Find |rx×rz|

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