   Chapter 16.7, Problem 13E

Chapter
Section
Textbook Problem

Evaluate the surface integral.13. ∫∫s z2dS, S is the part of the paraboloid x = y2 + z2 given by 0 ≤ x ≤ 1

To determine

To find: The value of Sz2dS .

Explanation

Given data:

x=y2+z2 and 0x1 .

Formula used:

Sf(x,y,z)dS=Df(r(u,v))|ru×rv|dA (1)

ry=xyi+yyj+zyk (2)

rz=xzi+yzj+zzk (3)

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

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

Substitute (y2+z2) for x ,

r(y,z)=(y2+z2)i+yj+zk

Find ry .

Substitute y2+z2 for x in equation (2),

ry=y(y2+z2)i+yyj+zyk=(2y+0)i+(1)j+(0)k=2yi+j

Find rz .

Substitute y2+z2 for x in equation (3),

rz=z(y2+z2)i+yzj+zzk=(0+2z)i+(0)j+(1)k=2zi+k

Find ru×rv .

ry×rz=(2yi+j)×(2zi+k)=|ijk2y102z01|=(10)i(2y0)j+(02z)k=i2yj2zk

Find |ry×rz| .

|ry×rz|=|i2yj2zk|=(1)2+(2y)2+(2z)2=1+4y2+4z2=1+4(y2+z2)

Find Sz2dS .

Modify equation (1) as follows.

Sz2dS=Dz2(|ry×rz|)dA

Apply limits and substitute 1+4(y2+z2) for |ry×rz| ,

Sz2dS=y2+z21z21+4(y2+z2)dA (4)

Let y=rcosθ , z=rsinθ and limits for r and θ are 0 to 1 and 0 to 2π .

Modify the equation (4) as follows

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