   Chapter 16.7, Problem 8E

Chapter
Section
Textbook Problem

Evaluate the surface integral.8. ∬s (x2 + y2) dS, S is the surface with vector equation r(u, v) = ⟨2uv, u2 − v2, u2 + v2⟩, u2 + v2 ≤ 1

To determine

To find: The value of S(x2+y2)dS .

Explanation

Given data:

r(u,v)=2uv,u2v2,u2+v2 and u2+v21 .

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)

Find ru .

Substitute 2uv for x , u2v2 for y and u2+v2 for z in equation (2),

ru=u(2uv)i+u(u2v2)j+u(u2+v2)k=(2v)i+(2u0)j+(2u+0)k=2vi+2uj+2uk=2v,2u,2u

Find rv .

Substitute 2uv for x , u2v2 for y and u2+v2 for z in equation (3),

rv=v(2uv)i+v(u2v2)j+v(u2+v2)k=(2u)i+(02v)j+(0+2v)k=2ui2vj+2vk=2u,2v,2v

Find ru×rv .

ru×rv=2v,2u,2u×2u,2v,2v=|ijk2v2u2u2u2v2v|=(4uv+4uv)i(4v24u2)j+(4v24u2)k=8uvi(4v24u2)j+(4v24u2)k

ru×rv=8uv,(4v24u2),(4v24u2)

Find |ru×rv| .

|ru×rv|=|8uv,(4v24u2),(4v24u2)|=(8uv)2+(4v24u2)2+(4v24u2)2=64u2v2+(16v432v2u2+16u4)+(16v4+32v2u2+16u4)=64u2v2+32v4+32u4

Simplify the equation.

|ru×rv|=32(2u2v2+v4+u4)=32(u2+v2)2 {(a+b)2=a2+b2+2ab}=42(u2+v2)

Find S(x2+y2)dS

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