Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS D 325
39. The surfaceS is given by z = f(x, y) = 6- 3x - 2y which intersects the xy-plane in the line 3x + 2y = 6, soD is the
triangular region given by { (x, y ) I 0:::; x:::; 2, 0 :::; v $ 3 - ~x }. By Formula 9, the surface area of S is
I
A (S)= !fv 1+ (~Y + (~Y dA
= ffv J1 + ( - 3) 2 + (- 2)~ dA = Ji4J.(v dA = v'I4A(D ) = v'I4 (~ · 2 · 3) = 3 y'l4.
41 . Here we can write z = f(x, y) = i --:- i x- jy and Dis the disk x 2 + y 2 $ 3, so by Fonnula 9 the area of the surface is
A(S) =
1 ~v
1 + (:~r + (:~r dA 1 ~v = J1 + (-t/ + ( - ~) 2 dA = -41 1 ~v dA
= :lP A(D) = fl · 1r(y'3) 2 = ViA1r
43. z = f (x, y) = ~(x 3 1 2 + y 3 1 2 ) and D = {(x, y) I 0 $ x $ 1, 0 $ y $ 1 }. Then f, = x 1 1 2 , f v = y 1 1 2 and
A(S) = ffv J1 + (..fi) 2 + (.jfj) 2 dA = f 0
1
J; y'1 + x + ydydx
= fo 1 [t Cx + y + 1) 312 J::: dx = j f 0
1
[(x+ 2)
312 - (x + 1)
3 1
2 ] dx
= j [Hx + 2)5/2 _ ~(x + 1)5/ 2]: = Ji(35/2 _ 2s; 2 _ 2s~ 2 + 1) = A-( 3 1>/2 _ 2112 + 1)
45. z = J(x, y) = xy with x 2 + y 2 $ 1, so f, = y, fu = x =>
A(S ) = ffv Jl + Y 2 + X 2 dA = J;"' J; -.fr2TI rdrd(} = J:"' [ ~ (r 2 + I?' 2J::: d(}
= J;"' i(2 V2 - 1) d8 = 2 ; (2 .j2 - 1)
47. A parametric representation of the surface is x = x, y = 4x + z 2 , z = z with 0 :::; x :::; 1: 0 :::; z :::; 1.
Hence r :x: x r : = (i + 4 j ) X (2z j + k) = 4 i - j + 2z k.
Note: In general, ify = f(x, z ) then r , ~ r: = :~ i-j + ¥z k and A (S) = JL 1 + ( :~y + ( ~~y dA. Then
A(S) = J; J; ) 17 + 4z 2 1
dxdz = J 0
) 17 + 4z 2 dz
= %(z ) 17 + 4z 2 + ¥ lni2z + v'4z 2 + 17 I)] ~ = 4I + 1 ,t(ln(2 + v'2I) - ln v'I7]
49. r ., = (2u, v , 0), r , = (0, u, v), and r u x r ,. = (v 2 , - 2uv, 2u 2 ). Then
1
A(S) = ffv lr u x r vl dA = J 0
J; .../v 4 + 4u 2 v 2 + 4u 4 dv du = J; J; J(v 2 + 2u2)2 dv du
1
= J J; 0 + (v2 2u 2 ) dv du = J; (tv 3 + 2u 2 1
vJ ::~ du = J 0 (J + 4u 2 ) du = [ ~u + ~ u 3 ]~ = 4
51. From Equation 9 we have A(S) = ffv J l + (1, ) 2 + Uu) 2 dA. But if Jf,J :::; 1 and If vi :::; 1 then 0 :::; (f.,? :::; 1,
0 $ U u) 2 $ 1 => 1 $ 1 + U:~Y + Cfu) 2 $ 3 => 1 $ J 1 + (!:,) 2 + Uu) 2 $ v'3. By Property 15.3 .1 1,
ffv1dA $ ffvJl+(f,.)2+(f 11 )2 dA $ ffv ../3dA => A(D) $ A(S)$../3A(D) => .
1rR 2 $ A(S) $ ../31rR 2 •
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