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

-I
SECTION 15.6 SURFACE AREA 0 267
15.6 Surface Area
1. Here z = f(x, y) = 2 + 3x + 4y and Dis the rectangle [0, 5) x [I, 4), so by Formula 2 the area of the surface is
A(S) = ffD J[f,(x,y)] 2 + [! 11 (x,y)]2 + I dA= ffo -../3 2 +4 2 +IdA= ../26ffDdA
= ../26 A( D) = ../26 (5)(3) = I5 ../26
3. 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 0:5 x :5 2,0:5 y :53- ~x}. Thus
A(S) = JJD J( -3) 2 + ( -2) 2 +IdA= .JI4 ffo dA = JI4 A( D ) = JI4 (~ · 2 · 3) = 3 .JI4
5. y 2 + z 2 = 9 =* z = ~ · fx = 0, fv = -y(9 - y 2 ) - 112 =*
A(S) = h412 Jo2 + [- y(9- y2)-1/2)2 + 1 dy dx = h4 h 2 J 9 ~2y2 + I dy dx
= t { 2 3
lo lo J9 - y 2 lo · 3 v=O
7. z = f(x, y) = y 2 - x 2 with I ::; x 2 + y 2 ::; 4. Then
dydx = 3 t [sin- 1 IL] 11 = 2 dx = 3 [(sin- 1 (~))x]~ = I 2sin- 1 (~)
A( S) = J f D J 1 + 4x 2 + 4y 2 2 2
dA = .{ 0
2" f 1 Vf+"4r2 r dr dO = J:.,. dO J 1
r -../I + 4r2 dr
= (0]~" u2(I + 4r 2 ) 3 1 2 J: = {f(17vT7 - 5v'5)
9. z = f(x, y) = xy with x 2 + y 2 :5 1, so / ex = y, f v =X =*
A(S) = JJ Jy 2 + x 2 + 1dA = t" r 1 -../r 2 +I r dr dO= J; 2 " [l(1·2 + 1) 3 1 2 ] r=l dO
. D 0 J o 0 3 r=O
= I:'lr H2-../2 - I) dO= 2 I(2-../2 - 1)
A(S) = !L
x2 +y2
2 2 2+ IdA
a - x - y
"/2 1 acos6 / r2
= /_ --+ 1rdrd0
- 1r/2 o a2 - r2
a coso ar
= drdO
/_
- .r/2 o · Va 2 - r 2
'lr f21
= /_"/2 [-a Ja2...:. r2 ] r=acose dO
- 1rj2 r=O
1r/2 1 -rr/2
= -a( J a2 - a 2 cos 2 0-a) dO = 2a 2 (I - Vl - cos 2 0) dO
/_
-1r/2 0
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