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

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-I<br />

SECTION 15.6 SURFACE AREA 0 267<br />

15.6 Surface Area<br />

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<br />

A(S) = ffD J[f,(x,y)] 2 + [! 11 (x,y)]2 + I dA= ffo -../3 2 +4 2 +IdA= ../26ffDdA<br />

= ../26 A( D) = ../26 (5)(3) = I5 ../26<br />

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<br />

{(x,y) I 0:5 x :5 2,0:5 y :53- ~x}. Thus<br />

A(S) = JJD J( -3) 2 + ( -2) 2 +IdA= .JI4 ffo dA = JI4 A( D ) = JI4 (~ · 2 · 3) = 3 .JI4<br />

5. y 2 + z 2 = 9 =* z = ~ · fx = 0, fv = -y(9 - y 2 ) - 112 =*<br />

A(S) = h412 Jo2 + [- y(9- y2)-1/2)2 + 1 dy dx = h4 h 2 J 9 ~2y2 + I dy dx<br />

= t { 2 3<br />

lo lo J9 - y 2 lo · 3 v=O<br />

7. z = f(x, y) = y 2 - x 2 with I ::; x 2 + y 2 ::; 4. Then<br />

dydx = 3 t [sin- 1 IL] 11 = 2 dx = 3 [(sin- 1 (~))x]~ = I 2sin- 1 (~)<br />

A( S) = J f D J 1 + 4x 2 + 4y 2 2 2<br />

dA = .{ 0<br />

2" f 1 Vf+"4r2 r dr dO = J:.,. dO J 1<br />

r -../I + 4r2 dr<br />

= (0]~" u2(I + 4r 2 ) 3 1 2 J: = {f(17vT7 - 5v'5)<br />

9. z = f(x, y) = xy with x 2 + y 2 :5 1, so / ex = y, f v =X =*<br />

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<br />

. D 0 J o 0 3 r=O<br />

= I:'lr H2-../2 - I) dO= 2 I(2-../2 - 1)<br />

A(S) = !L<br />

x2 +y2<br />

2 2 2+ IdA<br />

a - x - y<br />

"/2 1 acos6 / r2<br />

= /_ --+ 1rdrd0<br />

- 1r/2 o a2 - r2<br />

a coso ar<br />

= drdO<br />

/_<br />

- .r/2 o · Va 2 - r 2<br />

'lr f21<br />

= /_"/2 [-a Ja2...:. r2 ] r=acose dO<br />

- 1rj2 r=O<br />

1r/2 1 -rr/2<br />

= -a( J a2 - a 2 cos 2 0-a) dO = 2a 2 (I - Vl - cos 2 0) dO<br />

/_<br />

-1r/2 0<br />

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