Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] CÁC!" ### – #$%EGR&#S %R#'&S E(erc)cios *.5.
∫∫∫ xyz dV , onde T é o paralelepípedo retângulo 0,1 × 0, 2 × 1, 3 2
1. Calcule
T
Solu+,o: 1 2 3
1
2
3
dV = ∫ ∫ ∫ xyz dz dzdydx = ∫ xdx ∫ ydy ∫ z dz dz ∫∫∫ xyz dV 2
2
T
0 0 1
1
2
0
1
3
2
∫∫∫ ∫∫∫ T
0
z3 1 22 33 1 × = × × − 3 2 2 3 3 0 1 1 27 − 1 26 26 × 2 × = ∴ xyz2dV = xyz2dV = 3 2 3 3 T x2 y2 × xyz dV = 2 0 2
∫∫∫ ∫∫∫ T
2
∫∫∫
2. Calcule
∫∫∫ ∫∫∫ xdV , onde T é o tetraedro liitado pelo! plano! coordenado! e pelo plano x + 2y + z = " . T
Solu+,o:
-i. /0
1 #$%&'% #$%&'% C()' C()'% % * $%&(+ $%&(+ 62- 302/2 302/226 26 C()+ C()+ 6262- 216/ 216/66 66
-i. /1
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] y 2
z
="−x −
⇒0≤ z ≤"−x −
z
=0⇔"−x−
y 2
y 2
=0⇔
y 2
( $ig. 01)
= " − x ∴ y = −2x +
#!!i+
0 ≤ x ≤ " 0 y 2x ≤ ≤− + " −2x +
$ig. 02) ( $ig.
"− x −
y 2
∫∫∫ xdV = ∫ ∫ ∫ xdzdydx T
∫∫∫
0
0
0
"−x −
" −2x +
xdV
T
∫∫∫ ∫∫∫ xdV T
∫∫∫ xdV T
∫∫∫ xdV T
∫∫∫ xdV T
∫∫∫ ∫∫∫ xdV T
∫∫∫ xdV T
xz
∫∫∫ xdV = ∫ ( −"x T
0
∫∫∫ ∫∫∫ xdV = ∫∫∫ T
xdV
" −2 x +
− x − y dydx ∫0 0 ∫ 2 0 0 −2 x − + " −2x + " 2 2x xy x y dx = ∫ ∫ "x − x2 − dydx = ∫ "xy − x2y − ⋅ 2 2 2 0 0 0 0 −2x + " x 2 2 dx = ∫ "xy − x y − ⋅ y " 0 0 " x 2 = ∫ "x ( −2x + ) − x2 ( −2x + ) − ⋅ ( −2x + ) dx " 0 " x dx = ∫ ( −2x + ) "x − x2 − ( −2x + ) dx " 0 " x2 2 = ∫ ( −2x + ) "x − x + − 2x dx 2 0 " x2 = ∫ ( −2x + ) 2x − dx 2 0 =∫
"
T
y 2
x" "
dydx
=∫
x "
"
2
"x ) dx = ∫ ( x 3 + 16x − x 2 ) dx + x + 16x − "x
16x2 + 2
= 6" + 12 −
3
2
0
−
3 "
x 3
12 3
0
=
x" "
+ x −
= 12 −
2
12 3
=
x 3 3
"
=
"" "
∫∫∫
xdV
0
6" ∴ 3
T
+ ×" − 2
=
× "3 3
=
6" 3
2 #$%&'% #$%&'% C()' C()'% % * $%&(+ $%&(+ 62- 302/2 302/226 26 C()+ C()+ 6262- 216/ 216/66 66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected]
dxdydz
3. (xerc. 11- Calcule
∫∫∫ ( x + y + z + 1)
2
, onde T é o !lido deliitado pelo! plano! coordenado! e pelo plano
T
x + y + z = 2. Solu+,o:
-i. /0 -i. /1
z =2−x−y⇒0≤ z ≤2−x−y
( $ig. 01)
z = 0 ⇔ 2 − x − y = 0 ∴ y = −x + 2 #!!i+
0 ≤ x ≤ 2 0 ≤ y ≤ −x + 2
( $ig. 02)
3 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] 2
dxdydz
∫∫∫ ( x + y + z + 1)
2
T
4e !ol5endo
=∫
− x +2 2 − x − y
0
dzdydx
∫ ∫ ( x + y + z + 1 ) 0
2
0
dz
∫ ( x + y + z + 1)
2
+
$azendo + u
= x + y + z +1 →
du dz
= 1 ∴ dz = du
'u!tituindo + dz
∫ ( x + y + z + 1)
2
dz
∫ ( x + y + z + 1)
2
=∫ =−
du u2
= ∫ u du = −2
1 x+y
+z +1
u−1 ( −1)
1 u
+c=− +c
+c
" #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] 2
dxdydz
∫∫∫ ( x + y + z + 1)
2
=
T
dz
∫ ( x + y + z + 1)
2
dzdydx
∫ ∫ ∫ ( x + y + z + 1)
0
=−
− x +2 2 − x − y 0
2
0
1 x + y +z +1
+c
#!!i+ 2−x −y 1 ∫∫∫T ( x + y + z + 1) 2 = 0∫ 0∫ − x + y + z + 1 0 dydx 2 − x +2 dxdydz 1 1 = ∫ ∫ − − − dydx 2 ∫∫∫ + + + x y 0 1 + + − − + x y 2 x y 1 + + + x y z 1 ) T ( 0 0 2 − x +2 1 dxdydz 1 = − − − ∫∫∫T ( x + y + z + 1) 2 0∫ 0∫ 3 x + y + 1 dydx 2 − x +2 1 dxdydz 1 = − ∫∫∫T ( x + y + z + 1) 2 0∫ 0∫ x + y + 1 3 dydx 2 − x +2 dxdydz dy 1 −x +2 = − dy ∫∫∫T ( x + y + z + 1) 2 0∫ 0∫ x + y + 1 3 0∫ dx − x +2 2 dxdydz 1 ∫∫∫T ( x + y + z + 1) 2 = 0∫ ln ( x + y + 1) − 3 y 0 dx 2
dxdydz
− x +2
2
ln x − x + 2 + 1 − 1 −x + 2 − ln x + 0 + 1 + 1 ⋅ 0 = ) ( ) ) 3( 2 ∫∫∫ ∫ ( 3 + + + x y z 1 ) T ( 0 2 dxdydz ln 3 + 1 x − 2 − ln x + 1 dx = ∫∫∫T ( x + y + z + 1) 2 0∫ ( ) 3 3 ( ) dxdydz
∫∫∫ T
2
2
∫
∫
∫
∫
2
1 x2 2 = x ln ( 3) + ⋅ − x − ( x + 1) ln ( x + 1) − ( x + 1) 2 3 2 3 ( x + y + z + 1) 0 dxdydz
22 2 = 2ln ( 3) + − × 2 − ( 2 + 1) ln ( 2 + 1) − ( 2 + 1) − 1 2 6 3 ( x + y + z + 1)
∫∫∫ T
2
1 2 = + − ln 3 dx xdx dx − ln ( x + 1) dx ( ) 2 3 3 + + + x y z 1 ( ) 0 0 0 0
∫∫∫ T
2
dxdydz
dxdydz dxdydz
∫∫∫ ( x + y + z + 1)
2
= 2ln ( 3) + " − " − 3ln ( 3) + 3 − 1 = " − ln ( 3)
2
= " − ln ( 3)
6
T
∫∫∫
dxdydz
3
3
#$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected]
∫
"2S.: lnudu = u ⋅ lnu − u + c E(erc)cios *.3
∫∫∫ ( x
2
1. (xerc. 12- Calcular a integral
T
exterior ao cone
x2
+ y2 + z2 ) dV , !endo T a regio interior 8 e!9era x2 + y2 + z2 = e
+ y2 = z2 .
Solu+,o:
( coordenada! e!9érica!, teo! :ue+
x2
+ y2 + z 2 = ⇔ r 2!en 2θ co! 2 φ + r 2!en 2θ!en 2φ + r 2co! 2 θ = ⇔ r 2!en 2θ (co! 2 φ + !en 2φ ) + r 2co! 2θ =
#!!i+ r2!en2θ + r 2 co! 2 θ
= ⇔ r 2 ( co! 2 θ + !en 2θ ) = ⇔ r 2 = ∴ 0 ≤ r ≤ 3
e x2
+ y2 = z 2 ⇔ r 2!en 2θ co! 2 φ + r 2!en 2θ!en 2φ = r 2co! 2 θ ⇔ r 2!en 2θ (co! 2 φ + !en 2φ ) = r 2co! 2θ
#!!i + r2 !en2θ
= r 2 co!2 θ ⇔
!en2θ co!2 θ
= 1 ⇔ tan θ = ±1 ∴
3π "
≤θ≤
π "
)ogo +
0 ≤ r ≤ 3 3π π ;+ ≤θ≤ " " 0 ≤ φ ≤ 2π
6 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected]
'u!tituindo+
∫∫∫ ( x
2
+ y + z ) dV = 2
2
T
3π 2π " 3
2π
3π "
3
∫ ∫ ∫ r r !enθdrdθdφ = ∫ dφ ∫ !enθdθ∫ r dr 2 2
0
"
π 0
0
"
π
0
"
3
r 3π π 3 2 2 2 + + = φ − θ = π − − + − x y z dV co! 2 0 co! co! 0 ( ) ( ) ( ) π ∫∫∫ 0 " " " 0 T 2 2 2"3 "6 2π 2 2 2 + + = π × + × = x y z dV 2 ( ) ∫∫∫ 2 2 T "6 2π ∫∫∫ ( x2 + y2 + z2 ) dV = 2π
T
2. (xerc. 1"- Calcular
∫∫∫ dV , !endo T a ca!ca e!9érica deliitada por x T
2
+ y2 + z2 = e
Solu+,o:
( coordenada! e!9érica!, teo! :ue+
7 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
x2
+ y2 + z2 = 16 .
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] x2
+ y2 + z2 = ⇔ r 2!en 2θ co! 2 φ + r 2!en 2θ!en 2φ + r 2 co! 2 θ = ⇔ r 2!en 2θ (co! 2φ + !en 2φ ) + r 2co! 2θ =
#!!i+ r2!en2θ + r 2 co! 2 θ
= ⇔ r 2 ( co! 2 θ + !en 2θ ) = ⇔ r 2 = ∴ 0 ≤ r 1 ≤ 3
e x2
+ y2 + z2 = 16 ⇔ r 2!en 2θ co! 2 φ + r 2!en 2θ!en 2φ + r 2co! 2 θ = 16 ⇔ r 2!en 2θ (co! 2φ + !en 2φ ) + r 2co! 2 θ = 16
#!!i+ r2!en2θ + r 2 co! 2 θ
= 16 ⇔ r 2 ( co! 2 θ + !en 2θ ) = 16 ⇔ r 2 = 16 ∴ 0 ≤ r 2 ≤ "
)ogo +
3 ≤ r ≤ " ; + 0 ≤ θ ≤ π 0 ≤ φ ≤ 2π 'u!tituindo+ 2π
π "
2π
π
"
∫∫∫ dV = ∫ ∫ ∫ r !enθdrdθdφ = ∫ dφ∫ !enθdθ∫ r dr 2
T
2
0 0 3
0
0
3
r3 "3 33 = φ − θ = π − × − π + × dV co! 2 0 co! co! 0 ( )0 ( ) ( ) ( ) − ∫∫∫ 0 3 3 3 3 T 6" − 27 = "π × 37 = 1"π = π × + × dV 2 1 1 ( ) 3 ∫∫∫ 3 3 3 T 1"π dV = ∫∫∫
π
2π
3
T
3. (xerc. 17- Calcular
∫∫∫ xdV , !endo T a regio deliitada por x
2
T
2
Solu+,o: 1= Tran!9ora>o+
2
+ ( y − 3) + ( z − 2 ) = .
( x, y, z ) → ( u, 5, ?)
#$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] u = x ∴x = u dV =
5 = y − 3 ∴y = 5 + 3
? = z − 2 ∴z = ? + 2
∂ ( x,y,z ) dud5d? ∂ ( u,5,? )
%nde +
∂x ∂u ∂ ( x,y,z ) ∂y = ∂ ( u, 5, ? ) ∂u ∂z ∂u
∂x ∂5 ∂y ∂5 ∂z ∂5
∂x ∂? 1 0 0 ∂y = 0 1 0 =1 ∂? ∂z 0 0 1 ∂?
#!!i+
∫∫∫
xdV =
T
∫∫∫
u
T@
∂ ( x,y,z) dud5d? = ∂ ( u,5,? )
∫∫∫ udud5d? T@
%nde + T @ = u2 + 52 + ?2 = )ogo+
∫∫∫ xdV = ∫∫∫ udud5d? T
T @ + u2 + 52 + ?2 =
T@
→ ( r, φ, θ ) u = r!enθ co! φ 5 = r!enθ!enφ ? = r co! θ u2 + 52 + ?2 = ⇔ r2!en2 θ co!2 φ + r2 !en2 θ!en2φ + r2 co!2 θ = ⇔ r2 !en2θ (co!2 φ + !en2φ ) + r2 co!2 θ = 2= Tran!9ora>o + ( u,5,? )
#!!i+ r2!en2 θ + r2 co!2 θ
= ⇔ r2 ( co!2 θ + !en2θ ) = ⇔ r2 = ∴ 0 ≤ r ≤ 3
)ogo +
0 ≤ r ≤ 3 ; + 0 ≤ φ ≤ π 0 ≤ θ ≤ 2π
#$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] #!!i+
∫∫∫
xdV
∫∫∫
xdV
=
T
∫∫∫
=
udud5d?
T@
=
2π
∫
T
2π
π
3
∫ ∫∫
r!enθ co! φr2 !enθdrdθ dφ
0 0 0
π
3
0
0
∫
2π
∫
∫
=
π
3
0
0
∫
∫
!en2 θ dθ co! φ r3 dr
0
!en2 θdθ co! φ r3 dr
0
=
2π
1 − 1 co! 2θ dθπ co! φ dφ3 r3 dr ∫ 2 2 ( ) 0∫ ∫ 0 0
2π
1 θ − 1 !en 2θ × !enφ π × r " = xdV ( ) ( ) 0 ∫∫∫ 2 " 0 " 0 T 0 1 0 3" 1 1 − 0 = π × 0 × =0 xdV = ( 2π − 0) − !en ( "π ) − !en ( 0) × !en ( π ) − !en ( 0) × ∫∫∫ " " " T 2 ∫∫∫ xdV = 0 T
". (xerc. 1- Calcular
2 2 2 dV , onde T é elip!ide x + y + z = 1 . a2 2 c2
∫∫∫ T
Solu+,o: 1= Tran!9ora>o + ( x,y,z ) x
= au
y
= 5
z
→ ( u,5,? )
= c?
#!!i + x2 a2 e
+
y2 2
+
z2 c2
=1⇔
∂x ∂u x,y,z ∂( ) ∂y = ∂ ( u, 5, ? ) ∂u ∂z ∂u
∂x ∂5 ∂y ∂5 ∂z ∂5
a2u2 a2
+
2 52 2
+
c2 ?2 c2
= 1 ∴ T @ + u2 + 52 + ?2 = 1
∂x ∂? a 0 0 ∂y 0 0 ac = = ∂? ∂z 0 0 c ∂?
#!!i + dV
=
∂ ( x,y,z) dud5d? = acdud5d? ∂ ( u,5,?)
Ae!!a 9ora+
∫∫∫ dV = ∫∫∫ acdud5d? T
%nde + T @ + u2 + 52 + ?2 = 1
T@
10 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] → ( r, φ, θ ) u = r!enθ co! φ 5 = r!enθ!enφ ? = r co! θ u2 + 52 + ?2 = 1 ⇔ r2!en2 θ co!2 φ + r2 !en2θ !en2 φ + r2 co!2 θ = ⇔ r2 !en2 θ (co!2 φ + !en2 φ )+ r2 co!2 θ = 1 2= Tran!9ora>o + ( u, 5,? )
#!!i+ r2!en2 θ + r2 co!2 θ
= 1 ⇔ r2 ( co!2 θ + !en2θ ) = 1 ⇔ r2 = 1∴ 0 ≤ r ≤ 1
)ogo +
0 ≤ r ≤ 1 ; + 0 ≤ φ ≤ π 0 ≤ θ ≤ 2π #!!i+ 2π
π 3
π
2π
3
∫∫∫ dV = acdud5d? = ∫ ∫ ∫ acr !enθdrdθdφ = ac∫ !enθdθ ∫ dφ∫ r dr 2
T
2
0 0 0
π
2π
0
0
0
3
∫∫∫ dV = ac∫ !enθdθ ∫ dφ∫ r dr 2
T
0
0
0
r3 = × − θ × φ × dV ac co! ( ) ( ) 0 = ac × − co!( π) + co!( 0) × ( 2π − 0) ∫∫∫ 0 3 0 T 1 "π ∫∫∫ dV = ac × ( 1 + 1) × 2π × = ac π
2π
3
T
∫∫∫ dV = T
13 × − 0 3
3
"π ac 3
. (xerc. 20- Calcular
∫∫∫ ( x − 2y) dV , !endo T a regio deliitada por+ T
2
2
( x − 1 ) + ( y − 2 ) = 1, z = 0 e z = x + y . Solu+,o:
→ ( u, 5 ) 5 = y − 2 ∴y = 5 + 2
1= Tran!9ora>7o + ( x,y ) u
= x − 1∴ x = u + 1
z
=0
e
z
=u + 5 +3
#!!i+ 2
2
( x − 1) + ( y − 2 ) = 1 ⇔ u2 + 52 = 1 ∴ 4 = u2 + 52 = 1 e
∂x ∂ ( x, y ) ∂u = ∂ ( u, 5 ) ∂x ∂5
∂y ∂u = 1 0 = 1 ∴ d' = ∂ ( x, y) dud5 = dud5 ∂ ( u, 5) ∂y 0 1 ∂5
11 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] #!!i+
u + 5 + 3 ∫∫∫T ( x − 2y ) dV = ∫∫4 ∫0 ( u + 1 − 25 − ") dzdud5 u + 5 + 3 − = − − x 2y dV u 25 3 ) ) ∫ dz dud5 ∫∫∫T ( ∫∫4 ( 0 u+5+3 ∫∫∫ ( x − 2y ) dV = ∫∫ ( u − 25 − 3) dud5 ( z) 0 T
4
∫∫∫ ( x − 2y ) dV = ∫∫ ( u − 25 − 3) ( u + 5 + 3) dud5 T
4
Ba!+ u=
ρ co! φ
5
= ρ!enφ
0 4+ 0
≤ ρ ≤1 ≤ φ ≤ 2π
#!!i+
∂u ∂5 ∂ ( u, 5 ) ∂ρ ∂ρ co! φ !enφ = = = ρ co!2 φ + ρ!en2 φ ∂ ( ρ, φ) ∂u ∂5 −ρ!enφ ρ co! φ ∂φ ∂φ 1 ∂ ( u, 5 ) = ρ ( co!2 φ + !en2φ) = ρ ∂ ( ρ, φ) #!!i+ dud5
=
∂ ( u, 5 ) dρdφ = ρdρd φ ∂ ( ρ, φ)
'u!tituindo +
∫∫∫ ( x − 2y ) dV = ∫∫ ( u − 25 − 3) ( u + 5 + 3) dud5 T
4
2π 1
∫∫∫ ( x − 2y ) dV = ∫ ∫ ( ρ co! φ − 2ρ!enφ − 3) ( ρ co! φ + ρ!enφ + 3) ρd ρdφ < = ∫∫∫ ( x − 2y) dV T
0 0
T
2π 1
<
=
∫ ∫ ( ρ
2
co!2 φ + ρ2 co! φ!enφ + 3ρ co! φ
− 2ρ2!enφ co! φ − 2ρ2!en2φ − 6ρ!en φ − 3ρ co! φ − 3ρ!enφ − .) ρdρdφ
0 0
<
=
2π 1
∫ ∫ ( ρ
2
co!2 φ − ρ2 co! φ!enφ − 2ρ2!en2φ − .ρ!en φ − .) ρd ρd φ
0 0
4e!ol5endo a! !eguinte! integrai!+
12 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] 2π 1
<1 =
∫ ∫(ρ
2
co! φ ) ρd ρdφ = 2
0 0
2π 1
∫ ∫ ( co!
2
φ) ρ3 d ρd φ
0 0
2π
1
2π
ρ" 1 1 2 3 <1 = ∫ co! φdφ∫ ρ dρ = ∫ + co! ( 2 φ) d φ × 2 2 " 0 0 0 0 1
2π
0 1 π 1 1 π 1 1 <1 = φ + !en ( 2φ ) × − 0 = × 2π × = ∴ <1 = " " " " 2 2 0 "
<2 =
2π 1
∫ ∫ ( −ρ
2
1
0
0
∫
∫
co! φ!en φ) ρd ρd φ = − co! φ!en φd φ ρ3 d ρ
0 0
2π
1
ρ" !en2 φ <2 = − × " 0 2 0 2π 1
<3 =
2π
1 = " × 0 = 0 ∴ <2 = 0 2π 1
∫ ∫ ( −2ρ !en φ) ρdρdφ = −2 ∫ ∫ !en φρ dρd φ 2
2
0 0
2
3
0 0
1
2π
ρ" 2 π 1 1 3 2 <3 = −2∫ ρ dρ ∫ !en φdφ = −2 × × ∫ − co! ( 2 φ) dφ " 0 0 2 2 0 0 1
2π
0 π π 1 1 1 1 1 <3 = −2 × × φ − !en ( 2φ) = − × × 2π = − ∴ <3 = − " 2 " 2 2 2 2 0 2π 1
<" =
2π 1
∫ ∫ ( −ρ!enφ)ρdρdφ = − ∫ ∫ !en φρ d ρd φ 2
0 0
0 0
2π
1
ρ3 2π <" = −∫ ρ d ρ × ∫ !enφd φ = − × × ( − co! φ) 0 3 0 0 0 1
2
<" = − ×
1 × − co! ( 2π) + co! ( 0 ) = ( −3 ) × ( 0 ) = 0 ∴ <" = 0 3
13 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] 2π 1
2π 1
2π
1
0 0
0
0
∫ ∫ ( −)ρdρdφ = − ∫ ∫ ρdρdφ = − ∫ dφ∫ ρd ρ
< =
0 0
1
2π
< = − × ( φ ) 0
ρ2 × = − × 2π × 1 = − π ∴ < = − π 2 2 0
#!!i+ <=
2π 1
∫ ∫ ( ρ
2
co!2 φ − ρ2 co! φ!en φ − 2 ρ2 !en2 φ − ρ!en φ − ) ρd ρd φ
0 0
< = <1 + <2 + <3 + <" + < <=
π
+0−
" )ogo +
π
+ 0 − π = −
2
∫∫∫ ( x − 2y ) dV = − T
37π "
37π "
E(erc)cios *.9 1. (xerc.0"- Calcular o 5olue do !lido liitado por
z = − x2 − 2y2 , no prieiro octante.
Solu+,o: Teo! :ue+
z z
= − x2 − 2y 2 = 0+
−x
2
− 2y = 2
x2 0 ⇔ x + 2y = ∴ 4 + 2
2
+
y2 "
=1
= 0, y = 0 + z = − x 2 − 2y 2 ⇔ z = − 0 − 0 = ∴ z = x
#!!i+
∫∫∫ dV = ∫∫ dxdy ∫ dz = ∫∫ dxdy T
4
0
4
%nde+ 4+
x2
+
y2 "
=1
∫∫
4e !ol5endo dxdy, onde 4 + 4
x2
+
y2 "
=1
1" #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] 1= Tran!9ora>7o + ( x,y ) x = 2 2u #!!i+ x2 e
+
y2 "
y
→ ( u, 5)
= 25
=1 ⇔
∂x ∂ ( x, y ) ∂u = ∂ ( u, 5 ) ∂y ∂u
u2
+
"52 "
= 1 ∴ 4 @ + u2 + 52 = 1
∂x ∂5 = 2 2 0 = " 2 ∂y 0 2 ∂5
#!!i+ dxdy
=
∂ ( x, y ) dud5 = " 2dud5 ∂ ( u, 5 )
'u!tituindo+
∫∫
dxdy 4
= ∫∫ " 2dud5 onde 4 @ + u2 + 5 2 = 1 4@
2= Tran!9ora>o + ( u,5 )
→ ( ρ, φ)
= ρ co! φ 5 = ρ!enφ 0 ≤ ρ ≤ 1 4@+ π 0 ≤ φ ≤ 2 u
e
∂u ∂5 ∂ ( u, 5 ) co! φ !enφ ∂ρ ∂ρ dud5 = dρdφ = d ρd φ = d ρd φ ∂ ( ρ, φ ) ∂u ∂5 −ρ!enφ ρ co! φ ∂φ ∂φ dud5 = ( ρ co!2 φ + ρ!en2 φ ) d ρd φ = ρd ρd φ #!!i+ 1
π
∫∫
dxdy 4
π ρ 1 π = 32 2 ∫ ∫ ρdρdφ = 32 2 × × ( φ ) 02 = 32 2 × × = 2 π 2 2 2 0 0 0 2 1
2
)ogo+
∫∫∫ dV = ∫∫ dxdy = T
2π ∴ V = 2π
4
1 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected]
2. (xerc. 0- Calcular o 5olue do !lido acia do plano xy deliitado por
z = x2 + y2 e x2 + y2 = 16 .
Solu+,o: Teo! :ue+
= x2 + y 2 e x2 + y 2 = 16 = 0+ x2 + y2 = 0 ∴ ( 0,0, 0 ) x = 0, y = 0 + z = 16 ∴ ( 0, 0,16 ) z z
#!!i+ 16
∫∫∫ dV = ∫∫ dxdy ∫ dz = 16 ∫∫ dxdy T
4
0
4
%nde+ 4 + x2
+ y2 = 16
∫∫
4e ! ol5endo 16 dxdy, onde 4 + x 2
+ y 2 = 16
4
Tran!9ora>o + ( x, y )
→ ( ρ, φ )
= ρ co! φ y = ρ!en φ 0 ≤ ρ ≤ " 4@+ 0 ≤ φ ≤ π x
e
∂u ∂5 ∂ ( u, 5 ) co! φ !enφ ∂ρ ∂ρ dxdy = dρdφ = dρdφ = d ρdφ ∂ ( ρ, φ ) ∂u ∂5 −ρ!enφ ρ co! φ ∂φ ∂φ dxdy = ( ρ co!2 φ + ρ!en2φ ) d ρd φ = ρd ρd φ #!!i+ π "
π
"
ρ2 16∫∫ dxdy = 16 ∫ ∫ ρdρd φ = 16 ∫ d φ∫ ρd ρ = 16 × ( φ ) 0 × 2 0 4 0 0 0 0
∫∫
16 dxdy
= 16 × π ×
4
"2 2
"
π
= 12 π
)ogo+
∫∫∫ dV = 16∫∫ dxdy = 12π ∴ V = 12π T
4
16 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected]
3. (xerc. 06- Calcular o 5olue do !lido acia do paraolide
z = x2 + y2 e aaixo do cone z =
Solu+,o: Teo! :ue+
17 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
x2 + y2
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] z
= x2 + y2
z
= x2 + y2
#!!i+
∫∫∫ dV = ∫∫∫ dxdydz T
T
= r!enθ co! φ y = r!en θ!en φ z = r co! θ z = x2 + y2 ⇔ r co! θ = r 2!en2 θ co!2 φ + r 2!en2 θ!en2 φ x
r co! θ
= r 2!en2 θ ( co!2 φ + !en2 φ) ⇔ r co! θ = r 2!en2 θ ∴r =
co! θ !en2 θ
e z
= x2 + y2 ⇔ r co! θ = r 2!en2 θ co!2 φ + r 2!en2 θ!en2 φ
r co! θ
= r2!en2θ ( co!2 φ + !en2 φ) = r 2!en2 θ = r!en θ
#!!i+ r co! θ
= r!enθ ⇔
!enθ co! θ
= 1 ⇔ tan θ = 1 ∴ θ =
π "
#!!i+
0 ≤ θ ≤ π " co! θ T + 0 ≤ r ≤ !en2 θ 0 ≤ φ ≤ 2π 'u!tituindo + π co! θ 2 π " !en2θ
∫∫∫ dV = ∫∫∫ dxdydz = ∫ ∫ ∫ r !enθdrdθd φ T
T
0 0
co! θ
∫∫∫ dV = ( φ) T
2π 0
T
0
π
r3 !en θ " × × ∫ !enθdθ 3 0 0 2
3
π
∫∫∫
2
π
1 co! θ !en θd θ dV = 2π × × 3 0 !en2 θ "
∫
π
2π
"
∫∫∫ dV = 3 ∫ cot an θ co! !ec T
2
1
2π 3 − u du 3 0
∫
=
2π − 3
u" × "
1
0
∫
du θ − 2 co!!ec θ 2π 1 π π = − × = − ∴V =
θd θ =
0
∫∫∫ dV = T
3
2 π co! 3 θ dθ = 3 0 !enθ "
2π u3 co! !ec 2 3
∫
3
"
6
6
". (xerc. 11- Calcular o 5olue do !lido deliitado pela! !uper9ície!
x2 + y2 = 16, z = 2, x + z =
1 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] Solu+,o: Teo! :ue+
x2 + y2 = 16, z = 2, x + z = ∴ z = − x #!!i+ −x
∫∫∫ dV = ∫∫ ∫ dzdxdy = ∫∫ ( − x − 2) dxdy T
4
2
4
2π "
2π "
2π "
∫∫∫ dV = ∫ ∫ ( 7 − ρ co! φ) ρdρdφ = 7 ∫ ∫ ρdρdφ − ∫ ∫ ρ co! φd ρd φ 2
T
0 0
0 0
" ρ3 " 2π ρ2 dV = 7 × × ( φ ) 0 − ∫∫∫ 3 0 2 0 T
0 0
0 2π × ( !enφ) 0
#!!i+ "2 × 2π = 112π ∴ V = 112π dV = 7 × 2
∫∫∫ T
. (xerc.13- Calcular o 5olue do !lido deliitado pela! !uper9ície!
z = 2x2 + y2 e z = " − 3x2 − y2 .
Solu+,o: Teo! :ue+
z = 2x2 + y2 e z #!!i +
= " − 3x 2 − y 2
2x2 + y2 = " − 3x 2 Ai5idindo por " + x2 "
+
2y2 "
=1⇔
− y 2 ⇔ 2x 2 + 3x 2 + y 2 + y 2 = " ⇔ x 2 + 2y 2 = "
x2 "
+
y2 2
" − 3x2 − y2
∫∫∫ dV = ∫∫ ∫ T
4
2x2 + y2
dzdxdy
= 1 ∴4 +
x2 "
+
y2 2
=1
= ∫∫ ( " − 3x2 − y2 − 2x 2 − y 2 )dxdy 4
#!!i +
∫∫∫
dV
T
= ∫∫ ( " − x2 − 2y2 ) dxdy, on de 4 + 4
x2 "
+
y2 2
=1
1 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] 1= Tran!9ora>o ( x, y ) x
=
2 u
dxdy
=
y
→ ( u, 5 )
= 25
∂ ( x, y ) dud5 ∂ ( u, 5 )
%nde +
∂x ∂ ( x, y ) ∂u = ∂ ( u, 5 ) ∂x ∂5
∂y 2 ∂u = ∂y 0 ∂5
∫∫∫ dV = ∫∫ ( " − x T
0
=
2
− 2y 2 ) dxdy =
2
4
2 10
∫∫ ( " − "u
2
4@
− "5 2 )
2 10 dud5, onde 4 @ + u 2 + 5
2
=1
#!!i+
∫∫∫ dV = T
2 10
∫∫ ( " − "u
2
− "5 2 ) dud5, onde 4 @ + u 2 + 5 2 = 1
4@
2= Tran!9ora>o + ( u, 5 ) → ( ρ, φ ) u = ρ co! φ
5 = ρ!en φ
0 ≤ ρ ≤ 1 4@+ 0 ≤ φ ≤ 2π #!!i+
∫∫∫
dV =
T
∫∫∫
dV =
T
2 10 2 10
∫∫ ( " − "u
− "52 ) dud5
2
4@
2π 1
∫ ∫ ( " − "ρ
2
co!2 φ − "ρ2 !en2 φ) ρd ρd φ
0 0
#!!i+
∫∫∫ T
2π 1 2π 1 2 10 2 π 1 3 2 dV = "ρdρdφ − " ρ co! φdρd φ − " ρ3 !en2 φd ρd φ 0 0 0 0 0 0
∫∫
∫∫
∫∫
20 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected]
4e!ol5endo a!
∫∫∫ T
π 2 10 π " π 3 2 ρ co! φd ρd φ − " ρ3 !en2 φd ρd φ dV = "ρdρdφ − 0 0 00 0 0 2 1
2 1
∫∫
∫∫∫ dV = T
2 1
∫∫
∫∫
2 10 ( <1 + <2 + <3 )
%nde + 2π 1
2π
1
ρ2 12 = " π ∴ <1 = " π <1 = " ∫ ∫ ρdρdφ = " ∫ dφ ∫ ρd ρ = " × ( φ) 0 × = " × 2 π × 2 2 0 0 0 0 0 2π 1 1 2π " " 1 1 3 2 3 <2 = − ∫ ∫ ρ co! φd ρd φ = − × ∫ ρ d ρ × ∫ + co! ( 2φ ) dφ 00 0 2 2 0 1
2π
0 1 2 π " ρ 1 1 <2 = − × × φ + !en ( 2φ) = ( − " ) × × = −π ∴ <2 = −π " 0 2 " " 2 0 "
1
e 2π 1
<3 = −"
2π
1
1 1
∫ ∫ ρ !en φdρdφ = −" × ∫ ρ dρ × ∫ 2 − 2 co! ( 2φ) d φ 3
2
0 0
3
0
0
0 ρ 1 1 2 π 1 <3 = −" × × φ − !en ( 2φ) = ( − " ) × × = −π ∴ <3 = −π " 2 " " 2 0 0 "
1
'u!tituindo +
∫∫∫
dV =
T
2 10 2 10 2 10 " 10 " 10 × 2π = ∴V = ( <1 + <2 + <3 ) = ( "π − π − π ) =
21 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected]
6. (xerc. 10- Calcule o 5olue da parte da e!9era,
x2 + y2 + z2 = ,
entre o! plano! z 1D z 2. (!oce a
regio.
Solu+,o:
z = 1+
x2 + y2 + z2 = ⇔ x2 + y2 + 12 = ⇔ x2 + y2 + 1 = ∴ x2 + y2 =
z = 2+
x2 + y2 + z2 = ⇔ x2 + y2 + 22 = ⇔ x2 + y2 + " = ∴ x2 + y2 =
<- CElculo de V1+
22 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected]
Ftilizando coordenada! cilíndrica!+ 2 2π
V1
2π 3
= ∫ ∫ ∫ rdrdθdz + ∫ ∫ ∫ rdzdrd θ 0 0 0
V1
. −r2
=r
2
2
0
( ) × 2π × 2 = 2
0
0
2 2π 3
× " π + ∫ ∫ r . − r2 drd θ 0
3 0 1 3 3 3 1 1 2 V1 = 10π + 2π − ( . − r ) = 11 π − 2 π ( . − . ) − ( . − ) 3 3 3 = π + π × = π + 16π = "6π ∴ = "6π
V1
10
2
3
10
3
3
V1
3
<<- CElculo de V2+
23 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected]
Ftilizando coordenada! cilíndrica!+ 1 2π
V2
= ∫ ∫ ∫ rdrdθdz + ∫ ∫ ∫ rdzdrd θ 0 0 0
V2
. −r2
2π 3
=r
2
2
0
× 2π × 1 = 0
(
0
)
2
2
2π 3
× 2 π + ∫ ∫ r . − r2 drd θ 0
3 0 1 3 3 3 1 1 2 = π − 2 π ( . − . ) − ( . − ) V2 = π + 2 π − ( . − r ) 3 3 3 = π + π × 1 = π + 2π = 26π ∴ = 26π
V2
2
3
3
3
V2
3
#!!i+ V = V1 − V2 =
"6π 26π 20π 20π − = u.5. ∴ V = u.5. 3 3 3 3
4e!olu>o de
∫ r
− r2 dr
2" #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected]
∫ r
− r2 dr = G
$azendo + u = − r2 ⇒ u2 = − r2 → 2u
du = − 2r ∴ rdr = −udu dr
'u!tituindo + u3 ∫ r − r dr = ∫ u ( −udu) = − ∫ u du = − 3 + c #!!i+ 3 1 2 2 ∫ r − r dr = − 3 − r + c 2
2
(
)
2 #$%&'% C()'% * $%&(+ 62- 302/226 C()+ 62- 216/66
