Multivariate Calculus Assignment: Comprehensive Problem Solutions

This assignment provides detailed solutions to several multivariate calculus problems. The first problem involves computing the mass of a body bounded by a parabolic cylinder and several planes using a triple integral. The second problem uses spherical coordinates to calculate a triple integral over ...Read More

Running head: MULTIVARIATE CALCULUS 1
Multivariate Calculus
Name
Institution

MULTIVARIATE CALCULUS 2
Multivariate Calculus
Question 1
Parabolic cylinder z=4−x2
Planes x=0 , y=6 , z=0
The density of the body is 1. Therefore, mass=volume
z=4−x2 =0 , 4=x2 , x=2
0 ≤ x ≤ 2, 0 ≤ y ≤ 6 ,∧0 ≤ z ≤ 4−x2
Mass=volume=∫
0
2
∫
0
6
∫
0
4− x2
dzdydx
∫
0
4 −x2
dz= [ z ] 0
4− x2
=4−z2
∫
0
6
4−x2 dy = [ (4−x2 ) y ] 0
6
=6(4−x2)
∫
0
2
6 (4−x2 )dx=6 [(4 x− x
3
3
) ]0
2
6 (4 ( 2 ) −2
3
3
)=32
Question 2
Since the double cone is twice a single cone, we work with the region R( Z >0) then double the
result. The region formed is shown in the figure below.

MULTIVARIATE CALCULUS 3
We know, r2=x2 + y2 + z2
¿ 1 ≤r2 ≤ 4 we get 1 ≤ r ≤ 2.
0 ≤ θ ≤2 π ,0 ≤ ∅ ≤ π
4 ,∧z=rcos ∅
∭
V
❑
z2 dv=∫
1
2
∫
0
2 π
∫
0
π
4
r4 cos2 ∅ sin ∅ d ∅ dθdr
Let cos ∅=u , d ∅ = −1
sin ∅ du
∫
0
π
4
r 4 cos2 ∅ sin ∅ d ∅ =¿ r4
∫
0
π
4
u2 sin ∅ ( −1
sin ∅ du)¿
¿−r 4
∫
0
π
4
u2 du=−r4
|u3
3 |0
π
4

MULTIVARIATE CALCULUS 4
−r 4
| u3
3 |0
π
4
=−r4
| cos3 ∅
3 |0
π
4
= r 4
3 ( 1− 1
2 √ 2 )
∫
0
2 π
r4
3 ( 1− 1
2 √ 2 ) dθ
¿ r4
3 (1− 1
2 √2 )∫
0
π
4
dθ
¿ r4
3 (1− 1
2 √2 )|θ |0
2 π
= 2 π r4
3 (1− 1
2 √2 )
∫
1
2
2 π r 4
3 ( 1− 1
2 √ 2 ) dr
¿ 2 π
3 ( 1− 1
2 √ 2 ) ∫
1
2
r 4 dr
¿ 2 π
3 (1− 1
2 √2 )|r5
5 |1
2
¿ 2 π
3 (1− 1
2 √2 ) (32
5 − 1
5 )
¿ 2 π
3 ( 1− 1
2 √ 2 ) ( 31
5 )
¿ 62 π
15 ( 1− 1
2 √ 2 )
For a double cone

MULTIVARIATE CALCULUS 5
¿ 2 × 62 π
15 ( 1− 1
2 √ 2 )
¿ 124 π
15 (1− 1
2 √2 )=16.7885
Hence,
∭
V
❑
z2 dv=16.7885
Question 3
Part i
∫
C
❑
xydx +2 x2 ydy x=et , y=e−t
dx=et dt , dy=−e−t dt
∫
C
❑
xydx +2 x2 ydy= ∫
t=0
2
et (−e−t )et dt +2 ( et ) 2
e−t (−e−t dt )
¿ ∫
t =0
2
( −e−t +2 t ) dt−(2 e2 t −t −t )dt
¿ ∫
t =0
2
( −et ) dt − ( 2 e0 ) dt
¿−1 ∫
t =0
2
( et +2 ) dt =−|et +2 t|0
2
=e2+2 ( 2 ) =e2+ 4
¿ 11.389

MULTIVARIATE CALCULUS 6
Part ii
∫
C
❑ 2 xy
√1+ 2 y ds , x=t , y= t 2
2
dx=1 dt , dy= 2t
2 =t
dx2=(1 dt)2=dt2
dy2=(tdt )2=t2 dt2
ds= √ dx2 +dy2= √ dt2+ t2 dt2
¿ √ dt2( 1+t2 )= √ (1+t2)dt
∫
C
❑ 2 xy
√1+ 2 y ds=∫
t=0
t=1 2(t)( t2
2 )
√1+ 2(t2
2 )
× √(1+t2 )dt
¿ ∫
t =0
t =1
t3
√1+t 2 × √(1+t2) dt
∫
t =0
t =1
t3 dt=|t4
4 |0
1
= 1
4
Question 4
r ( t )=cost ~i+sint ~j+t ~
k
F ( x , y , z )= y ~
i+ x ~
j+¿5
~
k
∫ F . d r =∫ F (r ( t ) ) . r (t)'

MULTIVARIATE CALCULUS 7
r (t )'= d
dt (cost ~
i+ sint ~
j+t ~
k)
r (t )' =−sint ~
i+cost ~
j+1 ~
k
x=cost , y=sint , z =t
Given that F ( x , y , z )= y ~
i+x ~
j+¿5
~
k
F ¿
F ( r ( t ) ) .r ( t ) '= ( sint ~
i+cost ~
j+5 ~
k ) .(−sint ~
i+cost ~
j+1 ~
k)
¿−sin2 t+cos2 t +5=cos2 t−sin2 t+5
But, using the identity cos2 t−sin2 t=cos (2 t)¿
F ( r ( t ) ) .r ( t ) '=cos ( 2 t ) +5
∫ F . d r =∫ F (r ( t )) . r (t)'=∫(cos ( 2 t )+5)dt
¿ ∫
t =0
t =1
(cos ( 2 t ) +5)dt
¿| 1
2 sin ( 2 t ) +5 t|0
1
=0.4546+5=5.4546
∫ F . d r =5.4546
Question 5
∬
S
❑
( x+ z ) ds, y2 +z2=9 ( first octant ) . Planes , x=0 , x=4

MULTIVARIATE CALCULUS 8
For the given surface we use θ∧x as the parameters. The parametric equations are:
x=x , y =rcosθ , z=rsinθ
The volume lies in the first octant which means that θ lies between 0 and π
2
y2 + z2=9
r2 cos2 θ+r2 sin2 θ=9
r2 ( cos2 θ+sin2 θ ) =9
r2=9 ,r =3
x=x , y =3 cosθ , z=3 sinθ
Therefore, 0 ≤ θ ≤ π
2 ∧0 ≤ x ≤ 4
rθ ×r x=
| i j k
∂ x /∂ θ ∂ y /∂ θ ∂ z /∂θ
1 0 0 |
¿
|i j k
0 −3 sinθ 3 cosθ
1 0 0 |
¿− j (−3 cosθ )+ k ( 3 sinθ )=−3 cosθ j +3 sinθ k
|rθ ×r x|= √ 9 ¿ ¿
We have,

MULTIVARIATE CALCULUS 9
∬
S
❑
( x+z ) ds=∬
D
❑
( x + z ) |rθ ×r x| dA
¿∫
0
π
2
∫
0
4
( x+ z ) ( 3 ) dxd θ
¿ 3∫
0
π
2
∫
0
4
( x+3 sinθ ) dxdθ
¿ 3∫
0
π
2
| x2
2 +3 ( x ) sinθ|0
4
dθ
¿ 3∫
0
π
2
( 42
2 +3 ( 4 ) sinθ )dθ
¿ 3∫
0
π
2
( 8+12 sinθ ) dθ
¿ 3 |8 θ−12 cosθ|0
π
2
¿ 3 ( 8 ( π
2 )−12 cos ( π
2 ) −8 ( 0 ) +12 cos 0 )
¿ 3(4 π +12)
¿ 12 π +36=73.699 units