Multivariate Calculus

This article covers solved problems related to Multivariate Calculus including finding mass, volume, and surface area. It also includes solved problems related to double integrals and line integrals.

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