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Chapter 13 Vector Calculus
Section 13.1 Direction Fields
Suppose we are given a first order differential equation of the form dy / dx = y ′= F x y ( , )
where F x y ( , )is some expression in x y ,. Separable equations are the special case in
which F x y ( , )can be factored as a function of x times a function of y. In some cases it is hard or impossible to find a formula for the solution, but still we can visualize the solution curves by means of a direction field. If a solution curve passes through a point ( , x y ) then its slope at that point is dy / dx = y ′= F x y ( , ). If we draw short line segments
with slope dy / dx = y ′= F x y ( , )at several points ( , x y ) , the result is called a direction field or slope field. These line segments indicate the direction in which a solution curve is heading, so the direction field helps us visualize the general shape of these curves.
Definition : Let D be a set in ℝ 2. A vector field on ℝ^2 is a function F that assigns to each point ( , x y ) in D a two-dimensional vector F ( , x y ).
Definition : Let E be a subset of ℝ^3. A vector field on ℝ^3 is a function F that assigns to each point ( , x y z , ) in D a three-dimensional vector F ( , x y z , ).
Examples:
- Sketch the direction field for y ′^ = F x y ( , ) = x^2 + y^2 − 1 and sketch the solution curve that passes through the origin. Solution: We calculate the following points: x -2 -1 0 2 -2 -1 0 1 1 2 -- -- y 0 0 0 0 1 1 1 1 0 1 -- -- y ′ 3 0 -1 3 4 1 0 1 0 4 -- --
- Sketch the vector field F x y ( , ) = ( x − y i ) + xj by drawing a diagram.
x 0 0 1 -1 0 -- -- -- -- y 0 1 0 0 -1 -- -- -- -- F <0,0>^ <-1,0>^ <1,1>^ <-1,-1>^ <1,0>^ --^ --^ --^ --
- Sketch the vector field F x y ( , )= yj by drawing a diagram.
- Sketch the vector field 2 2
yi xj F x y x y
by drawing a diagram.
- Sketch the vector field ( , ) (^2 )
yi xj F x y x y
by drawing a diagram.
- Sketch the vector field F x y ( , )= zj by drawing a diagram.
- Sketch the direction field of the scalar function f ( , x y ) = sin( x + y )by drawing a diagram. Remember that F x y ( , )= ∇ f
- Find the gradient field of f ( , x y ) = ln( x + 2 y )and sketch it. Remember to find F x y ( , )= ∇ f
9. Sketch the gradient field of φ ( , x y )= x + y
Section 13.2 Line Integrals
The first goal of this section is to define what it means to integrate a function along a curve. To motivate the definition we will consider the problem of finding the mass of a thin wire whose linear density function (mass per unit length) is known. We assure that we can model the wire by a smooth curve C between two points P and Q in 3-space. Given any point ( , x y z , )on C we let f ( , x y z , ) denote the corresponding value of the
density function. To compute mass of the wire divide C into n small sections using succession of distinct partition points P = p 0 (^) , p 1 (^) , p 2 , ⋯, pn = Q and
( *^ , *^ , *)
△ M k ≈ f xk yk zk △ Sk , where △ Mk be the mass of the kth section and △ S k be the length
between pk and pk − 1. We then find ( , , )
C
M = ∫ f x y z dS. And in 2-D ( , )
C
M = ∫ f x y dS
If C is a smooth curve parameterized by r t ( ) = x t i ( ) + y t ( ) , j a ≤ t ≤ b then
( , ) ( ( ), ( )) ( )
b
C a
∫^ f^ x y dS^ =∫ f^ x t^ y t^ r^ ′ t dt
Examples:
- Evaluate the line integral ( , ) , ( ) 2 , 0 1 C
∫^ f^ x y dS r t^ =^ ti^ +^ tj^ ≤^ t ≤
Solution:
1 3 0
b
C a
∫^ f^ x y dS^ =^ ∫ f^ x t^ y t^ r^ ′ t dt^ =^ ∫ +^ t^ dt =
- Evaluate the line integral ( 3 ) C
∫^ xy^ + z^ dS from (1, 0, 0) to^ ( 1, 0,−^ π)along the helix
C that is represented by x = cos , t y = sin , t z = t , 0≤ t ≤ π, and
dS = ( dx / dt )^2 + ( dy / dt ) 2 + 1 dt
Solution: 3 3 4 0
( ) (cos sin ) 2 2 / 4 C
xy z dS t t t dt
π
∫ +^ =^ ∫ +^ =^ π
- Evaluate the line integral (2 2 ) C
∫ +^ x y dS where C is the upper half of the unit circle
x^2^ + y^2 =1.
- Evaluate the line integral a) sin cos C
∫^ xdx^ + ydy where C is the arc consisting the
top half of the circle x^2 + y^2 = 1 from (1, 0) to ( 1, 0)− and the line segment from ( 1, 0) to ( 2,3)− −
Solution: 1 2
sin cos sin cos sin cos C C C
∫^ xdx^ +^ ydy^ =^ ∫ xdx^ +^ ydy^ +^ ∫ xdx^ + ydy
0 2
1
sin cos sin(cos )( sin ) cos(sin ) cos
sin cos(3 3)( 3)
cos1 cos 2 sin 3
C
xdx ydy t t dt t tdt
xdx x dx
π
−
−
- Evaluate , : 2 , 3 , 2 , 0 1 C
∫^ zdx^ +^ xdy^ +^ ydz C^ x^ =^ t^ y^ =^ t^ z^ =^ t^ ≤^ t ≤
Solution:
1 2 2 2 3 0
C
∫^ zdx^ +^ xdy^ +^ ydz^ =^ ∫ t^ t dt^ +^ t^ t^ dt^ +^ t^ t dt =
- ; ( , , ) , ( ) sin cos , 0 C
∫^ F dr ⋅^^ F x y z^ =^ zi^ +^ yj^ −^ xk r t^ =^ ti^ +^ tj^ +^ tk^ ≤^ t ≤^ π
Solution: 0
C
F dr F r t r t dt
π
∫ ⋅^ =^ ∫ ⋅^ ′ =^ π, where
F = zi + yj − xk = i cos t + j sin t − kt r , ′( ) t =< 1, − cos , t − sin t >
- Find the mass and center of mass of a thin wire in the shape of a quarter circle x^2^ + y^2 = r^2 , x ≥ 0, y ≥ 0 , if the density function is ρ ( , x y ) = x + y
Solution: x = r cos , t y = r sin , 0 t ≤ t ≤ π/ 2, ds = x ɺ^2 + y dt ɺ^2 = rdt
/ 2 2 2 0
( ) (sin cos ) 2 C
m x y ds r t t dt r
π
C^8
r x x x y ds m
C^8
r y y x y ds m
Section 13.3 Fundamental Theorem for Line Integrals
Fundamental Theorem for line integrals:
- ( ) ( ) ( )
b
a
∫^ F^ ′^ x dx^ =^ F b^ − F a
- C is a smooth curve given by ( ), , ( ( )) ( ( )) C
r t a ≤ t ≤ b ∫∇ f ⋅ dr = f r b − f r a
- In general C 1 (^) C 2
∫ ∇^ f^^ ⋅^ dr^ ≠^ ∫ ∇ f^^ ⋅ dr. But when^ ∇ f^ is continuous then
C 1 (^) C 2
∫ ∇^ f^^ ⋅^ dr^ =^ ∫∇ f^^ ⋅ dr
- F = Pi + Qj on an open simply connected region D and
P Q
y x
, then F is
conservative.
- Given F = Pi + Qj on an open simply connected region D and
P Q
y x
, then F is
conservative.
Examples:
- Given that F x y z ( , , ) = y i^2 + (2 xy + e^3^ z^ ) j + 3 ye^3 zk , find f such that F = ∇ f
Answer: f ( , x y z , ) = xy^2 + ye^3 z + C
- Is F x y ( , ) = ( x^2 + 4 xy i ) + (4 xy − y^3 ) j conservative vector field? If it is, find f
such that F = ∇ f
Solution: 4 , 4
P Q
x y y x
, not conservative
- Is F x y ( , ) = (1 + 2 xy + ln x i ) + ( x^2 ) j conservative vector field? If it is, find f such
that F = ∇ f
Solution: 2
P Q
x y x
. F is conservative.
Now F = ∇ f =< 1 + 2 xy + ln x x ,^2 >=< fx , fy >
Now you try to find that f ( , x y ) = x y^2 + x ln x + K , where K is a constant.
- Is
2 2 ( , ) 1 2 2 arctan , : ( ) 2 , 0 1 y F x y i j y x C r t t i tj t x
, conservative vector
field? If it is, find f such that F = ∇ f and evaluate C
∫^ F dr ⋅
Solution: a) (^2)
P Q y y x x
. F is conservative.
Now find that f ( , x y ) = y^2 arctan x + K
C C
F dr f dr f r b f r a
f r f r f f π
sin cos 0 C D
Q P
ydx x ydy dA x y
- Let D be a region bounded by a simple closed path C in the xy-plane. Use Green’s theorem to prove that the coordinates of the centroid ( , x y )of D are given as (^1 2) , (^12) (^2) C (^2) C
x x dy y y dx A A
= ∫ = − ∫ , where A is the area of D.
Solution: By Green’s theorem 2
2 C 2 D D
x dy xdA xdA x A A A
∫ =^ ∫∫ =^ ∫∫ = and
2 C 2 D D
y dy y dA ydA y A A A
- Find the centroid of a semicircular region of radius a
Solution: For the semicircle 2
A = π a. Now
2 2 2 2 2 2 0
cos cos 0
a
C a
x x dy x dy a t a tdt a a
π
2 2 2 2 2 0
0 sin ( sin 3
a
C a
a y y dx dx a t a tdt a a
π
- Evaluate the line integral 2 3 C
∫^ xy dx^ + x dy , where C is the rectangle with vertices
(0,0), (2,0), and (0,3) by a) directly, b) Green’s theorem Solution: a) Directly
1 2 3 4
2 3 2 3
3 2
0 0
C C C C C
xy dx x dy xy dx x dy
dt t dt
where
1 2 3 2
C x t y t C x y t t C x t y t C x y t t
b) By Green’s theorem
2 3 2 3 2 0 0
C
∫^ xy dx^ +^ x dy^ =^ ∫ ∫ x^ −^ xy dydx =
- Evaluate the line integral C
∫^ ydx^ − xdy , where C is the circle^ x^2^^ +^ y^2 =^1 by a)
directly, b) Green’s theorem
Solution: a) Directly: use x = cos , t y = sin t then 2 2 2 0
sin cos 2 C
ydx xdy tdt tdt
π
∫ −^ =^ ∫−^ −^ = −^ π
b) Green’s theorem: 2 2 (1)^2 C D
∫^ ydx^ −^ xdy^ =^ ∫∫−^ dA = −^ π
- Evaluate the line integral C
∫^ xdx^ + ydy , where C consists of line segments from
(0,1) to (0,0) and from (0,0) to (1,0)and the parabola y = 1 − x^2 from (1,0) to (0,1) by a) directly, b) Green’s theorem
a)
1 1 1 2 0 0 0
C
∫^ xdx^ +^ ydy^ =^ ∫ t^ −^ dt^ +^ ∫ tdt^ +^ ∫ t^ −^ dt^ +^ t^ −^ t^ −^ t dt =
b) Green’s theorem: 0 0 C D
∫^ xdx^ +^ ydy^ =^ ∫∫ dA =
- Use Green’s theorem to evaluate C
∫^ F dr ⋅. Check the orientation before applying
Green’s theorem. Given that F =< y^2 cos x x ,^2 + 2 y sin x > and C is the triangle from (0,0) to (2,6) to (2,0) to (0,0). Answer: -
Solution: Hint: Use 2 cos ( 2 2 sin ) C C
F dr y xdx x y x dy −
∫ ⋅^ = −^ ∫ +^ + and apply Green’s
theorem.
- A particle starts at the point (-2,0), moves along the x-axis to (2,0), and then along
the semicircle y = 4 − x^2 to the starting point. Use Green’s theorem to find the work done on this particle by the force field F =< x x ,^3 + 3 xy^2 >
Solution:
2
3 2
2 4 2 2 2 0 2 3 0 0
C C x
W F dr xdx x xy dy
x y dydx
r drd
π
−
−
- Given that F x y z ( , , ) = e x^ sin y i + e x cos y j + z k. Find curl F and div F. Solution: Use the definition of curl and div to get Curl F = O , a zero vector and div F = 1 a scalar.
- Is F x y z ( , , ) = zy i + xz j + xy k a conservative field? If it is find the scalar function f. Solution: Verify that Curl F = O and so F is conservative. And we have now that ∇ f = F =< yz xz xy , , > Use f (^) x = yz ⇒ f ( , x y z , ) = xyz + g y z ( , ) by integrating both sides with respect to x And again by differentiation with respect to y we have f (^) y = xz + g ′ ( ,^ y z ) ⇒ g ′( , y z ) = 0 ⇒ g y z ( , ) = h z ( ). Thus we have f ( , x y z , ) = xyz + h z ( ) ⇒ f (^) z ( , x y z , ) = xy + h ′( ) z ⇒ h z ( ) = K a constant. Finally we have f ( , x y z , )= xyz + K
- Is there a vector field G on ℝ^3 such that curl G = xy i^2 + yz^2 j + zx k^2? Explain. Solution: use the theorem that for F x y z ( , , )= P i + Q j + R k a vector field on ℝ^3 we have div curl F ( ) = 0 ⇒ ∇ ⋅ ∇ ×( F ) = 0. So we find that div curl G ( ) = y^2^ + z^2 + x^2 ≠ 0. Thus there is no such function G for which curl G = xy i^2 + yz^2 j + zx k^2.
- For a given vector field F if ∇ ⋅ F = 0 , the field F is incompressible. Show that the field F = f ( y z i , ) + g x z ( , ) j + h x y k ( , ) is incompressible. Solution: Just show that div F = 0
- Let r = x i + y j + z k , r = r the magnitude of r = x i + y j + z k , show that ∇ ⋅ ( rr ) = 4 r Solution: ∇ ⋅ ( rr ) = ( i ∂ / ∂ x + j ∂ / ∂ ∂ + y k ∂ / ∂ z ) ( ⋅ rxi + ryj + rzk )and now do the rest.
Section 13.6 Parametric surfaces and their areas We described a vector function of single variable r ( t ) which represents a space curve in vector form. Likewise we will describe a surface by a vector function r ( u , v ) of two parameters u and v. Suppose that r u v ( , ) = x u v i ( , ) + y u v ( , ) j + z u v k ( , ) is a vector valued
function defined on a region D in the uv -plane. We have x , y , z the components of r , are functions of two variables u and v with domain D. The set of all points ( x , y , z ) in ℝ^3 such that the parametric equations x = x u v ( , ), y = y u v ( , ), z = z u v ( , )and ( , ) u v varies throughout D, is called a parametric surface S.
Tangent planes: The tangent plane to a parametric surface S traces out by a vector function r u v ( , ) = x u v i ( , ) + y u v ( , ) j + z u v k ( , ) at a point p 0 with position vector
r 0 (^) ( u 0 (^) , v 0 )is given as the plane containing the tangent vectors ru , rv and the vector ru × rv is
the normal vector to the tangent plane. Remember that
ru = xu ( u 0 (^) , v 0 (^) ), yu ( u 0 (^) , v 0 (^) ), zu ( u 0 (^) , v 0 )
Examples
- Find a parametric representation of the surface represented by x^2 + y^2 + z^2 = a^2 Solution: Let us use spherical coordinates as
x = ρ sin φ cos θ , y = ρ sin φ sin θ , z = ρ cosφ, where ρ= a , 0 ≤ θ ≤ 2 π , 0≤ φ ≤π
The corresponding vector representation is
r ( , φ θ ) = ρ sin φ cos θ i + ρ sin φ sin θ j + ρ cosφ k.
- Find a parametric representation of the cylinder x^2 + y^2 = 4, 0 ≤ z ≤ 1 Solution: use cylindrical coordinates
x = r cos θ, y = r sin θ, z = z r ; = 2, 0 ≤ θ ≤ 2 π, 0 ≤ z ≤ 1
find the tangent plane to the surface with parametric equations x = u^2 , y = v^2 , z = u + 2 v at the point (1, 1, 3). Solution: ru = 2 , 0,1 u and rv = 0, 2 , 2 , v ru × rv = −4 , u −2 , 4 v uv. The tangent plane at the given point has the equation x + 2 y − 2 z + 3 = 0. Observe that the point (1, 1, 3) corresponds to u = 1, and v = 1.
Section 13.7 Surface Integrals
Formulas for this section:
1. ( , , ) ( , , ( , )) 1 ( / )^2 ( / )^2
S S
∫∫^ f^ x y z dS^ =^ ∫∫ f^ x y g x y^ + ∂ z^^ ∂ x^^ + ∂ z^^ ∂ y^ dA
- ( , , ) ( ( , )) (^) u v S S
∫∫^ f^ x y z dS^ =^ ∫∫ f^ r u v^ r^ × r dA
S S
∫∫^ F x y z^ ⋅^ dS^ =^ ∫∫ F x y z^ ⋅ n dS
Examples:
- Evaluate S
∫∫^ y dS , where S is the surface^ z^ =^ x^ +^ y^^2 , 0^ ≤^ x^ ≤^ 1, 0^ ≤^ y ≤^2
Solution: We have ∂ z / ∂ x = 1, ∂ z / ∂ y = 2 y and then 1 2 2 2 0 0
S D
∫∫^ ydS^ =^ ∫∫ y^ +^ +^ y^ dA^ =^ ∫ ∫ y^ +^ y dydx =
- Evaluate 2 S
∫∫^ x dS , where S is the unit sphere^ z^^2 +^ x^^2 +^ y^2 =^1
S S D
∫∫^ F dr ⋅^^ =^ ∫∫ curl F dS ⋅^^ =^ ∫∫ curl F^ ⋅ kdA
Examples
- Evaluate ,^2 S
∫∫^ F dr ⋅^^ F^ = −^ y i^ +^ xj^ + z k , where C is the curve of intersection of the
plane y + z = 2 and the cylinder x^2 + y^2 = 1
Solution: Find that curl F = (1 +2 ) y k 2 1
0 0
(1 2 ) (1 2 sin ) S S D D
F dr curl F dS curl F kdA y dA r rdrd
π
∫∫ ⋅^ =^ ∫∫ ⋅^ =^ ∫∫ ⋅^ =^ ∫∫ +^ =^ ∫ ∫ +^ θ^ θ^ =π
- F x y z ( , , ) =< ax^3^ − 3 xz^2 , x y^2 + by^3 , cz^3 > , where C is the curve of intersection of the hyperbolic paraboloid z = y^2 − x^2 ,and the cylinder x^2 + y^2 = 1 oriented counter clockwise as viewed from above, and C is the boundary of S. find a, b, c for which S
∫∫^ F dS ⋅ is independent of the choice of S.
Solution: Suppose G be any vector field such that F = curl G , then
S D
∫∫^ F dS ⋅^^ =^ ∫∫ curl G dS ⋅ depends only on the values of S and div curl G = 0.
So we have
2 2 2
P Q R
divF x y z a x by c z a b c
Section 16.9 Divergence Theorem
Let E be a simple solid region and let S be the boundary surface of E, given with positive (upward) orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E. Then
S E
∫∫^ F^ ⋅^ dS^ =∫∫∫ div F dV
Example 1. Find the flux of the vector field F x y z ( , , )= zi + yj + xk over the unit sphere
x^2^ + y^2 + z^2 = 1
Solution: 1 4 / 3 S E E
∫∫^ F dS ⋅^^ =^ ∫∫∫ div F dV^ =^ ∫∫∫ dV =^ π
