Calculus III
Philippe Rukimbira
Department of Mathematics
Florida International University
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13.3 Partial Derivatives
Example:
f (x, y ) = (x 2 + 2xy )5
The partial derivative of f with respect to x is:
∂f
(x, y ) = 5(x 2 + 2xy )4 (2x + 2y ).
∂x
What is
∂f
∂y ?
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13.3 Partial Derivatives
Example:
f (x, y ) = (x 2 + 2xy )5
The partial derivative of f with respect to x is:
∂f
(x, y ) = 5(x 2 + 2xy )4 (2x + 2y ).
∂x
What is
∂f
∂y ?
The answer is:
5(x 2 + 2xy )4 (2x).
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The second order partial derivative of f with respect to y is:
∂2f
(x, y ) = 40(x 2 + 2xy )3 (2x)(2x).
∂y 2
∂2f
∂x∂y
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∂ ∂
=
∂x ( ∂y f )
∂
2
4
=
∂x (10x(x + 2xy ) )
= 10(x 2 + 2xy )4 + 40x(x 2 + 2xy )3 (2x)
=
20(x 2 + 2xy )3 (5x 2 + xy )
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What is
∂2f
?
∂x 2
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13.4 Differentiability, Differentials and local linearity
Denote by ∆f = f (x0 + ∆x, y0 + ∆y ) − f 9X0 , Y0 ).
f is said to be differentiable at (x0 , y0 ) if:
∆f =
∂f
∂f
(x0 , y0 )∆x +
(x0 , y0 ) + 1 ∆x + 2 ∆y
∂x
∂y
where
lim
(∆x,∆y )→(0,0)
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1 = 0 =
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lim
(∆x,∆y )→(0,0)
2 .
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Example
f (x, y ) = x 2 + y 2 , at (x0 , y0 ) = (2, 3).
∆f
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=
f (2 + ∆x, 3 + ∆y ) − f (2, 3)
=
(2 + ∆x)2 + (3 + ∆y )2 − 4 − 9
= 4 + 4∆x + (∆x)2 + 9 + 6∆y + (∆y )2 − 4 − 9
=
4∆x + 6∆y + (∆x)2 + ∆y )2 .
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Let 1 = ∆x, 2 = ∆
lim
(∆x,∆y )→(0,0)
1 = 0;
lim
(∆x,∆y )→(0,0)
2 = 0
∂f
∂f
(2, 3) = 4, and
(2, 3) = 6
∂x
∂y
as expected!
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If all first order partial derivatives of f exist and are continuous at a
point, then f is differentiable at that point.
Differentials:
∆f ' fx (x0 , y0 )∆x + fy (x0 , y0 )∆y
(∆f ' fx ∆x + fy ∆y + fz ∆z)
these are approximations for ∆f
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If z = f (x, y ), then the differential dz of f is
dz = fx (x0 , y0 )dx + fy (x0 , y0 )dy .
The same way,
dw = fx dx + fy dy + fz dz.
dz or dw are called the Total differentials of f at (x0 , y0 ).
dz = df is an approximation for ∆z = ∆f .
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Use dz to approximate ∆z where z = xy 2 , from its value at (0.5, 1.0)
to its value at (0.503, 1.004).
Definition (Linear Approximation)
L(x, y ) = f (x0 , y0 ) + fx (x0 , y0 ) + fy (x0 , y0 )dy
Or
L(x, y ) = f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ).
L(x, y )
is known as the local linear approximation for f (x, y ) at the point
(x0 , y0 ).
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Example
f (x, y ) =
q
x2 + y2
Find L(x, y ) at (1, 2). The formula is similar for functions of three
variables f (x, y , z).
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13.5 The Chain Rules: Chain Rule number 1
z = f (x(t), y (t))
dz
∂f dx
∂f dy
=
+
.
dt
∂x dt
∂y dt
Example:
z = ln(2x 2 + y ), x(t) =
dz
dt
=
=
=
2
t, y (t) = t 3 .
dy
1
1
4x dx
dt + 2x 2 +y dt
2x 2 y
√
1
1
1 2 − 31
t 2√
+
2 4
2 3t
t
2t+t 3
2t+t 3
2
2
+
2t+t 3
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√
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2
2
1
3[2t 2 +t 3 ]t 3
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The Chain Rules: Chain Rule number 2
z = f (x(u, v ), y (u, v ))
∂z
∂f ∂x
∂f
=
+
∂u
∂x ∂u ∂y
∂z
∂f ∂x
∂f
=
+
∂v
∂x ∂v
∂y
∂y
∂u
∂y
∂v
Example:
z = 8x 2 y − 2x + 3y , x = uv , y = u − v
∂z
∂u
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∂z ∂x
∂z ∂y
=
∂x ∂u + ∂y ∂u
=
(16xy − 2)v + (8x 2 + 3)(−v )
= (16uv (u − v ) − 2)v − v (8u 2 v 2 + 3)
=
16u 2 v 2 − 24uv 3 − 5v
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Compute
∂z
.
∂v
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13.6 Directional derivatives and gradients
Let ~u be a unit vector. There are infinitely many curves C(t) with
dC
~
dt (0) = u .
Let C(t) = (x(t), y (t), z(t))
Given a function f (x, y , z),
Definition
The directional derivative of f in the direction of ~u is by definition:
d
f (x(t), y (t), z(t))(0).
dt
It is denoted by
~u f .
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Definition
If f (x, y , z) is a function, the gradient of f , ∇f , is the vector defined by :
∇f =
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∂f ~ ∂f ~ ∂f ~
i+
j+
k.
∂x
∂y
∂z
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Proposition
~u f = ~u .∇f .
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Proof
~u f
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d
= dt
f (x(t), y (t), z(t))(0)
∂f dx
∂f dy
∂f dz
= ∂x dt + ∂y
dt + ∂z dt
∂f
∂f
∂f
u1 + ∂y
u2 + ∂z
u3
= ∂x
= ~u .∇f
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Property (I)
The component of ∇f in any given direction gives the directional
derivative in that direction.
Proof.
Let ~u be a direction (a unit vector). Then
~u f = ~u .∇f = k~u kk∇f k cos θ = k∇f k cos θ.
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Property (II)
At each point, the vector ∇f points in the direction of the maximum rate
of increase of the function f .
Proof.
~u f = ~u .∇f
takes its maximum when ~u and ∇f are parallel, and θ = 0.
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Property (III)
The magnitude of ∇f equals the maximum rate of increase of f per unit
distance.
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Property (IV)
~ there passes a level
Through any point (x0 , y0 , z0 ) where ∇f 6= O,
surface f (x, y , z) = f (x0 , y0 , z0 ). The vector ∇f is normal to this surface
at the point (x0 , y0 , z0 ).
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Proof:
Let C(t) be a curve on the level surface f (x, y , z) = f (x0 , y0 , z0 ) such
~
that C(0) = (x0 , y0 , z0 ) and dC
dt (0) = u a unit tangent vector. On one
hand,
d
~u f = dt
f (x(t), y (t), z(t))(0)
d
= dt f (x0 , y0 , z0 )(0)
= 0
on the other hand,
~u f = ~u .∇f
So, we see that
~u .∇f = 0
therefore ∇f is perpendicular to any vector tangent to the level surface;
equivalently, ∇f is perpendicular to the level surface.
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Example
Given f (x, y , z) = x 2 + y 2 + z 2 , find the maximum value of the rate of
change of f at (3, 0, 4).
Solution
We know the ∇f gives the direction of maximum rate of change.
∇f (3, 0, 4) = (2x, 2y , 2z)(3, 0, 4) = (6, 0, 8).
But also that k∇f k =
change.
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√
36 + 64 = 10 gives the maximum rate of
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13.7 Tangent Planes and Normal Lines
Theorem
Let P0 = (x0 , y0 , z0 ) be a point on the surface z = f (x, y ). If f (x, y ) is
differentiable at (x0 , y0 ), then the surface has a tangent plane at P0 .
This plane has equation:
∂f
∂f
(x0 , y0 )(x − x0 ) +
(x0 , y0 )(y − y0 ) − (z − z0 ) = 0.
∂x
∂y
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Proof
∂f ∂f
To prove the theorem, it is enough to show that < ∂x
, ∂y , −1 > is a
normal vector to the surface. To that end, let C = (x(t), y (t), z(t)) be
any curve on the surface such that C(t0 ) = P0 . We need only to show
∂f ∂f
, ∂y , −1 > is perpendicular to to c, that is, perpendicular to
that < ∂x
the tangent vector of C, which is <
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dx dy dz
dt , dt , dt
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> at t = t0 .
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Proof
∂f ∂f
To prove the theorem, it is enough to show that < ∂x
, ∂y , −1 > is a
normal vector to the surface. To that end, let C = (x(t), y (t), z(t)) be
any curve on the surface such that C(t0 ) = P0 . We need only to show
∂f ∂f
, ∂y , −1 > is perpendicular to to c, that is, perpendicular to
that < ∂x
the tangent vector of C, which is <
f (x, y ) − z = 0, we get
dx dy dz
dt , dt , dt
∂f dx
∂f dy
dz
0 = ∂x
dt + ∂y dt − dt
∂f ∂f
= << ∂x
, ∂y , −1 > . <
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> at t = t0 . From
dx dy dz
dt , dt , dt
>
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Take Notice: If f (x, y ) is differentiable at x0 , y0 ), then the vector
N =<
∂f
∂f
(x0 , y0 ),
(x0 , y0 ), −1 >
∂x
∂y
is a normal vector to the surface z = f (x, y ) at P0 = (x0 , y0 , z0 ).
The line through P0 parallel to N is called the normal line to the surface
at P0 .
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Example
Find the equation for the tangent plane and the normal line to the
surface
z = 4x 3 y 2 + 2y
at P = (1. − 2, 12).
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Example
Find the equation for the tangent plane and the normal line to the
surface
z = 4x 3 y 2 + 2y
at P = (1. − 2, 12).
A normal vector at P is
(12x 2 y 2 , 8x 3 y + 2, −1)(1, −2) = (48, −14, −1).
So the tangent plane has equation:
48(x − 1) − 14(y + 2) − (z − 12) = 0
and the normal line has parametric equations:
x = 1 + 48t
y = −2 − 14t
z = 12 − t
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Exercise
Find a point on the surface z = 3x 2 − y 2 at which the tangent plane is
parallel to the plane 6x + 4y − z = 5.
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Exercise
Find a point on the surface z = 3x 2 − y 2 at which the tangent plane is
parallel to the plane 6x + 4y − z = 5.
We need to find points where the normal to the surface is parallel to
(6, 4, −1). That is, we need to solve
(6x, −2y , −1) = k (6, 4, −1)
Necessarily, k = 1, and y = −2, x = 1.
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So at (1, −2, −1), the tangent plane is parallel to 6x + 4y − z = 5.
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13.8 Maxima and Minima of Functions of two variables
Theorem
If f has a relative extremum at a point (x0 , y0 ) and if the first order
partial derivatives of f exist at this point, then
fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0
A point (x0 , y0 ) where fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0 is called a
critical point.
So, we will remember that local extrema occur at critical points. But not
every critical point leads to a local extremum as the last of the following
examples shows:
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Examples
z = f (x, y ) = x 2 + (y − 1)2 + 5
f (0, 1) = 5 is a local minimum. Notice that both partial derivatives
of f vanish at (0, 1)
z = g(x, y ) = −x 2 − (y − 1)2 + 2
f (0, 1) = 2 is a local maximum. Observe that both partial
derivatives of g vanish at (0, 1).
z = h(x, y ) = x 2 − (y − 1)2 + 10
Notice that both partial derivatives of h at (0, 1) are equal to zero,
but f (0, 1) = 10 is neither a local maximum, nor a local minimum.
A point like (0, 1, h(0, 1)) = (0, 1, 10) is called a saddle point for
the graph of h(x, y ).
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Second Order Partial Derivatives Test for Local
Extrema
Theorem
Let f (x, y ) be a differentiable function of 2 variables and assume (a, b)
is a critical point for f .
Let A = fxx (a, b), B = fxy (a, b), C = fyy (a, b) and D = AC − B 2 .
1. If D > 0 and A > 0, then f (a, b) is a local minimum.
2. If D > 0 and A < 0, then f (a, b) is a local maximum.
3. If D < 0, then the graph of f has a saddle point at (a, b, f (a, b)).
4. If D = 0, the test is inconclusive!
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Examples
1. Locate all local maxima, local minima and saddle points if any, for
f (x, y ) = x 3 − 3xy − y 3 .
2. Same for
f (x, y ) = y sin x
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Absolute Extrema
To find absolute extrema on a closed bounded set R,
1. Find critical points in R
2. Find all boundary points at which the absolute extrema can occur
(including corners)
3. Compare values and decide
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Example
f (x, y ) = x 2 − 3y 2 − 2x + 6y ,
R is the rectangle with vertices (0, 0), (2, 0), (2, 2), and (0, 2).
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13.9 Lagrange Multipliers (Constrained Optimization)
∇f is zero at any local extremum of a continuously differentiable
~ are called critical points.
function f . Points at which ∇f = O
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Exercise
Find the maximum value of
f (x, y , z) = x 4 − 3y 2 − 4z 4 + 2z 2
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Exercise
Find the maximum value of
f (x, y , z) = x 4 − 3y 2 − 4z 4 + 2z 2
Since
lim
k(x,y ,z)k→+∞
f (x, y , z) = −∞
we know that f has a maximum value.
Also, since a maximum value will occur at a critical point, let us see
what the critical points are:
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−4x 3 = 0
−4y = 0
3
−16z + 4z = 0
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x =0
y =0
z = 0 or z = ± 21
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1
1
1
1
f (0, 0, 0) = 0, f (0, 0, ) = , f (0, 0, − ) =
2
4
2
4
Thus, the maximum of f is 14 ; it is attained at the points (0, 0, − 12 ) and
(0, 0, 12 ).
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Suppose g(x, y , z) is a continuously differentiable function and the
surface g(x, y , z) = 0 divides all space into two regions:
g(x, y , z) < 0 and g(x, y , z) > 0.
Problem: Maximize f (x, y , z) subject to the constraint g(x, y , z) = 0.
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Suppose (x0 , y0 , z0 ) is a point on the surface g(x, y , z) = 0 and
f (x0 , y0 , z0 ) is maximum among all points on the surface g(x, y , z) = 0.
If ∇f were not perpendicular to the surface, then it would have a
nonzero tangential component along the surface, that would be a
direction of increase for f along the surface, contradicting the
maximality of f (x0 , y0 , z0 ). So the only possibility is that ∇f is
proportional to ∇g, that is, for some constant λ,
∇f + λ∇g = 0.
This is a necessary condition for a constrained maximum (minimum).
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The parameter λ is called the Lagrange Multiplier.
Example; Find the maximum of f (x, y , z)) = −x 4 − 2y 2 − 4z 4 + 2z 2
subject to the constraint g(x, y , z) = x + y − 2 = 0
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The parameter λ is called the Lagrange Multiplier.
Example; Find the maximum of f (x, y , z)) = −x 4 − 2y 2 − 4z 4 + 2z 2
subject to the constraint g(x, y , z) = x + y − 2 = 0
∇f = (−4x 3 , −4y , 4z − 16z 3 )
∇g = (1, 1, 0)
λ(1, 1, 0) + (−4x 3 , −4y , 4z − 16z 3 ) = 0
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y = x 3 , z = 0 or z = ±
x + x 3 − 2 = 0,
1
and x + y = 0
2
implies (x − 1)(x 2 − x + 2) = 0
x =1
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So the constrained extrema can be found only at
1
1
(1, 1, ), (1, 1, − ) and (1, 1, 0)
2
2
f (1, 1, ± 21 = − 11
4 and f (1, 1, 0) = −3, so the maximum is
1
− 11
=
f
(1,
1,
±
4
2 ).
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Generalized Lagrange Multipliers
Maximize f (x, y , z) = z, subject to x 2 + y 2 = 1 and 2x + 2y + z = 0.
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Generalized Lagrange Multipliers
Maximize f (x, y , z) = z, subject to x 2 + y 2 = 1 and 2x + 2y + z = 0.
∇f + λ1 ∇g1 + λ2 ∇g2 = 0
Complete the solution!
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14.1 Double Integrals
Motivated by the volume problem: Given the graph of a function f (x, y )
of two variables, find the volume under the graph and over some
region R in the X -Y plane.
Riemann sums are here again!
The region R is divided into subregions Rij with dimensions ∆xi by ∆yj
and area
∆Aij = ∆xi × ∆yj
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Pick (ui , vj ) ∈ Rij and write the Riemann sum:
Sn =
mi
l X
X
f (ui , vj )∆Aij ,
i=1 j=1
where m1 + m2 + ... + ml = n.
The volume, if defined, should be
V
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= limmax(∆xi ,∆yj )→0 Sn
P P i
= limn→∞ li=1 m
j=1 f (ui , vj )∆Aij
Pl
P i
= liml→∞ i=1 limmi →∞ m
j=1 f (ui , vj )∆yj ∆xi
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When the above limit exists, it is called the double integral of f over R
and denoted by
Z Z
f (x, y )dA
R
where
dA = dxdy or dydx
A geometric interpretation of the double integral is that for f > 0,
RR
R fdA represents the volume under the graph of f , above the region
R in X , Y plane,
Observation
If f = 1, then the volume
area of R.
RR
R
dA is numerically equal to the surface
Example: Compute the double integral
rectangle 1 ≤ x ≤ 3, 2 ≤ y ≤ 4.
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RR
R (30 − xy )dA
where R is the
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When the above limit exists, it is called the double integral of f over R
and denoted by
Z Z
f (x, y )dA
R
where
dA = dxdy or dydx
A geometric interpretation of the double integral is that for f > 0,
RR
R fdA represents the volume under the graph of f , above the region
R in X , Y plane,
Observation
If f = 1, then the volume
area of R.
RR
R
dA is numerically equal to the surface
Example: Compute the double integral
rectangle 1 ≤ x ≤ 3, 2 ≤ y ≤ 4.
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R3R4
2313
(30 − xy )dydxMAC=
RR
R3
R (30 − xy )dA
where R is the
2 4
(30y − x y ) dx
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14.2 Double Integrals over non-rectangular regions
For double integration, we identify two types of region to integrate over!
The X regions and the Y regions. Some regions are of both types,
looking at them from one type to the other is related to reversing the
order of integration.
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X-Regions
An X -region R is one that can be described by:
a≤x ≤b
g(x) ≤ y ≤ h(x)
for some functions g and h and constants a and b.
A double integral over such a region can be written as :
Z Z
b
Z
Z
h(x)
fdA =
f (x, y )dydx
R
a
Example: Compute the integral
Z Z
g(x)
(x 2 − y 2 )dA
R
where R is the region bounded by y = x and y = x 2 .
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The region R is described by:
0≤x ≤1
x2 ≤ y ≤ x
The integral is given by
RR
R (x
2
− y 2 )dA =
=
=
=
=
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R1Rx
2
2
x 2 (x − y )dydx
y 3 y =x
2
0 x y − 3 y =x 3 dx
R1 2 3
x6
4
!dx
0 3x − x +
3
x4
x5
x 7 1
6 − 5 + 21 0
1
1
1
6 − 5 + 21
R01
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Y-Regions
A Y -region R is one that can be described by:
c≤y ≤d
h(y ) ≤ x ≤ g(y )
for some constants c and d and function h and g.
An integral over such a region can be written as:
Z Z
d
Z
Z
g(y )
fdA =
R
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f (x, y )dxdy
c
h(y )
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Reversing the order of integration
Example: Evaluate
Z
0
1
2
Z
1
2
ey dydx
2x
Notice that the first integration is not easy!
The region R over which it is being integrated is described by:
1
2
2x ≤ y ≤ 1
0≤x ≤
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After sketching the region R, we observe that an alternate description
of R is
0≤y ≤1
y
0≤x ≤
2
Using the alternate description of R, we see that the same integral can
be written as:
Z
0
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y
2
1Z
2
ey dxdy
0
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Now we are able to start the integration process! We have reversed
the order of integration. Finish the double integration!
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14.3 Double Integrals in Polar Coordinates
Consider the region R described in polar coordinates by:
α≤θ≤β
a≤r ≤b
Divide the region r into subregions Rij , where each Rij is described as
θi−1 ≤ θ ≤ θi
rj−1 ≤ r ≤ rj
Proposition
Denoting by ∆ri = ri − ri−1 and ∆θj = θj − θj−1 , the area ∆Aij of Rij is
given by:
1
∆Aij = (ri + ri−1 )∆ri ∆θj
2
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Proof
∆Aij is the difference between areas of sectors with angle ∆θj and
radii ri and ri−1 . Therefore,
∆Aij =
PR (FIU)
ri2
2 ∆θj
−
2
ri−1
2 ∆θj
=
=
=
1 2
2
2 (ri − ri−1 )∆θj
1
2 (ri + ri−1 )(ri − ri−1 )∆θj
1
2 (ri + ri−1 )∆ri ∆θj
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Choose r¯i = 12 (ri + ri−1 ) and θ̄j in Rij , then
∆Aij = r¯i ∆ri ∆θj
and the Riemann sum associated to a function f (r , θ) and a partition P
over the region R will be
S(P, f ) =
XX
j
f (r¯i , θ̄j )∆Aij =
i
By definition, the double integral
given by
lim
kPk→0
PR (FIU)
XX
j
XX
i
RR
R
f (r̄i , θ̄j )r¯i ∆ri ∆θj
j
fdA of f over the region R is
f (r̄i , θ̄j )r¯i ∆ri ∆θj =
Z
β
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b(θ)
f (r , θ)rdrdθ
α
i
Z
a(θ)
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Example
Evaluate the integral
√
Z
0
1Z
4−x 2
(x 2 + y 2 )dydx
0
using polar coordinates.
From the sketch, the region can be described as
0≤r ≤2
π
2
Recall: In polar coordinates dA = rdrdθ.
0≤θ≤
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So
R 2 R √4−x 2
0
0
(x 2 + y 2 )dydx
=
R
=
R
π
2
0π
2
R0π2
R2
0
r4
r 2 rdrdθ
2
dθ
4 0
= 0 4dθ
= 2π
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Example 2
The cardioid is given by
r = a(1 − cos θ)
Find its enclosed surface area.
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Example 2
The cardioid is given by
r = a(1 − cos θ)
Find its enclosed surface area.
Sketch the cardioid on the X Y plane.
Its description is
0 ≤ θ ≤ 2π
0 ≤ r ≤ a(1 − cos θ)
The surface area is given by
Z
π
Z
a(1−cos θ)
A=2
rdrdθ = 3π
0
0
a2
.
2
Compute the integral in details!
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14.4 Parametric Surfaces; Surface area
A parameterized curve C is a vector valued function
r : R → R3 : t 7→ r (t) = (x(t), y (t), z(t)).
A parameterized surface S is a vector valued function
r : R2 → R3 : (u, v ) 7→ r (u, v ) = (x(u, v ), y (u, v ), z(u, v )).
Example:
r (u, v ) = (u, v , 4 − u 2 − v 2 )
Eliminating u, v from the parametric equations shows that
z = 4 − x2 − v2
The surface S is a piece of a paraboloid of revolution (Circular
paraboloid).
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Tangent Plane to Parameterized Surfaces
Let r (u, v ) be a parameterized surface. Its tangent plane at any point is
∂r
∂r
parallel to ∂u
and ∂v
, that is, parallel to
(
∂x ∂y ∂z
∂x ∂y ∂z
,
,
) and ( ,
,
).
∂u ∂u ∂u
∂v ∂v ∂v
Therefore,
~ = ( ∂r ) × ( ∂r )
N
∂u
∂v
is a normal vector to the surface.
Definition
~n =
~
N
~
kNk
is called the principal normal vector.
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~n =
∂r
∂u
∂r
k ∂u
×
×
∂r
∂v
∂r
∂v k
Example:
x = uv , y = u, z = v 2 .
Find an equation of the tangent plane at the point where u = 2 and
v = −1.
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Surface Area of a Parameterized Surface
Let Rij be a rectangle in the subdivision od the domain of
parametrization D. Its edges are (ui , vj ), ui + ∆ui , vj ), (ui , vj + ∆vj ) and
(ui + ∆ui , vj + ∆vj ). The image of Rij is denoted by σij . Its surface area
∆Sij is approximated by the area of the parallelogram spanned by
r (ui + ∆ui − r (ui , vj ) and r (ui , vj + ∆vj ) − r (ui , vj ).
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In turn, from the Mean Value Theorem,
PR (FIU)
r (ui , vj + ∆vj − r (ui , vj ) '
∂r
∆vj
∂v
r (ui + ∆ui , vj ) − r (ui , vj ) '
∂r
∆ui
∂u
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Therefore,
∆Sij ' k
∂r
∂r
∂r
∂r
∆ui ×
∆vj k = k
×
k∆ui ∆vj
∂u
∂v
∂u ∂v
The surface area of S is approximated by the Riemann sum
Sn =
X ∂r
∂r
×
k∆Aij
k
∂u ∂v
i,j
The surface area S is given by
Z Z
k
S = lim Sij =
n→∞
PR (FIU)
D
MAC 2313
∂r
∂r
×
kdA.
∂u ∂v
68 / 104
In the special case where the surface is the graph of
z = f (x, y )
then a parametrization is
r (x, y ) = (x, y , f (x, y ))
and the surface area S is given by
Z Z s
∂f
∂f
( )2 + ( )2 + 1dA.
S=
∂x
∂y
D
Example: 1. r (u, v ) = (u, v , 4 − u 2 − v 2 ), D = {u 2 + v 2 ≤ 4}.
Compute the surface area.
2. Find the surface area of the p
portion of the sphere x 2 + y 2 + z 2 = 8
that is cut out by the cone z = x 2 + y 2 .
3. Find the surface area of the paraboloid portion z = 9 − x 2 − y 2 that
lies between the planes z + 0 and z = 8.
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1.
2. The domain of parametrization is a disc of radius 2.
−2 ≤ x ≤ 2
p
p
− 4 − x2 ≤ y ≤ 4 − x2
The surface area is given by
s
Z Z √
4−x 2
2
S=
√
−2
−
4−x 2
x2
y2
+
+ 1dydx.
8 − x2 − y2 8 − x2 − y2
We can use polar coordinates to compute this integral.
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Z
0
PR (FIU)
2 Z 2π
0
s
√
r2
+ 1rdrdθ = (16 − 8 2)π
2
8−r
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For example 3, the domain of parametrization is an annulus 1 ≤ r ≤ 3.
Z
3 Z 2π
S=
1
PR (FIU)
p
4r 2 + 1rdrdθ
0
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Think of the surface area S as
Z
ds
S=
S
One defines the surface integral of any function f on the surface as
Z
Z Z
fdS =
S
f (x(u, v ), y (u, v ))k
D
∂r
∂r
×
kdA
∂u ∂v
where
dA = dudv or dvdu.
Examples will be provided later!
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14.5 Triple Integrals
An example of a 3-dimensional region R:
The region bounded by
x = 0, y = 0, z = 0
and
x + 2y + 3z = 6.
After sketching the region, it can be described as follows:
0 ≤ x
0 ≤ y
0 ≤ z
PR (FIU)
≤ 6
≤ 6−x
2
≤ 6−x−2y
3
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There are up to 5 more different descriptions of R. Any triple integral
RRR
R fdV could be written as an iterated integral
Z
0
PR (FIU)
6Z
0
6−x
2
Z
6−x−2y
3
f (x, y , z)dzdydx.
0
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Example
Find the volume integral of f (x, y , z) = x + y + z over the box bounded
by x = 1, y = 2, z = 1 + x and the coordinate planes.
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Example
Find the volume integral of f (x, y , z) = x + y + z over the box bounded
by x = 1, y = 2, z = 1 + x and the coordinate planes.
After sketching the region, it can be described as follows:
0 ≤ x
0 ≤ y
0 ≤ z
PR (FIU)
≤ 1
≤ 2
≤ 1+x
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The integral is written as follows:
R 1 R 2 R 1+x
0
0
0
(x + y + z)dzdydx
=
=
=
=
=
PR (FIU)
z 2 1+x
0 0 (x + y )z + 2 0 dydx
R1R2
2
(1 + x)(x + y ) + (1+x)
dydx
2
0
0
R1
y 2 2
2
0R(1 + x)(xy + 2 ) 0 + (1 + x) dx
1
3 0 (1 + x)2 dx
(1 + x)3 |10 = 7
R1R2
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Use a triple integral to find the volume of the solid bounded by the
surface y = x 2 , the plane y + z = 4 and z = 0.
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Use a triple integral to find the volume of the solid bounded by the
surface y = x 2 , the plane y + z = 4 and z = 0.
After sketching, the region is described as follows:
−2 ≤ x
x2 ≤ y
0 ≤ z
PR (FIU)
≤ 2
≤ 4
≤ 4−y
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The volume is given by
R 2 R 4 R 4−y
−2 x 2 0
dzdydx
=
=
R2
−2 (8
−
4x 2
+
x4
2 )dx
R2 R4
− y )dydx
y 2 4
−2 [4y − 2 ] x 2 dx
R−2
2
= 16 −
PR (FIU)
MAC 2313
x 2 (4
32
3
+
32
10
+ 16 −
32
3
+
32
10
79 / 104
14.7 Change of variables in multiple integrals:
Jacobians
Motivation: Recall the u substitution technique of integration.
Proposition
From
Z
b
f (x)dx,
a
make a substitution x = g(u), dx = g 0 (u)du.
The integral becomes:
Z
g −1 (b)
0
Z
β
f (g(u)g (u)du =
g −1 (a)
PR (FIU)
f (g(u)|g 0 (u)|du, α < β
α
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Proof
If g −1 (a) < g −1 (b), then g 0 (u) > 0 and thus, g 0 (u) = |g 0 (u)|.
b
Z
Z
g −1 (b)
f (x)dx =
f (g(u))|g 0 (u)|du.
g −1 (a)
a
If g −1 (b) < g −1 (a), then g 0 (u) < 0, thus g 0 (u) = −|g 0 (u)|.
Z
b
Z
g −1 (b)
f (x)dx = −
f (g(u))|g 0 (u)|du =
g −1 (a)
a
Z
g −1 (a)
f (g(u))|g 0 (u)|du.
g −1 (b)
In each case, the integral is given by
Z
β
f (g(u))|g 0 (u)|du, α < β.
α
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Generalization to higher dimensions
Under a reparameterization (x, y ) = r (u, v ), from a domain S into a
domain R, a small rectangle with vertices
(u0 , v0 ), (u0 + ∆u, v0 ), (u0 , v0 + ∆v ), (u0 + ∆u, v0 + ∆v ) transforms
into R, approximated by a rectangle based on r (u0 , v0 ) and edges
a'
r (u0 + ∆u, v0 ) − r (u0 , v0 )
∂r
∆u '
∆u
∆u
∂u
and similarly,
b'
∂r
∆v
∂v
The elementary surface ∆A in the x, y parameters is approximately
∂r
∆A = ∆x∆y ' k ∂u
∆u ×
∂r
∂v ∆v k
∂r
= k ∂u
×
= |det
PR (FIU)
MAC 2313
∂r
∂v k∆u∆v
!
∂y
∂x
∂u
∂u
∆u∆v
∂y
∂x
∂v
∂v
82 / 104
The change of variable formula in double integrals is therefore: when
(x, y ) = r (u, v ),
Z Z
Z Z
∂(x, y )
f (x, y )dAx,y =
f (u, v )|det(
)|dAu,v .
∂(u,
v)
R
S
∂(x,y )
The determinant det( ∂(u,v
) ) is called the Jacobian of the coordinate
change.
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Example
Compute the integral
Z Z
exy dA
R
where R is determined by
y=
1
2
1
x, y = x, y = , y =
2
x
x
using the transformation:
u=
PR (FIU)
y
, v = xy
x
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The domain S is described as
u=
1
, u = 1, v = 1, v = 2
2
We need the coordinates changes as
r
√
v
x=
, y = uv
u
Then the Jacobian is
Z Z
exy dAx,y
R
PR (FIU)
∂(x, y )
1
=− .
∂(u, v )
2u
Z Z
Z Z
1 2 11 v
v 1
=
e
dAu,v =
e dudv .
|2u|
2 1 1 u
S
2
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In triple integrals, the Jacobian is a 3 by 3 determinant
det
Z Z Z
∂(x, y , z)
.
∂(u, v , w)
Z Z Z
f (x, y , z)dVx,y ,z =
R
f (u, v , w)|
S
∂(x, y , z)
dVu,v ,w .
∂(u, v , w)
Example: Find the volume of
x2 +
y 2 z2
+
=1
4
9
using the transformation
x = u, y = 2v , z = 3w
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14.6 Triple Integrals in Cylindrical and Spherical
Coordinates
Example: Evaluate the integral using spherical coordinates:
Z 2 Z √4−y 2 Z √4−x 2 −y 2
2z
p
dzdxdy .
x2 + y2
0
0
0
PR (FIU)
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15.1 Vector fields
A vector field F is a vector valued function.
F : R3 → R3 .
F (x, y , z) = (F1 (x, y , z), F2 (x, y , z), F3 (x, y , z)).
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Examples
Given a differentiable function
f : R3 → R,
∇f is a vector field, the gradient vector field of f .
∇f
PR (FIU)
:
R3
→ R3
∂f ∂f ∂f
, ∂y , ∂z )
(x, y , z) 7→ ∇f (x, y , z) = ( ∂x
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Definition
A curve C(t) = (x(t), y (t), z(t)) such that c 0 (t) = F (C(t)) is called a
flow line for the vector field F .
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Examples
Find the equation of flow lines for F (x, y ) = (−y , x).
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Examples
Find the equation of flow lines for F (x, y ) = (−y , x).
By definition, C(t) − (x(t), y (t)) is a flow line if
dx
dt
dy
dt
PR (FIU)
= −y
= x
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Solving the system leads to:
d 2x
dy
= −x
=−
dt
dt 2
That is:
d 2x
+ x = 0.
dt 2
The solution is :
x(t) = a cos t + b sin t
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From
dy
dt
= x, one obtains
y (t) = a sin t − b cos t + C
with C = 0 since
PR (FIU)
dx
dt
= −y .
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C(t) = (a cos t + b sin t, a sin t − b cos t)
Eliminating t shows that the flow lines are circles of radius
centered at (0, 0).
PR (FIU)
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√
a2 + b 2 ,
94 / 104
Example 2
F (x, y , z) = (
x2
x
y
, 2
).
2
+ y x + y2
C(t) = (x(t), y (t))
dx
dt
dy
dt
PR (FIU)
=
=
x
x 2 +y 2
y
x 2 +y 2
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x2
1 dy
1
1 dx
=
=
2
x dt
y dt
+y
1 dx
1 dy
−
=0
x dt
y dt
x
d
(ln ) = 0
dt
y
x
=C
y
or
x = Cy
The flow lines are straight lines through the origin, minus the origin
itself.
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15.2 Line Integrals
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15.7 The divergence Theorem
Definition of divergence as flux density
The divergence of a vector field is the measure of "flux out" of a closed
surface. The divergence or flux density of a vector field F is defined by
~ (x, y , z) =
div F
~
~ .dS
F
volume(σ)→0 (volume enclosed by σ)
R
σ
lim
In rectangular coordinates:
~ = ∂F1 + ∂F2 + ∂F3
div F
∂x
∂y
∂z
Observation:
~ =0
div curl F
This can be shown directly (do the calculation!)
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~ = 0,
On a doubly connected region, one has the converse: If div G
~
~
~
then G = curl F for some F .
Examples of divergence free vector fields, or solenoidal vector fields,
~ Maxwell’s equation states exactly that:
are the magnetic fields B.
~ = 0.
div B
Another example: For a magnetic dipole (or a current loop), with
constant dipole moment µ
~,
~
(~
µ.~r )~r
~ =− µ
B
+3
, ~r =
6 ~0
3
~
kr k
k~r k5
~ = 0.
Show that div B
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~ =<
Alternate notation:With ∇
∂
∂
∂
∂x , ∂y , ∂z
>
~ = ∇.
~
~ F
div F
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The divergence Theorem expresses the total flux as the integral of the
flux density.
Z
S
~ .~n dS =
F
Z
~ dV
div F
W
where S is the boundary of the region W , oriented by outer unit
normals.
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Example: Use the divergence Theorem to find the flux of
F (x, y , z) = (x 2 y , −xy 2 , z + 2) across the surface σ of the solid
bounded above by the plane z = 2x and below by the paraboloid
z = x 2 + y 2.
PR (FIU)
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16.8: Stokes’ Theorem
~ as the integral of
Stokes’ Theorem expresses the total circulation of F
the circulation density:
Z
~ .~n dS =
curl F
Z
~ .dr
~
F
∂σ
σ
A special case of Stokes Theorem is Green’s Theorem:
~ =< f , g, 0 >
F
PR (FIU)
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Example: Verify Stokes’ Theorem for
~ (x, y , z) =< x, y , z >
F
p
where σ the upper hemisphere z = a2 − x 2 − y 2 with upward
orientation.
PR (FIU)
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