Calculus III

Philippe Rukimbira

Department of MathematicsFlorida International University

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13.3 Partial Derivatives

Example:f (x , y) = (x2 + 2xy)5

The partial derivative of f with respect to x is:

∂f∂x

(x , y) = 5(x2 + 2xy)4(2x + 2y).

What is ∂f∂y ?

The answer is:5(x2 + 2xy)4(2x).

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13.3 Partial Derivatives

Example:f (x , y) = (x2 + 2xy)5

The partial derivative of f with respect to x is:

∂f∂x

(x , y) = 5(x2 + 2xy)4(2x + 2y).

What is ∂f∂y ?

The answer is:5(x2 + 2xy)4(2x).

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The second order partial derivative of f with respect to y is:

∂2f∂y2 (x , y) = 40(x2 + 2xy)3(2x)(2x).

∂2f∂x∂y = ∂

∂x ( ∂∂y f )

= ∂∂x (10x(x2 + 2xy)4)

= 10(x2 + 2xy)4 + 40x(x2 + 2xy)3(2x)

= 20(x2 + 2xy)3(5x2 + xy)

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What is∂2f∂x2 ?

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13.4 Differentiability, Differentials and local linearity

Denote by ∆f = f (x0 + ∆x , y0 + ∆y)− f9X0,Y0).

f is said to be differentiable at (x0, y0) if:

∆f =∂f∂x

(x0, y0)∆x +∂f∂y

(x0, y0) + ε1∆x + ε2∆y

wherelim

(∆x ,∆y)→(0,0)ε1 = 0 = lim

(∆x ,∆y)→(0,0)ε2.

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Example

f (x , y) = x2 + y2, at (x0, y0) = (2,3).

∆f = 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 ε1 = ∆x , ε2 = ∆

lim(∆x ,∆y)→(0,0)

ε1 = 0; lim(∆x ,∆y)→(0,0)

ε2 = 0

∂f∂x

(2,3) = 4, and∂f∂y

(2,3) = 6

as expected!

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If all first order partial derivatives of f exist and are continuous at apoint, 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 = fxdx + fydy + fzdz.

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 = xy2, 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) =√

x2 + y2

Find L(x , y) at (1,2). The formula is similar for functions of threevariables f (x , y , z).

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13.5 The Chain Rules: Chain Rule number 1

z = f (x(t), y(t))

dzdt

=∂f∂x

dxdt

+∂f∂y

dydt.

Example:

z = ln(2x2 + y), x(t) =√

t , y(t) = t23 .

dzdt = 1

2x2y 4x dxdt + 1

2x2+ydydt

= 1

2t+t23

4√

t 12√

t+ 1

2t+t23

23 t−

13

= 2

2t+t23

+ 2

3[2t2+t23 ]t

13

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The Chain Rules: Chain Rule number 2

z = f (x(u, v), y(u, v))

∂z∂u

=∂f∂x

∂x∂u

+∂f∂y

∂y∂u

∂z∂v

=∂f∂x

∂x∂v

+∂f∂y

∂y∂v

Example:z = 8x2y − 2x + 3y , x = uv , y = u − v

∂z∂u = ∂z

∂x∂x∂u + ∂z

∂y∂y∂u

= (16xy − 2)v + (8x2 + 3)(−v)

= (16uv(u − v)− 2)v − v(8u2v2 + 3)

= 16u2v2 − 24uv3 − 5vPR (FIU) MAC 2313 13 / 104

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) withdCdt (0) = ~u.

Let C(t) = (x(t), y(t), z(t))

Given a function f (x , y , z),

DefinitionThe directional derivative of f in the direction of ~u is by definition:

ddt

f (x(t), y(t), z(t))(0).

It is denoted by~uf .

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DefinitionIf f (x , y , z) is a function, the gradient of f , ∇f , is the vector defined by :

∇f =∂f∂x~i +

∂f∂y~j +

∂f∂z~k .

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Proposition

~uf = ~u.∇f .

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Proof

~uf = ddt f (x(t), y(t), z(t))(0)

= ∂f∂x

dxdt + ∂f

∂ydydt + ∂f

∂zdzdt

= ∂f∂x u1 + ∂f

∂y u2 + ∂f∂z u3

= ~u.∇f

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Property (I)The component of ∇f in any given direction gives the directionalderivative in that direction.

Proof.

Let ~u be a direction (a unit vector). Then

~uf = ~u.∇f = ‖~u‖‖∇f‖ cos θ = ‖∇f‖ cos θ.

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Property (II)At each point, the vector ∇f points in the direction of the maximum rateof increase of the function f .

Proof.

~uf = ~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 unitdistance.

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Property (IV)

Through any point (x0, y0, z0) where ∇f 6= ~O, there passes a levelsurface f (x , y , z) = f (x0, y0, z0). The vector ∇f is normal to this surfaceat 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) suchthat C(0) = (x0, y0, z0) and dC

dt (0) = ~u a unit tangent vector. On onehand,

~uf = ddt f (x(t), y(t), z(t))(0)

= ddt f (x0, y0, z0)(0)

= 0

on the other hand,~uf = ~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) = x2 + y2 + z2, find the maximum value of the rate ofchange 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 ‖∇f‖ =√

36 + 64 = 10 gives the maximum rate ofchange.

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13.7 Tangent Planes and Normal Lines

TheoremLet P0 = (x0, y0, z0) be a point on the surface z = f (x , y). If f (x , y) isdifferentiable at (x0, y0), then the surface has a tangent plane at P0.This plane has equation:

∂f∂x

(x0, y0)(x − x0) +∂f∂y

(x0, y0)(y − y0)− (z − z0) = 0.

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Proof

To prove the theorem, it is enough to show that < ∂f∂x ,

∂f∂y ,−1 > is a

normal vector to the surface. To that end, let C = (x(t), y(t), z(t)) beany curve on the surface such that C(t0) = P0. We need only to showthat < ∂f

∂x ,∂f∂y ,−1 > is perpendicular to to c, that is, perpendicular to

the tangent vector of C, which is < dxdt ,

dydt ,

dzdt > at t = t0.

Fromf (x , y)− z = 0, we get

0 = ∂f∂x

dxdt + ∂f

∂ydydt −

dzdt

= << ∂f∂x ,

∂f∂y ,−1 > . < dx

dt ,dydt ,

dzdt >

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Proof

To prove the theorem, it is enough to show that < ∂f∂x ,

∂f∂y ,−1 > is a

normal vector to the surface. To that end, let C = (x(t), y(t), z(t)) beany curve on the surface such that C(t0) = P0. We need only to showthat < ∂f

∂x ,∂f∂y ,−1 > is perpendicular to to c, that is, perpendicular to

the tangent vector of C, which is < dxdt ,

dydt ,

dzdt > at t = t0. From

f (x , y)− z = 0, we get

0 = ∂f∂x

dxdt + ∂f

∂ydydt −

dzdt

= << ∂f∂x ,

∂f∂y ,−1 > . < dx

dt ,dydt ,

dzdt >

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Take Notice: If f (x , y) is differentiable at x0, y0), then the vector

N =<∂f∂x

(x0, y0),∂f∂y

(x0, y0),−1 >

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 surfaceat P0.

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Example

Find the equation for the tangent plane and the normal line to thesurface

z = 4x3y2 + 2y

at P = (1.− 2,12).

A normal vector at P is

(12x2y2,8x3y + 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 + 48ty = −2− 14tz = 12− t

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Example

Find the equation for the tangent plane and the normal line to thesurface

z = 4x3y2 + 2y

at P = (1.− 2,12).

A normal vector at P is

(12x2y2,8x3y + 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 + 48ty = −2− 14tz = 12− t

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Exercise

Find a point on the surface z = 3x2 − y2 at which the tangent plane isparallel 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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Exercise

Find a point on the surface z = 3x2 − y2 at which the tangent plane isparallel 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

TheoremIf f has a relative extremum at a point (x0, y0) and if the first orderpartial 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 acritical point.

So, we will remember that local extrema occur at critical points. But notevery critical point leads to a local extremum as the last of the followingexamples shows:

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Examples

z = f (x , y) = x2 + (y − 1)2 + 5

f (0,1) = 5 is a local minimum. Notice that both partial derivativesof f vanish at (0,1)

z = g(x , y) = −x2 − (y − 1)2 + 2

f (0,1) = 2 is a local maximum. Observe that both partialderivatives of g vanish at (0,1).

z = h(x , y) = x2 − (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 forthe graph of h(x , y).

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Second Order Partial Derivatives Test for LocalExtrema

TheoremLet 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 − B2.

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) = x3 − 3xy − y3.

2. Same forf (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) = x2 − 3y2 − 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 differentiablefunction f . Points at which ∇f = ~O are called critical points.

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Exercise

Find the maximum value of

f (x , y , z) = x4 − 3y2 − 4z4 + 2z2

Sincelim

‖(x ,y ,z)‖→+∞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 seewhat the critical points are:

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Exercise

Find the maximum value of

f (x , y , z) = x4 − 3y2 − 4z4 + 2z2

Sincelim

‖(x ,y ,z)‖→+∞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 seewhat the critical points are:

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−4x3 = 0 x = 0−4y = 0 y = 0

−16z3 + 4z = 0 z = 0 or z = ±12

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f (0,0,0) = 0, f (0,0,12

) =14, f (0,0,−1

2) =

14

Thus, the maximum of f is 14 ; it is attained at the points (0,0,−1

2) and(0,0, 1

2).

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Suppose g(x , y , z) is a continuously differentiable function and thesurface 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 andf (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 anonzero tangential component along the surface, that would be adirection of increase for f along the surface, contradicting themaximality of f (x0, y0, z0). So the only possibility is that ∇f isproportional 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)) = −x4 − 2y2 − 4z4 + 2z2

subject to the constraint g(x , y , z) = x + y − 2 = 0

∇f = (−4x3,−4y ,4z − 16z3)

∇g = (1,1,0)

λ(1,1,0) + (−4x3,−4y ,4z − 16z3) = 0

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The parameter λ is called the Lagrange Multiplier.

Example; Find the maximum of f (x , y , z)) = −x4 − 2y2 − 4z4 + 2z2

subject to the constraint g(x , y , z) = x + y − 2 = 0

∇f = (−4x3,−4y ,4z − 16z3)

∇g = (1,1,0)

λ(1,1,0) + (−4x3,−4y ,4z − 16z3) = 0

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y = x3, z = 0 or z = ±12

and x + y = 0

x + x3 − 2 = 0, implies (x − 1)(x2 − x + 2) = 0

x = 1

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So the constrained extrema can be found only at

(1,1,12

), (1,1,−12

) and (1,1,0)

f (1,1,±12 = −11

4 and f (1,1,0) = −3, so the maximum is−11

4 = f (1,1,±12).

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Generalized Lagrange Multipliers

Maximize f (x , y , z) = z, subject to x2 + y2 = 1 and 2x + 2y + z = 0.

∇f + λ1∇g1 + λ2∇g2 = 0

Complete the solution!

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Generalized Lagrange Multipliers

Maximize f (x , y , z) = z, subject to x2 + y2 = 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 someregion 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 =l∑

i=1

mi∑j=1

f (ui , vj)∆Aij ,

where m1 + m2 + ...+ ml = n.

The volume, if defined, should be

V = limmax(∆xi ,∆yj )→0 Sn

= limn→∞∑l

i=1∑mi

j=1 f (ui , vj)∆Aij

= liml→∞∑l

i=1 limmi→∞∑mi

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 Rand denoted by ∫ ∫

Rf (x , y)dA

wheredA = dxdy or dydx

A geometric interpretation of the double integral is that for f > 0,∫ ∫R fdA represents the volume under the graph of f , above the region

R in X ,Y plane,

ObservationIf f = 1, then the volume

∫ ∫R dA is numerically equal to the surface

area of R.

Example: Compute the double integral∫ ∫

R(30− xy)dA where R is therectangle 1 ≤ x ≤ 3, 2 ≤ y ≤ 4.

∫ 31

∫ 42 (30− xy)dydx =

∫ 31 (30y − x y2

2 )∣∣42dx

=∫ 3

1 (60− 6x)dx= (60x − 3x2)

∣∣31

= 96

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When the above limit exists, it is called the double integral of f over Rand denoted by ∫ ∫

Rf (x , y)dA

wheredA = dxdy or dydx

A geometric interpretation of the double integral is that for f > 0,∫ ∫R fdA represents the volume under the graph of f , above the region

R in X ,Y plane,

ObservationIf f = 1, then the volume

∫ ∫R dA is numerically equal to the surface

area of R.

Example: Compute the double integral∫ ∫

R(30− xy)dA where R is therectangle 1 ≤ x ≤ 3, 2 ≤ y ≤ 4.

∫ 31

∫ 42 (30− xy)dydx =

∫ 31 (30y − x y2

2 )∣∣42dx

=∫ 3

1 (60− 6x)dx= (60x − 3x2)

∣∣31

= 96

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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 theorder 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 :

∫ ∫R

fdA =

∫ b

a

∫ h(x)

g(x)f (x , y)dydx

Example: Compute the integral∫ ∫R

(x2 − y2)dA

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The region R is described by:

0 ≤ x ≤ 1

x2 ≤ y ≤ x

The integral is given by∫ ∫R(x2 − y2)dA =

∫ 10

∫ xx2(x2 − y2)dydx

=∫ 1

0 x2y − y3

3

∣∣y=xy=x3dx

=∫ 1

023x3 − x4 + x6

3 !dx= x4

6 −x5

5 + x7

21

∣∣10

= 16 −

15 + 1

21

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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:

∫ ∫R

fdA =

∫ d

c

∫ g(y)

h(y)f (x , y)dxdy

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Reversing the order of integration

Example: Evaluate ∫ 12

0

∫ 1

2xey2

dydx

Notice that the first integration is not easy!

The region R over which it is being integrated is described by:

0 ≤ x ≤ 12

2x ≤ y ≤ 1

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After sketching the region R, we observe that an alternate descriptionof R is

0 ≤ y ≤ 1

0 ≤ x ≤ y2

Using the alternate description of R, we see that the same integral canbe written as:

∫ 1

0

∫ y2

0ey2

dxdy

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Now we are able to start the integration process! We have reversedthe 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

PropositionDenoting by ∆ri = ri − ri−1 and ∆θj = θj − θj−1, the area ∆Aij of Rij isgiven by:

∆Aij =12

(ri + ri−1)∆ri∆θj

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Proof

∆Aij is the difference between areas of sectors with angle ∆θj andradii ri and ri−1. Therefore,

∆Aij =r2i2 ∆θj −

r2i−12 ∆θj = 1

2(r2i − r2

i−1)∆θj

= 12(ri + ri−1)(ri − ri−1)∆θj

= 12(ri + ri−1)∆ri∆θj

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Choose r̄i = 12(ri + ri−1) and θ̄j in Rij , then

∆Aij = r̄i∆ri∆θj

and the Riemann sum associated to a function f (r , θ) and a partition Pover the region R will be

S(P, f ) =∑

j

∑i

f (r̄i , θ̄j)∆Aij =∑

i

∑j

f (r̄i , θ̄j)r̄i∆ri∆θj

By definition, the double integral∫ ∫

R fdA of f over the region R isgiven by

lim‖P‖→0

∑j

∑i

f (r̄i , θ̄j)r̄i∆ri∆θj =

∫ β

α

∫ b(θ)

a(θ)f (r , θ)rdrdθ

PR (FIU) MAC 2313 59 / 104

Example

Evaluate the integral

∫ 1

0

∫ √4−x2

0(x2 + y2)dydx

using polar coordinates.

From the sketch, the region can be described as

0 ≤ r ≤ 2

0 ≤ θ ≤ π

2Recall: In polar coordinates dA = rdrdθ.

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So ∫ 20

∫√4−x2

0 (x2 + y2)dydx =∫ π

20

∫ 20 r2rdrdθ

=∫ π

20

r4

4

∣∣20dθ

=∫ π

20 4dθ

= 2π

PR (FIU) MAC 2313 61 / 104

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 is0 ≤ θ ≤ 2π0 ≤ r ≤ a(1− cos θ)

The surface area is given by

A = 2∫ π

0

∫ a(1−cos θ)

0rdrdθ = 3π

a2

2.

Compute the integral in details!

PR (FIU) MAC 2313 62 / 104

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 is0 ≤ θ ≤ 2π0 ≤ r ≤ a(1− cos θ)

The surface area is given by

A = 2∫ π

0

∫ a(1−cos θ)

0rdrdθ = 3π

a2

2.

Compute the integral in details!PR (FIU) MAC 2313 62 / 104

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− u2 − v2)

Eliminating u, v from the parametric equations shows that

z = 4− x2 − v2

The surface S is a piece of a paraboloid of revolution (Circularparaboloid).

PR (FIU) MAC 2313 63 / 104

Tangent Plane to Parameterized Surfaces

Let r(u, v) be a parameterized surface. Its tangent plane at any point isparallel to ∂r

∂u and ∂r∂v , that is, parallel to

(∂x∂u,∂y∂u,∂z∂u

) and (∂x∂v,∂y∂v,∂z∂v

).

Therefore,~N = (

∂r∂u

)× (∂r∂v

)

is a normal vector to the surface.

Definition

~n =~N‖~N‖

is called the principal normal vector.

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~n =∂r∂u ×

∂r∂v

‖ ∂r∂u ×

∂r∂v ‖

Example:x = uv , y = u, z = v2.

Find an equation of the tangent plane at the point where u = 2 andv = −1.

PR (FIU) MAC 2313 65 / 104

Surface Area of a Parameterized Surface

Let Rij be a rectangle in the subdivision od the domain ofparametrization 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 byr(ui + ∆ui − r(ui , vj) and r(ui , vj + ∆vj)− r(ui , vj).

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In turn, from the Mean Value Theorem,

r(ui , vj + ∆vj − r(ui , vj) '∂r∂v

∆vj

r(ui + ∆ui , vj)− r(ui , vj) '∂r∂u

∆ui

PR (FIU) MAC 2313 67 / 104

Therefore,

∆Sij ' ‖∂r∂u

∆ui ×∂r∂v

∆vj‖ = ‖ ∂r∂u× ∂r∂v‖∆ui∆vj

The surface area of S is approximated by the Riemann sum

Sn =∑i,j

‖ ∂r∂u× ∂r∂v‖∆Aij

The surface area S is given by

S = limn→∞

Sij =

∫ ∫D‖ ∂r∂u× ∂r∂v‖dA.

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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

S =

∫ ∫D

√(∂f∂x

)2 + (∂f∂y

)2 + 1dA.

Example: 1. r(u, v) = (u, v ,4− u2 − v2), D = {u2 + v2 ≤ 4}.

Compute the surface area.

2. Find the surface area of the portion of the sphere x2 + y2 + z2 = 8that is cut out by the cone z =

√x2 + y2.

3. Find the surface area of the paraboloid portion z = 9− x2 − y2 thatlies between the planes z + 0 and z = 8.

PR (FIU) MAC 2313 69 / 104

1.

2. The domain of parametrization is a disc of radius 2.

−2 ≤ x ≤ 2

−√

4− x2 ≤ y ≤√

4− x2

The surface area is given by

S =

∫ 2

−2

∫ √4−x2

−√

4−x2

√x2

8− x2 − y2 +y2

8− x2 − y2 + 1dydx .

We can use polar coordinates to compute this integral.

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∫ 2

0

∫ 2π

0

√r2

8− r2 + 1rdrdθ = (16− 8√

2)π

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For example 3, the domain of parametrization is an annulus 1 ≤ r ≤ 3.

S =

∫ 3

1

∫ 2π

0

√4r2 + 1rdrdθ

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Think of the surface area S as

S =

∫S

ds

One defines the surface integral of any function f on the surface as

∫S

fdS =

∫ ∫D

f (x(u, v), y(u, v))‖ ∂r∂u× ∂r∂v‖dA

wheredA = dudv or dvdu.

Examples will be provided later!

PR (FIU) MAC 2313 73 / 104

14.5 Triple Integrals

An example of a 3-dimensional region R:

The region bounded by

x = 0, y = 0, z = 0

andx + 2y + 3z = 6.

After sketching the region, it can be described as follows:

0 ≤ x ≤ 60 ≤ y ≤ 6−x

20 ≤ z ≤ 6−x−2y

3

PR (FIU) MAC 2313 74 / 104

There are up to 5 more different descriptions of R. Any triple integral∫ ∫ ∫R fdV could be written as an iterated integral

∫ 6

0

∫ 6−x2

0

∫ 6−x−2y3

0f (x , y , z)dzdydx .

PR (FIU) MAC 2313 75 / 104

Example

Find the volume integral of f (x , y , z) = x + y + z over the box boundedby x = 1, y = 2, z = 1 + x and the coordinate planes.

After sketching the region, it can be described as follows:

0 ≤ x ≤ 10 ≤ y ≤ 20 ≤ z ≤ 1 + x

PR (FIU) MAC 2313 76 / 104

Example

Find the volume integral of f (x , y , z) = x + y + z over the box boundedby x = 1, y = 2, z = 1 + x and the coordinate planes.

After sketching the region, it can be described as follows:

0 ≤ x ≤ 10 ≤ y ≤ 20 ≤ z ≤ 1 + x

PR (FIU) MAC 2313 76 / 104

The integral is written as follows:∫ 10

∫ 20

∫ 1+x0 (x + y + z)dzdydx =

∫ 10

∫ 20 (x + y)z + z2

2

∣∣1+x0 dydx

=∫ 1

0

∫ 20 (1 + x)(x + y) + (1+x)2

2 dydx=

∫ 10 (1 + x)(xy + y2

2 )∣∣20 + (1 + x)2dx

= 3∫ 1

0 (1 + x)2dx= (1 + x)3|10 = 7

PR (FIU) MAC 2313 77 / 104

Use a triple integral to find the volume of the solid bounded by thesurface y = x2, the plane y + z = 4 and z = 0.

After sketching, the region is described as follows:

−2 ≤ x ≤ 2x2 ≤ y ≤ 40 ≤ z ≤ 4− y

PR (FIU) MAC 2313 78 / 104

Use a triple integral to find the volume of the solid bounded by thesurface y = x2, the plane y + z = 4 and z = 0.

After sketching, the region is described as follows:

−2 ≤ x ≤ 2x2 ≤ y ≤ 40 ≤ z ≤ 4− y

PR (FIU) MAC 2313 78 / 104

The volume is given by∫ 2−2

∫ 4x2

∫ 4−y0 dzdydx =

∫ 2−2

∫ 4x2(4− y)dydx

=∫ 2−2[4y − y2

2 ]∣∣4x2dx∫ 2

−2(8− 4x2 + x4

2 )dx= 16− 32

3 + 3210 + 16− 32

3 + 3210

PR (FIU) MAC 2313 79 / 104

14.7 Change of variables in multiple integrals:Jacobians

Motivation: Recall the u substitution technique of integration.

PropositionFrom ∫ b

af (x)dx ,

make a substitution x = g(u), dx = g′(u)du.The integral becomes:∫ g−1(b)

g−1(a)f (g(u)g′(u)du =

∫ β

αf (g(u)|g′(u)|du, α < β

PR (FIU) MAC 2313 80 / 104

Proof

If g−1(a) < g−1(b), then g′(u) > 0 and thus, g′(u) = |g′(u)|.∫ b

af (x)dx =

∫ g−1(b)

g−1(a)f (g(u))|g′(u)|du.

If g−1(b) < g−1(a), then g′(u) < 0, thus g′(u) = −|g′(u)|.∫ b

af (x)dx = −

∫ g−1(b)

g−1(a)f (g(u))|g′(u)|du =

∫ g−1(a)

g−1(b)f (g(u))|g′(u)|du.

In each case, the integral is given by∫ β

αf (g(u))|g′(u)|du, α < β.

PR (FIU) MAC 2313 81 / 104

Generalization to higher dimensions

Under a reparameterization (x , y) = r(u, v), from a domain S into adomain R, a small rectangle with vertices(u0, v0), (u0 + ∆u, v0), (u0, v0 + ∆v), (u0 + ∆u, v0 + ∆v) transformsinto R, approximated by a rectangle based on r(u0, v0) and edges

a ' r(u0 + ∆u, v0)− r(u0, v0)

∆u∆u ' ∂r

∂u∆u

and similarly,

b ' ∂r∂v

∆v

The elementary surface ∆A in the x , y parameters is approximately

∆A = ∆x∆y ' ‖ ∂r∂u ∆u × ∂r

∂v ∆v‖ = ‖ ∂r∂u ×

∂r∂v ‖∆u∆v

= |det

(∂x∂u

∂y∂u

∂x∂v

∂y∂v

)∆u∆v

PR (FIU) MAC 2313 82 / 104

The change of variable formula in double integrals is therefore: when(x , y) = r(u, v),∫ ∫

Rf (x , y)dAx ,y =

∫ ∫S

f (u, v)|det(∂(x , y)

∂(u, v))|dAu,v .

The determinant det(∂(x ,y)∂(u,v) ) is called the Jacobian of the coordinate

change.

PR (FIU) MAC 2313 83 / 104

Example

Compute the integral ∫ ∫R

exydA

where R is determined by

y =12

x , y = x , y =1x, y =

2x

using the transformation:

u =yx, v = xy

PR (FIU) MAC 2313 84 / 104

The domain S is described as

u =12, u = 1, v = 1, v = 2

We need the coordinates changes as

x =

√vu, y =

√uv

Then the Jacobian is∂(x , y)

∂(u, v)= − 1

2u.∫ ∫

RexydAx ,y =

∫ ∫S

ev 1|2u|

dAu,v =12

∫ 2

1

∫ 1

12

1u

ev dudv .

PR (FIU) MAC 2313 85 / 104

In triple integrals, the Jacobian is a 3 by 3 determinant

det∂(x , y , z)

∂(u, v ,w).

∫ ∫ ∫R

f (x , y , z)dVx ,y ,z =

∫ ∫ ∫S

f (u, v ,w)| ∂(x , y , z)

∂(u, v ,w)dVu,v ,w .

Example: Find the volume of

x2 +y2

4+

z2

9= 1

using the transformation

x = u, y = 2v , z = 3w

PR (FIU) MAC 2313 86 / 104

14.6 Triple Integrals in Cylindrical and SphericalCoordinates

Example: Evaluate the integral using spherical coordinates:∫ 2

0

∫ √4−y2

0

∫ √4−x2−y2

0

2z√x2 + y2

dzdxdy .

PR (FIU) MAC 2313 87 / 104

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)).

PR (FIU) MAC 2313 88 / 104

Examples

Given a differentiable function

f : R3 → R,

∇f is a vector field, the gradient vector field of f .

∇f : R3 → R3

(x , y , z) 7→ ∇f (x , y , z) = ( ∂f∂x ,

∂f∂y ,

∂f∂z )

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DefinitionA curve C(t) = (x(t), y(t), z(t)) such that c′(t) = F (C(t)) is called aflow line for the vector field F .

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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

dxdt = −ydydt = x

PR (FIU) MAC 2313 91 / 104

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

dxdt = −ydydt = x

PR (FIU) MAC 2313 91 / 104

Solving the system leads to:

d2xdt2 = −dy

dt= −x

That is:

d2xdt2 + x = 0.

The solution is :x(t) = a cos t + b sin t

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From dydt = x , one obtains

y(t) = a sin t − b cos t + C

with C = 0 since dxdt = −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√

a2 + b2,centered at (0,0).

PR (FIU) MAC 2313 94 / 104

Example 2

F (x , y , z) = (x

x2 + y2 ,y

x2 + y2 ).

C(t) = (x(t), y(t))

dxdt = x

x2+y2

dydt = y

x2+y2

PR (FIU) MAC 2313 95 / 104

1x2 + y2 =

1x

dxdt

=1y

dydt

1x

dxdt− 1

ydydt

= 0

ddt

(lnxy

) = 0

xy

= C

orx = Cy

The flow lines are straight lines through the origin, minus the originitself.

PR (FIU) MAC 2313 96 / 104

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 closedsurface. The divergence or flux density of a vector field F is defined by

div ~F (x , y , z) = limvolume(σ)→0

∫σ~F . ~dS

(volume enclosed by σ)

In rectangular coordinates:

div ~F =∂F1

∂x+∂F2

∂y+∂F3

∂z

Observation:div curl ~F = 0

This can be shown directly (do the calculation!)PR (FIU) MAC 2313 98 / 104

On a doubly connected region, one has the converse: If div ~G = 0,then ~G = curl ~F for some ~F .

Examples of divergence free vector fields, or solenoidal vector fields,are the magnetic fields ~B. Maxwell’s equation states exactly that:

div ~B = 0.

Another example: For a magnetic dipole (or a current loop), withconstant dipole moment ~µ,

~B = − ~µ

‖~r‖3+ 3

(~µ.~r)~r‖~r‖5

, ~r 6= ~0

Show that div ~B = 0.

PR (FIU) MAC 2313 99 / 104

Alternate notation:With ~∇ =< ∂∂x ,

∂∂y ,

∂∂z >

div ~F = ~∇.~F

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The divergence Theorem expresses the total flux as the integral of theflux density.

∫S

~F .~n dS =

∫W

div ~FdV

where S is the boundary of the region W , oriented by outer unitnormals.

PR (FIU) MAC 2313 101 / 104

Example: Use the divergence Theorem to find the flux ofF (x , y , z) = (x2y ,−xy2, z + 2) across the surface σ of the solidbounded above by the plane z = 2x and below by the paraboloidz = x2 + y2.

PR (FIU) MAC 2313 102 / 104

16.8: Stokes’ Theorem

Stokes’ Theorem expresses the total circulation of ~F as the integral ofthe circulation density:

∫σ

curl ~F .~n dS =

∫∂σ

~F . ~dr

A special case of Stokes Theorem is Green’s Theorem:

~F =< f ,g,0 >

PR (FIU) MAC 2313 103 / 104

Example: Verify Stokes’ Theorem for

~F (x , y , z) =< x , y , z >

where σ the upper hemisphere z =√

a2 − x2 − y2 with upwardorientation.

PR (FIU) MAC 2313 104 / 104