Engineering Analysis 2 : Multivariate Functions

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Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Engineering Analysis 2 : Multivariate functions

P. Rees, O. Kryvchenkova and P.D. Ledger,

[email protected]

College of Engineering, Swansea University, UK

PDL (CoE) SS 2017 1/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Outline

1 Multivariate functions

2 Partial Differentiation

3 Higher Order Partial Derivatives

4 Total Differentiation

5 Line Integrals

6 Surface Integrals

PDL (CoE) SS 2017 2/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Introduction

Recall from EG189: analysis of a function of single variable: Differentiation d f (x) d x

Expresses the rate of change of a function f with respect to x. Integration Z f (x) d x

Reverses the operation of differentiation and can used to work out areas under curves, and as we have just seen to solve ODEs.

We’ve seen in vectors we often need to deal with physical ﬁelds, that depend on more than one variable.

We may be interested in rates of change to each coordinate (e.g. x, y, z), time t, but sometimes also other physical ﬁelds as well.

PDL (CoE) SS 2017 3/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Introduction (Continue ...)

Example 1: the area A of a rectangular plate of width x and breadth y can be calculated

A = xy

The variables x and y are independent of each other.

In this case, the dependent variable A is a function of the two independent variables x and y as A = f (x, y) or A ≡ A(x, y)

PDL (CoE) SS 2017 4/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Introduction (Continue ...)

Example 2: the volume of a rectangular plate is given by

V = xyz

where the thickness of the plate is in z-direction.

In this case: V — is the dependent variable x, y, and z — are the independent variables We write V = f (x, y, z) or V ≡ V(x, y, z).

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Visualisation of Functions of Two and Three Variables

In general a function f can be a function of n independent variables. But, here we restrict ourselves to functions of two or three independent variables.

If n = 2 then we can write

f (x, y)

This is a function of two independent variables x and y. How to visualise such a function: 1. Use level curves which trace in the x, y domain on which the function f (x, y) has a constant value. 2. Plot the points corresponding to (x, y, z) where we choose z = f (x, y) in a rectangular coordinate system. ie the values of z are chosen to be the values of the function for a given x, y (rather than the z coordintates) and are shown as a surface.

PDL (CoE) SS 2017 6/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Visualisation of Functions of Two and Three Variables

Example: We wish to visualise the function

f (x, y) = x2 + y2

Solution: By using MATLAB, we can make a level surface plot and/or a surface plot of this functions as:

−2

−4

−6

−8

−10 −10 −8 −6 −4 −2 0 2 4 6 8 10

(a) The level surface plot (b) The surface plot

PDL (CoE) SS 2017 7/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Outline

1 Multivariate functions

2 Partial Differentiation

3 Higher Order Partial Derivatives

4 Total Differentiation

5 Line Integrals

6 Surface Integrals

PDL (CoE) SS 2017 8/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Partial Differentiation

1st Semester: the derivative of a function f (x) measures the slope of the tangent to the graph of the function.

If we have a function f (x, y) of two variables, slope has no sense because z = f (x, y) deﬁnes a surfaces in 3D. Simplest surface f (x, y) = 0 is ﬂat in both the x and y directions.

0.5

−0.5

−1 5 5 0 0

−5 −5 Figure: Visualisation surfaces for f (x, y) = 0

PDL (CoE) SS 2017 9/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Partial Differentiation

Consider f (x, y) = x + 2y i if we move along a line of fixed y and increasing x the slope is equal to 1. ii if we move along a line for which x is fixed and y is increasing then we find that the slope is equal to 2.

Visualisation of surface for f (x, y) = x + 2y

PDL (CoE) SS 2017 10/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Partial Differentiation

For a general surface the slope will be different depending on which direction we move in.

To ﬁnd out rates of change for functions of more than 1 variable we need a new kind of derivative —called a partial derivative.

The partial derivative of f (x, y) with respect to x is deﬁned as

f (x + ∆x, y) − f (x, y) lim ∆x→0 ∆x i.e. we differentiate: f (x, y) with respect to x while keeping y constant (ﬁxed). The partial derivative of f (x, y) with respect to x is the same as measuring the slope in the x direction. ∂f or ∂f /∂x or fx ∂x

PDL (CoE) SS 2017 11/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Partial Differentiation

In writing, care must be taken to distinguish between:

df ∆f ∂f , and dx ∆x ∂x Similarly, the partial derivative of f (x, y) with respect to y is

∂f f (x, y + ∆y) − f (x, y) = lim ∂y ∆x→0 ∆y

Differentiating f (x, y) with respect to y by keeping x constant. This partial derivative is the same as measuring the slope in the y direction.

PDL (CoE) SS 2017 12/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Partial Differentiation

Directional derivative

∂f ∂f If we know and , we know the surface for any direction. ∂x ∂y

If we deﬁne α — an angle to the x axis then the slope is:

∂f ∂f cos α + sin α ∂x ∂y

PDL (CoE) SS 2017 13/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Rules of Partial Differentiation

The rules of partial differentiation are very similar to those of ordinary differentiation Rule 1 (scaler multiplication rule) If k is a constant then ∂ ∂f ∂ ∂f (kf ) = k (kf ) = k ∂x ∂x ∂y ∂y

Rule 2 (sum rule) If u = f (x, y) and v = g(x, y) then

∂ ∂u ∂v ∂ ∂u ∂v (u + v) = + (u + v) = + ∂x ∂x ∂x ∂y ∂y ∂y

Rule 3 (product rule) If u = f (x, y) and v = g(x, y) then

∂ ∂v ∂u ∂ ∂v ∂u (uv) = u + v (uv) = u + v ∂x ∂x ∂x ∂y ∂y ∂y

PDL (CoE) SS 2017 14/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Partial Differentiation

Example: Given the function f (x, y) = x2y3 + 3y + x, determine its partial derivative with respect to x and y. Hence determine its directional derivative for a direction at angle α = π/4 to the x axis. Solution: To ﬁnd the partial derivative of f (x, y) with respect to x, we differentiate f (x, y) and keep y constant ∂f = 2xy3 + 1 ∂x The partial derivative of f (x, y) with respect to y, by differentiating f (x, y) while keeping x constant ∂f = 3x2y2 + 3 ∂y The directional derivative is √ 2 (2xy3 + 1) cos α + (3x2y2 + 3) sin α = (2xy3 + 3x2y2 + 4) 2

PDL (CoE) SS 2017 15/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Partial Differentiation

PDL (CoE) SS 2017 15/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Rules of Partial Differentiation Continued

Rule 4 (quotient rule) If u = f (x, y) and v = g(x, y) then

∂u ∂v ∂u ∂v ∂ u v ∂x − u ∂x ∂ u v ∂y − u ∂y = = ∂x v v2 ∂y v v2

Rule 5 (simple composite–functions or simple chain rule) If u = g(f (x, y)) = g(w) then

∂u dg ∂w ∂u dg ∂w = = ∂x dw ∂x ∂y dw ∂y

we will later extend this concept to where g can not be written just in terms of a single vairable w.

PDL (CoE) SS 2017 16/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Partial Differentiation

Example: Determine ∂f /∂x and ∂f /∂y when f (x, y) is

a) x2y2 + 3xy − x + 2 b) sin(x2 − 3y)

Solution: a) For f (x, y) = x2y2 + 3xy − x + 2 we have

∂f ∂f = 2xy2 + 3y − 1 = 2x2y + 3x ∂x ∂y

b) For f (x, y) = sin(x2 − 3y) = sin w with w = x2 + y2 we have

∂f df ∂w ∂ = = cos(x2 − 3y) (x2 − 3y) = 2x cos(x2 − 3y) ∂x dw ∂x ∂x ∂f = −3 cos(x2 − 3y) ∂y

PDL (CoE) SS 2017 17/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Generalising Partial Differentiation

We’ve already seen partial differentiation applied to functions of two variables.

The concept may be extended to functions of many variables.

For a function f (x1, x2, ··· xn) of n variables, the partial derivative with respect to xi is

∂f = ∂xi f (x , x , ··· , xi + ∆xi, x , ··· , xn) − f (x , x , ··· , xi, x , ··· , xn) lim 1 2 i+1 1 2 i+1 ∆xi→0 ∆xi

Differentiating the function with respect to xi while keeping all other n − 1 variables constant.

PDL (CoE) SS 2017 18/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Generalising Partial Differentiation

Example: Determine ∂f /∂x, ∂f /∂y and ∂f /∂z when

f (x, y, z) = xyz2 + 3xy − z

Solution: We obtain that ∂f = yz2 + 3y ∂x ∂f = xz2 + 3x ∂y ∂f = 2xyz − 1 ∂x

PDL (CoE) SS 2017 19/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Generalising the Chain Rule

We’ve already seen a particular case of the the chain rule in partial differentiation, now lets generalise it.

Let z = g(v, w) and v and w are themselves functions of two independent variables x and y. z is a function of x and y, say G(x, y). We want to differentiate z with respect to x or y ∂z ∂z ∂u ∂z ∂w ∂z ∂z ∂u ∂z ∂w = + = + ∂x ∂v ∂x ∂w ∂x ∂y ∂v ∂y ∂w ∂y In matrix notation ∂z ! ∂v ∂w ! ∂z ∂x ∂x ∂x ∂v ∂z = ∂v ∂w ∂z ∂y ∂y ∂y ∂w This is the chain rule for a function of a two variables.

If z = g(w) and w = f (x, y) this reduces to the simpler case previously presented.

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Chain RuleI

Example: Find ∂T/∂r and ∂T/∂θ when

T(x, y) = x2 + 2xy + y3x2

and x = r cos θ and y = r sin θ Solution: By the chain rule ∂T ∂T ∂x ∂T ∂y = + ∂r ∂x ∂r ∂y ∂r In this example ∂T ∂T = 2x + 2y + 2xy3 = 2x + 3x2y2 ∂x ∂y and ∂x ∂y = cos θ = sin θ ∂r ∂r so that ∂T = (2x + 2y + 2xy3) cos θ + (2x + 3x2y2) sin θ ∂r = (2r cos θ + 2r sin θ + 2r4 cos θ sin3 θ) cos θ + (2r cos θ + 3r4 cos2 θ sin2 θ) sin θ

PDL (CoE) SS 2017 21/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Chain RuleII

Similarly

∂T = −(2x + 2y + 2xy3)r sin θ + (2x + 3x2y2)r cos θ ∂θ = −(2r cos θ + 2r sin θ + 2r4 cos θ sin3 θ)r sin θ + (2r cos θ + 3r4 cos2 θ sin2 θ)r cos θ

PDL (CoE) SS 2017 22/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Chain RuleI

Example: Find dR/ ds when

R(x, y) = cosh(x2 + 3y)

and x(s) = s2 + 3s and y(s) = sin s. Solution: For this example, x and y are functions of s only so

dR ∂R dx ∂R dy = + ds ∂x ds ∂y ds

which gives

dR = 2x(2s + 3) sinh(x2 + 3y) + 3 cos s sinh(x2 + 3y) ds = 2(s2 + 3s)(2s + 3) sinh((s2 + 3s)2 + 3 sin s) + 3 cos s sinh((s2 + 3s)2 + 3 sin s) = 2(2s3 + 9s2 + 9s) sinh((s2 + 3s)2 + 3 sin s) + 3 cos s sinh((s2 + 3s)2 + 3 sin s)

PDL (CoE) SS 2017 23/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Outline

1 Multivariate functions

2 Partial Differentiation

3 Higher Order Partial Derivatives

4 Total Differentiation

5 Line Integrals

6 Surface Integrals

PDL (CoE) SS 2017 24/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Higher order partial derivatives

∂f ∂f We have considered functions like f (x, y) and found its partial derivatives ∂x and ∂y . If the partial derivatives are also functions of x and y, they can also be differentiated with respect to x and y. We deﬁne higher order partial derivatives as follows

∂2f ∂ ∂f = ∂x2 ∂x ∂x ∂2f ∂ ∂f = ∂y2 ∂y ∂y ∂2f ∂ ∂f = ∂x∂y ∂x ∂y ∂2f ∂ ∂f = ∂y∂x ∂y ∂x

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Higher Order Partial Derivatives

∂f ∂f ∂2f ∂2f If , , and exist and are continuous, then it follows that ∂x ∂y ∂x∂y ∂y∂x

∂2f ∂2f = ∂x∂y ∂y∂x

If the conditions are not fulﬁlled these so called mixed partial derivatives are not equal. ∂2f ∂2f 6= ∂x∂y ∂y∂x

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Higher Order Partial DerivativesI

Example: For the function f (x, y) = sin x cos y + x3ey ﬁnd all the second order partial derivatives.

Solution: First we ﬁnd the ﬁrst order partial derivatives

∂f ∂f = cos x cos y + 3x2ey = − sin x sin y + x3ey ∂x ∂y

Then by differentiating these expressions again we can ﬁnd the second order derivatives

∂2f ∂2f = − sin x cos y + 6xey = − sin x cos y + x3ey ∂x2 ∂y2 ∂2f ∂2f = − cos x sin y + 3x2ey = ∂x∂y ∂y∂x

∂2f ∂2f In this case, we have that = ∂x∂y ∂y∂x

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Higher Order Partial DerivativesI

Example: Find all second partial derivatives of z = sin(xy).

Solution: First of all the ﬁrst partial derivatives are found.

∂2z ∂2z = y cos(xy) = x cos(xy) ∂x ∂y Then each of these is differentiated with respect to x:

∂2z ∂ ∂z = = −y2 sin(xy) ∂x2 ∂x ∂x ∂2z ∂ ∂z = = −xy sin(xy) + cos(xy) ∂x∂y ∂x ∂y

Note here the need to use the product rule to differentiate x cos(xy) with respect to x. Also ∂2z ∂ ∂z ∂2z = = −xy sin(xy) + cos(xy) = ∂y∂x ∂y ∂x ∂x∂y ∂2z ∂ ∂z = = −y2 sin(xy) ∂y2 ∂y ∂y

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Outline

1 Multivariate functions

2 Partial Differentiation

3 Higher Order Partial Derivatives

4 Total Differentiation

5 Line Integrals

6 Surface Integrals

PDL (CoE) SS 2017 29/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Total differentiation

Let us deﬁne: Function z = f (x, y) — function of two variables x and y. ∆x is a small change in x, ∆y a small change in y and ∆z a small change in z. It follows that ∆z = f (x + ∆x, y + ∆y) − f (x, y)

Rewrite this as the sum of two terms: 1. the ﬁrst of which shows the change in z due to a change in x 2. the second which shows the change in z due to a change in y

∆z = [f (x + ∆x, y + ∆y) − f (x, y + ∆y)] + [f (x, y + ∆y) − f (x, y)]

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Total Differentiation

Multiply the ﬁrst term by ∆x/∆x = 1 and the second term by ∆y/∆y = 1

[f (x + ∆x, y + ∆y) − f (x, y + ∆y)] ∆z = ∆x+ ∆x [f (x, y + ∆y) − f (x, y)] + ∆y ∆y

By letting ∆x, ∆y and ∆z → 0, we get

∂f ∂f dz = dx + dy ∂x ∂y

In this expression dx, dy and dz are called differentials.

PDL (CoE) SS 2017 31/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Total Differentiation

If z = f (x) so that it is a function of one variable, the formula takes the form

df dz = dx dx

If w = f (x, y, z) is a function of three variables we have

∂f ∂f ∂f dw = dx + dy + dz ∂x ∂y ∂z

We can use differentials to calculate errors. If z = f (x, y) and ∆x and ∆y are errors in x and y, then the error in z is approximately given by

∂f ∂f ∆z ≈ ∆x + ∆y ∂x ∂y

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Total DifferentiationI

Example: We want to estimate p(3.01)2 + (3.97)2

p 2 2 Solution: Let z = f (x, y) = x + y . √ If we set x = 3 and y = 4 we can easily compute z = 32 + 42 = 5. Now p(3.01)2 + (3.97)2 is z when x is increased by ∆x = 0.01 and when y is decreased by 0.03, i.e., ∆y = −0.03

∂f ∂f ∆z ≈ ∆x + ∆y ∂x ∂y 1 1 = 2x(x2 + y2)−1/2∆x + 2y(x2 + y2)−1/2∆y 2 2 x y = ∆x + ∆y px2 + y2 px2 + y2 3 4 = × 0.01 + × (−0.03) = −0.018 5 5

So p(3.01)2 + (3.97)2 ≈ 5 + ∆z = 5 − 0.018 ≈ 4.98.

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Total DifferentiationI

Example: The height of a cylinder is under measured by 3% and the raduis is over measured by 2% we wish to estimate the percentage error in the volume.

Solution: The volume of a cylinder is given by V = πr2h as ∂V ∂V = 2πrh = πr2 ∂r ∂h The error in the volume may be written as ∂V ∂V ∆V ≈ ∆r + ∆h ∂r ∂h = 2πrh∆r + πr2∆h

As we are interested in the percentage error, we divide this by V

2πrh πr2 ∆V ≈ ∆r + ∆h πr2h πr2h 2∆r ∆h = + r h ∆r 2 ∆h 3 ∆V 1 From the question we know that = and = − giving = . This r 100 h 100 V 100 means that the volume is overestimated by 1%.

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Total DifferentiationI

Example: Find the total differential coefﬁcient of

x2y

with respect to x, when x and y are connected by the relation:

x2 + xy + y2 = 1

Solution: Let deﬁne u = x2y, then the total differential is

∂u ∂u du = dx + dy ∂x ∂y

Thus the total differential coefﬁcient of u with respect to x is

du ∂u ∂u dy = + dx ∂x ∂y dx du dy = 2xy + x2 dx dx

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Total DifferentiationII

2 2 ∂f From the implicit relation f = x + xy + y = 1, we ﬁrst ﬁnd fx = ∂x = 2x + y and ∂f fy = ∂y = 2y + x and see that f 2x + y x = fy x + 2y but, also using the techniques of implicit differentiation from EG189 we have in this case dy dy dy 2x − y fx 2x + y + x + 2y = 0 = − = − dx dx dx x + 2y fy so that du dy 2x + y = 2xy + x2 = 2xy + x2 − dx dx x + 2y du x2(2x + y) = 2xy − . dx x + 2y

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Outline

1 Multivariate functions

2 Partial Differentiation

3 Higher Order Partial Derivatives

4 Total Differentiation

5 Line Integrals

6 Surface Integrals

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Integration

In science and engineering: three types of integrals commonly arsie: line integrals, surface integrals and volume integrals.

Line integrals

Are used for computing induced voltage in a coil, mass of wire, work done by a force moving in a vector ﬁeld...

The basic form of a line integral is Z f (x, y, z) ds C here s is measured around the contour (possibly a curve or a straight line) C over which the integral is to be performed.

The contour can exist at some arbitrary location in space and rather than integrate than Z b integrate with respect to x as in EG189 in f (x)dx we need to integrate f (x, y, z) with a respect to s along C.

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Line Integrals

We don’t usually specify limits for line integrals as the starting and end points for the integral are deﬁned by the choice of contour C, which can be arbitrary:

x When performing contour integrals, we are working out the area below the function when restricted to the curve C as illustrated below for the case of f (x, y)

f(x,y)

y Function C

s Contour

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Line Integrals

p Note that in Cartesian coordinates ds = dx2 + dy2 + dz2 and in two–dimensions this simpliﬁes to

s 2 s 2 p dy dx ds = dx2 + dy2 = + 1dx = 1 + dy dx dy

provided s increases as x (resp. y) increases, with the different arrangements offering advantages for lines aligned with the x or y axes repectively. Example Compute the line integral of f (x, y) = x2 + y along the contour deﬁned by the line y = 2x between the points (−1, −2) and (1, 2). Solution Z Z q f (x, y)ds = x2 + yds ds = (2)2 + 1dx C C since dy/dx = 2. Thus

Z 1 √ √ x3 1 √ x2 + 2x 5dx = 5 + x2 = 2 5 −1 3 −1

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Working with Parameterisations r dy 2 However, particular care must be exercised with setting ds = dx + 1dx if s r 2 dx increasing as x is decreasing and similarly with ds = 1 + dy dy with s increasing and y decreasing. in such cases we need the negative root

Example R Find C ds for the curve y = x between a line starting at (1, 1) and ﬁnishing at (0, 0) Solution √ The solution is just the length of the line between the two points i.e. 2. √ A ﬁrst approach might be to say dy = dx and so ds = 2dx and since we go from (1, 1) to (0, 0) then to say Z Z 0 √ √ √ 0 ds = 2dx = − 2[x]1 = − 2 C 1 but this has the sign!

The trick√ is to use the negative root (as s is increasing as x is decreasing) so that ds = − 2dx √ Z 2 Z 0 √ Z 1 √ √ √ 1 ds = − 2dx = 2dx = 2[x]0 = 2 0 1 0

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The trick√ is to use the negative root (as s is increasing as x is decreasing) so that ds = − 2dx √ Z 2 Z 0 √ Z 1 √ √ √ 1 ds = − 2dx = 2dx = 2[x]0 = 2 0 1 0

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Line Integrals R Example: Evaluate C xy ds from (1, 0) to (0, 1) along the curve C that is the portion of x2 + y2 = 1 in the ﬁrst quadrant. y (0,1) C

(1,0) x

Solution: dy dy x By implicit differentiation 2x + 2y = 0 so = − . Thus dx dx y s s x2 1 ds = − + 1dx = − dx y2 1 − x2

since s is increasing as x is decreasing. If we integrate wrt s we go from (1, 0) to (0, 1) thus s 1 Z (0,1) Z 0 p 1 Z 1 x2 1 xy ds = − x 1 − x2 dx = xdx = = 2 (1,0) 1 1 − x 0 2 0 2

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Line Integrals R Example: Evaluate C xy ds from (1, 0) to (0, 1) along the curve C that is the portion of x2 + y2 = 1 in the ﬁrst quadrant. y (0,1) C

(1,0) x

Solution: dy dy x By implicit differentiation 2x + 2y = 0 so = − . Thus dx dx y s s x2 1 ds = − + 1dx = − dx y2 1 − x2

since s is increasing as x is decreasing. If we integrate wrt s we go from (1, 0) to (0, 1) thus s 1 Z (0,1) Z 0 p 1 Z 1 x2 1 xy ds = − x 1 − x2 dx = xdx = = 2 (1,0) 1 1 − x 0 2 0 2

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Line Integrals

Example: Evaluate the integral Z I = [(x2 + 2y) ds + (x + y2) ds] C from (0, 1) to (2, 3) along the curve C deﬁned by y = x + 1 Solution: √ Since y = x + 1 then dy = dx and ds = 2dx since in this case as s is increasing x is increasing.

Z 2 √ I = [(x2 + 2(x + 1)) + (x + (x + 1)2)] 2dx 0 √ √ Z 2 √ 2 5 2 64 2 = 2 (2x2 + 5x + 3) dx = 2 x3 + x2 + 3x = 0 3 2 0 3

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Line Integrals

Z 2 √ I = [(x2 + 2(x + 1)) + (x + (x + 1)2)] 2dx 0 √ √ Z 2 √ 2 5 2 64 2 = 2 (2x2 + 5x + 3) dx = 2 x3 + x2 + 3x = 0 3 2 0 3

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Line Integrals

R We’ve already seen line integrals in the form C f (x, y) ds for Cartesian coordinates. Other forms include:

r 2 r 2 dy dx ds = 1 + dx dx = 1 + dy dy in 2D Cartesian form

r 2 2 dx dy ds = dt + dt dt in 2D parametric form

r 2 2 dr ds = r + dθ dθ in 2D polar form (z constant) If a line integral is such that the integration is performed around a closed (simple) H curve, then we denote this type of integral by C ds — is evaluated by travelling around C in an anticlockwise direction.

Another form found when dealing with vector ﬁelds e.g. F(x, y, z) is the line integral R R C F · ds = C F · τ ds where ds ≡ τ ds and τ is a unit tangent to C whose orientation changes with position.

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Line Integrals

Example: Evaluate Z (z + x2 − 2y)ds C along the curve C which is speciﬁed parametrically as x = t2, y = t2 and z = 2t2 from (0, 0, 0) to (1, 1, 2). Solution: On the curve C, we generalise ds to s dx 2 dy 2 dz 2 p √ ds = + + dt = 4t2 + 4t2 + 16t2dt = 24tdt dt dt dt

since dx = 2tdt, dy = 2t dt and dz = 4t dt. Also at the point (0, 0, 0) t = 0 and at the point (1, 1, 2) t = 1. In this case as s is increasing as t increasing and so the positive root is correct. Z √ Z 1 (z + x2 − 2y)ds = 24 (2t2 + t4 − 2t2 )tdt C 0 √ Z 1 = 24 t5 dt 0 √ √ t6 1 24 = 24 = 6 0 6

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Line IntegralsI

I ds Example: Evaluate the integral p where C is the unit square with vertices C x2 + y2 (1, 1), (−1, 1), (−1, −1), (1, −1).

y (−1,1) (1,1) B A

x DE (−1,−1) (1,−1)

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Line IntegralsII

Solution: We can break the integral into four parts I Z Z Z Z f (x, y)ds = f (x, y)ds + f (x, y)ds + f (x, y)ds + f (x, y)ds C CAB CBD CDE CEA and use the parameterisation approach: 1 1 Along CAB y = 1 and x(t) = 2 (1 − t) − 2 (1 + t) = −t, −1 ≤ t ≤ 1 dx dy q = −1, = 0, ds = (−1)2 + 02dt = dt dt dt

Z ds Z 1 dt √ = √ 2 2 CAB x + 1 −1 t + 1

Along CBD x = −1 and y(t) = −t, −1 ≤ t ≤ 1

Z ds Z 1 dt = √ p 2 2 CBD y + 1 −1 t + 1

Along CDE y = −1 and x(t) = t, −1 ≤ t ≤ 1

Z ds Z 1 dt √ = √ 2 2 CDE x + 1 −1 t + 1 PDL (CoE) SS 2017 47/ 67 Multivariate functions Partial Differentiation Higher Order Partial Derivatives Total Differentiation Line Integrals Surface Integrals

Line Integrals III

Along CEA x = 1 and y(t) = t, −1 ≤ t ≤ 1

Z ds Z 1 dt = √ p 2 2 CEA y + 1 −1 t + 1

Note that each part the pameterisation t increases as s increases. Thus the integral becomes

I Z 1 1 ds dt h −1 i p = 4 √ = 4 sinh t C x2 + y2 −1 1 + t2 −1 = 8 sinh−1 1

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Outline

1 Multivariate functions

2 Partial Differentiation

3 Higher Order Partial Derivatives

4 Total Differentiation

5 Line Integrals

6 Surface Integrals

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Surface Integrals

Z b Recall f (x)dx is equal to the area under the curve f (x) between x = a and x = b a

f(x)

* f(x r )

x x* 0 x0 x1 r−1 r xr xn x a b

Figure: Integral of a function of a single variable

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Surface integrals

We would like have integrals of functions of more than one variable. Let’s consider z = f (x, y) and a region R of the xy plane:

Δ A i

Figure: Integral of a function of two variables.

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Surface integrals

We deﬁne the integral of f (x, y) over R

n ZZ X f (x, y) dA = lim f (¯xi, ¯yi)∆Ai R n → ∞ i=1 ∆Ai → 0

where ∆Ai is an elemental area of R and (¯xi, ¯yi) is a point in ∆Ai.

f (x, y) represents a surface and f (¯xi, ¯yi)∆Ai = ¯zi∆Ai is the volume between the z = 0 and z = ¯zi whose base cross-section is ∆Ai.

The integral is the limit of the sum of all such volumes and so it is the volume under the surface of z = f (x, y) above R.

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Surface Integrals

If we introduce a series of lines which are parallel to the x and y axis: y

Δ yι

Δ x xι

Figure: Lines introduced for double integrals.

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Surface Integrals

We can write ∆Ai = ∆xi∆yi, giving

n ZZ ZZ X f (x, y) dA = f (x, y) dx dy = lim f (¯xi, ¯yi)∆xi∆yi n→∞ R R i=1 RR Integrals of the type R f (x, y) dx dy can be evaluated as repeated single integrals in x and y. They are usually called double integrals. RR For the particular case of the integral R dA — this equals to the area of region R, ie integration can be used to ﬁnd out the area of any shape!

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Surface Integrals

Two alternatives to evaluating double integrals. If data is given as y = g(x), i.e., y is some function of x, then we work out the integral by ﬁrst performing the integration with respect to y and then with respect to x, i.e. " # ZZ Z b Z x=g2(x) f (x, y) dA = f (x, y) dy dx R a y=g1(x)

Alternatively, if we have that x is expressed as some function of y, e.g.,x = h(y), then we ﬁrst perform the integration with respect to x and then integrate with respect to y " # ZZ Z d Z x=h2(y) f (x, y) dA = f (x, y) dx dy R c x=h1(y)

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Surface Integrals

In the particular case where the region R is a rectangle, then the limits of the integration are constant and so it does not matter whether integrate x or y ﬁrst.

ZZ Z d Z b Z b Z d f (x, y) dA = f (x, y) dx dy = f (x, y) dy dx R c a a c Other elemental surface differentials include dA = dxdy Cartesian coordinates, z constant. dA = dρdz Cylinderical coordinates, φ constant dA = ρdρdφ Cylinderical coordinates, z constant dA = r2 sin θdθdφ Spherical coordinates, r constant dA = r sin θdrdφ Spherical coordinates, θ constant dA = rdrdθ Spherical coordinates, φ constant and apart from simple rectangular/cylindrical/spherical geometries we need to take care with the order of integration.

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Surface IntegralsI

Example: Evaluate the integral

Z 1 Z 3 (x2 + y2) dx dy 0 1 Solution: If we integrate with respect to x ﬁrst, then we obtain

Z 1 Z 3 Z 1 1 x=3 (x2 + y2) dx dy = x3 + y2x dy 0 1 0 3 x=1 Z 1 26 26 2 1 28 = + 2y2 dy = y + y3 = 0 3 3 3 0 3 Alternatively with respect to y ﬁrst

Z 1 Z 3 Z 3 1 y=1 (x2 + y2) dx dy = x2y + y3 dx 0 1 1 3 y=0 Z 3 1 28 = x2 + dx = 1 3 3

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Surface IntegralsI

RR 2 2 Example: Evaluate R(x + y ) dA over a triangle with vertices (0, 0), (2, 0) and (1, 1). Solution:

y y y 1 1 1 y=x y=2−x

1 2 x 1 2 x 1 2 x

First, integrating with respect to x ﬁrst gives

ZZ Z 1 Z x=2−y (x2 + y2) dA = (x2 + y2) dx dy R 0 x=y Z 1 1 x=2−y Z 1 8 8 4 = x3 + y2x dy = − 4y + 4y2 − y3 dy = 0 3 x=y 0 3 3 3

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Surface IntegralsII

Next integrating with respect to y ﬁrst

ZZ Z 1 Z y=x (x2 + y2) dA = (x2 + y2) dy dx+ R 0 y=0 Z 2 Z y=2−x + (x2 + y2) dy dx 1 y=0

Here the integrals are

Z 1 Z y=x Z 1 1 y=x (x2 + y2) dy dx = x2y + y3 dx = 0 y=0 0 3 y=0 Z 1 4 1 = x3 dx = 0 3 3 Z 2 Z y=2−x Z 2 1 y=2−x (x2 + y2) dy dx = x2y + y3 dx = 1 y=0 1 3 y=0 Z 2 8 4 = − 4x + 4x2 − x3 dx = 1 1 3 3

RR 2 2 1 4 So R(x + y ) dA = 1 + 3 = 3 .

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Surface IntegralsI

Example: By using a surface integral, check that that surface area of a sphere of radius r = a is 4πa2 Solution: Z In this case we want to ﬁnd dA and is best to use spherical coordinates:

z θ y r

On A r = a is ﬁxed and so dA = r2 sin θdθdφ = a2 sin θdθdφ so

Z Z 2π Z π Z 2π Z π area = dA = a2 sin θdθdφ = a2 sin θdθ dφ 0 0 0 0 Z 2π Z 2π 2 π 2 2 =a [− cos θ]0 dφ = a 2dφ = 4πa 0 0 since for a complete sphere 0 ≤ φ ≤ 2π and 0 ≤ θ ≤ π.

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Surface Integrals

Example: Z 8 Z y Change the integral xdxdy in to an equivalent polar integral and evaluate 0 0 Solution:

8 x=y

x The region of integration is a triangular region. Recall the transformations x = ρ cos φ y = ρ sin φ dA = ρdρdφ For this region we need to integrate from φ = π/4 to φ = π/2, but the radial integration depends on the angle φ, for each φ we need to integrate from ρ = 0 to 8/ sin φ.

Z 8 Z y Z π/2 Z 8/ sin φ Z π/2 ρ3 8/ sin φ xdxdy = ρ cos φρdρdφ = cos φ dφ 0 0 π/4 0 π/4 3 0 Z π/2 512 cos φ 512 h iπ/2 256 = φ = − −2 φ = 3 d sin π/4 3 sin φ 6 π/4 3

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Surface Integrals

Example: Z 8 Z y Change the integral xdxdy in to an equivalent polar integral and evaluate 0 0 Solution:

8 x=y

Z 8 Z y Z π/2 Z 8/ sin φ Z π/2 ρ3 8/ sin φ xdxdy = ρ cos φρdρdφ = cos φ dφ 0 0 π/4 0 π/4 3 0 Z π/2 512 cos φ 512 h iπ/2 256 = φ = − −2 φ = 3 d sin π/4 3 sin φ 6 π/4 3

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Volume Integrals

Volume integrals — three successive integrals. Volume integrals are of the form: ZZZ dV V and are called triple integrals.

Evaluated in the same way as double integrals — we start by evaluating the inner integral and work outwards.

The main difﬁculty — to determine the correct limits for the integration — make a sketch of the region.

The differential dV is dV = dxdydz for Cartesian coordinates dV = ρdρdφdz for Cylindrical coordinates dV = r2 sin θdrdφdθ for Spherical coordinates In the Cartesian case, if the integrals are evaluated in the order x, y, z then the limits on the y integral may depend on z but not on x with similar restrictions for cylindrical and spherical coordinates.

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Volume IntegralsI

Example: A cube 0 ≤ x, y, z ≤ 1m has a variable density given by ρ = 1 + x + y + z kg/m3, what is the total mass of the cube Solution: The total mass is given by

ZZZ M = ρ dV = V Z 1 Z 1 Z 1 = (1 + x + y + z) dx dy dz = 0 0 0 Z 1 Z 1 x2 1 = x + + xy + xz dy dz 0 0 2 0 Z 1 Z 1 3 = + y + z dy dz = 0 0 2 Z 1 3y y2 1 = + + yz dz 0 2 2 0 Z 1 z2 1 5 = (2 + z) dz = 2z + = kg 0 2 0 2

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Volume Integrals

Example: A tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1) (m) in Cartesian coordinates has density ρ = 10x kg/m3, what is its total mass? Solution: Consider the sketches

z z

(0,0,1)

1−x−y

y y

(0,1,0) (0,1,0)

(0,0,0) 1−x (0,0,0)

(1,0,0) (1,0,0) x x

For the triangular base, we integrate x from 0 to 1 − y and y from 0 to 1. For z we integrate from 0 to 1 − x − y.

This is since the plane containing the points (1, 0, 0), (0, 1, 0) and (0, 0, 1) is

n · w = n · c n = a × b

and a = −1e1 + 1e3 and b = −1e2 + 1e3 are two vectors in the plane and c = e3 a vector pointing to a point in the plane. Thus n = e1 + e2 + e3 and n · w = x + y + z = 1 z = 1 − x − y

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Volume Integrals

Example: A tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1) (m) in Cartesian coordinates has density ρ = 10x kg/m3, what is its total mass? Solution: Consider the sketches

z z

(0,0,1)

1−x−y

y y

(0,1,0) (0,1,0)

(0,0,0) 1−x (0,0,0)

(1,0,0) (1,0,0) x x

For the triangular base, we integrate x from 0 to 1 − y and y from 0 to 1. For z we integrate from 0 to 1 − x − y.

This is since the plane containing the points (1, 0, 0), (0, 1, 0) and (0, 0, 1) is

n · w = n · c n = a × b

and a = −1e1 + 1e3 and b = −1e2 + 1e3 are two vectors in the plane and c = e3 a vector pointing to a point in the plane. Thus n = e1 + e2 + e3 and n · w = x + y + z = 1 z = 1 − x − y

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Volume Integrals

Thus Z 1 Z 1−y Z 1−x−y Z 1 Z 1−y Z 1−x−y mass = ρdzdxdy = 10 x dz dx dy 0 0 0 0 0 0 Z 1 Z 1−y Z 1 2 1−y! 1−x−y x =10 x [z]0 dx dy = 10 x − − yx dy 0 0 0 2 0 Z 1 (1 − y)2 (1 − y)3 1 10 =10 dy = 10 − = kg 0 2 6 0 6

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Volume Integrals

Z Example: Evaluate the integral zdV where H is the upper half of the sphere H x2 + y2 + z2 = 1. Solution: This is best done in spherical coordinates.

z θ y r

The transformations from spherical to Cartesian are

x = r sin θ cos φ y = r sin θ sin φ z = r cos θ

and dV = r2 sin θdrdθdφ. For the upper half of the sphere x2 + y2 + z2 = 1 π 0 ≤ r ≤ 1 0 ≤ φ ≤ 2π 0 ≤ θ ≤ 2

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Volume Integrals

Z Example: Evaluate the integral zdV where H is the upper half of the sphere H x2 + y2 + z2 = 1. Solution: This is best done in spherical coordinates.

z θ y r

The transformations from spherical to Cartesian are

x = r sin θ cos φ y = r sin θ sin φ z = r cos θ

and dV = r2 sin θdrdθdφ. For the upper half of the sphere x2 + y2 + z2 = 1 π 0 ≤ r ≤ 1 0 ≤ φ ≤ 2π 0 ≤ θ ≤ 2

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Volume Integrals

Thus

Z Z 1 Z 2π Z π/2 Z 1 Z 2π Z π/2 zdV = (4r cos θ)(r2 sin θ)dθdφdr = 4 r3 cos θ sin θdθdφdr H 0 0 0 0 0 0 Z 1 Z 2π 2 π/2 Z 1 4 1 3 sin θ 3 2π r =4 r dφdr = 2 r [φ]0 = 4π = π 0 0 2 0 0 4 0

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