Second Year Mathematics Revision€¦ · Integral Calculus Change of Variables Jacobians In single...

Second Year Mathematics RevisionCalculus - Part 1

Written by: Ethan Brown, Kabir AgrawalPresented by: Alex Zhu

UNSW Mathematics Society

24 Apr 2020

Ethan Brown, Kabir Agrawal Second Year Mathematics Revision

Integral Calculus

Table of Contents

1 Integral CalculusDouble IntegralsTriple IntegralsChange of VariablesCentre of MassLeibniz Rule


Integral Calculus Integrals

Integration

Definition ∫ x=b

x=af (x)dx = lim

∆x→0

x=b∑x=a

f (x∗i )∆x

By definition, this means that to compute an area under a curve, weconstruct rectangles of height f (x∗i ) and width ∆x over the intervala ≤ x ≤ b, and add the area of the rectangles and take the limit asthe rectangles become super tiny.


Integral Calculus Double Integrals

Double Integration

Double integration is no different, except we integrate over a 2dimensional region.

Definition

∫ y=d

y=c

∫ x=b

x=af (x , y)dxdy = lim

∆y→0lim

∆x→0

y=d∑y=c

x=b∑x=a

f (x∗i , y∗i )∆x∆y

When integrating over a more complicated region, the process isprecisely the same. The notation is more generally described as:

V =

∫∫Rf (x , y)dxdy =

∫∫Rf (x , y)dA



Evaluating integrals

Fubini’s Theorem

Suppose f (x , y) is a continuous function on R = [a, b]× [c , d ], then∫ y=d

y=c

∫ x=b

x=af (x , y)dxdy =

∫ x=b

x=a

∫ y=d

y=cf (x , y)dydx

Thus, it is possible to reorder the integral signs to make life easier.



Double Integration Examples

Examples

Evaluate the following integrals:

1∫∫

R(2x − 4y3)dA, where R = [−5, 4]× [0, 3]

2∫∫

R xexydA, where R = [−1, 2]× [0, 1]

3∫∫

D exy dA, where D = (x , y)|1 ≤ y ≤ 2, y ≤ x ≤ y3



Solution Q1

∫ x=4

x=−5

∫ y=3

y=0(2x − 4y3)dydx =

∫ y=4

x=−5

[2xy − y4

]y=3

y=0

dx

=

∫ x=4

x=−5

((6x − 81)− (0− 0)

)dx

=

∫ x=4

x=−5(6x − 81)dx

=[3x2 − 81x

]x=4

x=−5

= −756



Solution Q2

∫ y=1

y=0

∫ x=2

x=−1xexydxdy =

∫ x=2

x=−1

∫ y=1

y=0xexydydx

=

∫ x=2

x=−1

[exy]y=1

y=0dx

=

∫ x=2

x=−1(ex − e0)dx

=[ex − x

]x=2

x=−1

= e2 − 2− (e−1 − (−1))

= e2 − e−1 − 3



Solution Q3

∫ y=2

y=1

∫ x=y3

x=ye

xy dxdy =

∫ y=2

y=1

[ye

xy

]x=y3

x=y

dy

=

∫ y=2

y=1yey

2 − ye1dy

=

∫ y=2

y=1(yey

2 − ey)dy

=

[1

2ey

2 − e

2y2

]y=2

y=1

=1

2e4 − e

2· 4−

(1

2e − e

2

)=

1

2e4 − 2e



Generalised Fubini’s Theorem

To change the order of integration, first draw R and thenreconstruct the limit.

General rules

1 Draw the region before working out the new region

2 The outer most integral should be bounded by constants, whilethe inner integral will typically be dependent on the othervariable.



Examples

Changing order of integration [MATH2011 Q102]

Evaluate the following integrals by changing the order of integration:

1∫ 1

0

∫ 1y2 2√xex

2dx dy

2∫ 1

0

∫ 1y sin(x2)dx dy



Solution Q1

∫ 1

0

∫ 1

y2

2√xex

2dx dy =

∫ x=1

x=0

∫ y=√x

y=02√xex

2dydx

=

∫ x=1

x=02√xex

2[y ]

y=√x

y=0 dx

=

∫ 1

02xex

2dx

=[ex

2]1

0

= e − 1



Solution Q2

∫ 1

0

∫ 1

ysin(x2)dx dy =

∫ x=1

x=0

∫ y=x

y=0sin(x2)dydx

=

∫ x=1

x=0sin(x2)

[y]x

0dx

=

∫ 1

0x sin(x2)dx

=1

2sin 1


Integral Calculus Triple Integrals

Triple Integration

The intuition for triple integrals is precisely the same, except thereneeds to be 3 sets of limits, one for the x direction, one for the yand one for the z directions.



Examples

Examples

1 Find the volume of a tetrahedron bounded by xa + y

b + zc = 1 in

the first octant.

2 Find the column common to the 2 cylinders x2 + y2 ≤ a2 andy2 + z2 ≤ a2

3 Evaluate∫∫∫

S x2dxdydz where S is the region bounded by

4x2 + y2 = 4, z + x = 2, z = 0



Solution Q1 [MATH2011 Q115]

∫∫∫SdV =

∫ x=a

x=0

∫ y=b(

1− xa

)y=0

∫ z=c(

1− xa− y

b

)z=0

dz dy dx

= c

∫ x=a

x=0

∫ y=b(

1− xa

)y=0

1− x

a− y

bdy dx

= c

∫ x=a

x=0

[(1− x

a

)y − 1

2by2

]y=b(

1− xa

)y=0

dx

= c

∫ x=a

x=0b(

1− x

a

)2− b

2

(1− x

a

)2dx

=1

2bc

∫ x=a

x=0

(1− x

a

)2dx

=abc

6




The region in consideration is the circle x2 + y2 = a2 in thexy -plane. So we are measuring the integral:∫∫∫

SdV =

∫ x=a

x=−a

∫ y=√a2−x2

y=−√a2−x2

∫ z=√

a2−y2

z=−√

a2−y2

dz dy dx

=

∫ y=a

y=−a

∫ x=√

a2−y2

x=−√

a2−y2

2√a2 − y2dx dy

(changing order of integration)

=

∫ y=a

y=−a2√a2 − y2

[x]x=√

a2−y2

x=−√

a2−y2dy

=

∫ y=a

y=−a4(a2 − y2)dy

=16

3a3




∫∫∫Sx2dV =

∫ x=1

x=−1

∫ y=√

4−4x2

y=−√

4−4x2

∫ z=2−x

z=0x2dz dy dx

=

∫ x=1

x=−1

∫ y=√

4−4x2

y=−√

4−4x2

x2(2− x)dy dx

=

∫ 1

−1x2(2− x) · 2

√4− 4x2dx

= 4

∫ 1

−1x2(2− x)

√1− x2dx

= 4

∫ 1

−12x2√

1− x2dx − 4

∫ 1

−1x3√

1− x2dx

The first integral can be solved using u = sin(x), the secondintegral is just 0 because the function is odd.


Integral Calculus Change of Variables

Integration by substitution

There are 3 main types of substitutions:

1 Polar co-ordinates

2 Cylindrical co-ordinates

3 Spherical co-ordinates

Each are basic changes of variable, the more general case will becovered in time.



Jacobians

In single variable integration:∫ x=b

x=af (x)dx =

∫ u=b′

u=a′f (u)

du

dxdx

So there is an extra factor that influences the rate at which thefunction changes. Similarly, when employing any change of variable,we use the determinant of the Jacobian to adjust the infinitesimalquantities.



Jacobians for polar co-ordinates

Polar co-ordinates

In polar system, we set x = r cos θ, y = r sin θ based on thegeometry of the space.

J = det

(∂x∂r

∂x∂θ

∂y∂r

∂y∂θ

)= det

(cos θ −r sin θsin θ r cos θ

)= r



Jacobian for cylindrical co-ordinates

Cylindrical co-ordinates

The system is exactly the same, except we add a 3rd dimension:x = r cos θ, y = r sin θ, z = z

J = det

∂x∂r

∂x∂θ

∂x∂z

∂y∂r

∂y∂θ

∂y∂z

∂z∂r

∂z∂θ

∂z∂z

= det

cos θ −r sin θ 0sin θ r cos θ 0

0 0 1

= r



Jacobian for Spherical co-ordinates

Spherical co-ordinates

We let the following change of variable occur:x = ρ sinϕ cos θ, y = ρ sinϕ sin θ, z = ρ cosϕ

J = det

∂x∂ρ

∂x∂ϕ

∂x∂θ

∂y∂ρ

∂y∂ϕ

∂y∂θ

∂z∂ρ

∂z∂ϕ

∂z∂θ

= det

sinϕ cos θ sinϕ sin θ cosϕρ cosϕ cos θ ρ cosϕ sin θ −ρ sinϕ−ρ sinϕ sin θ ρ sinϕ cos θ 0

= ρ2 sinϕ



General Change of Variables

Integration by Substitution

Consider the substitutions x = g(u, v), y = h(u, v). Then thefollowing holds:∫∫

Ωf (x , y)dx dy =

∫∫Ω′

f (g(u, v), h(u, v))∣∣∣∂(x , y)

∂(u, v)

∣∣∣du dv



Examples for polar co-ordinates

Examples

1 Use polar co-ordinates to find∫∫

Ω x2y3dR whereΩ = (x , y)|x2 + y2 ≤ 1, x ≥ 0, y ≥ 0

2 Find the area of the shape defined by the inequalitiesy ≥ 0, y ≥ −x , x2 + y2 ≤ 3

√x2 + y2 − 3x



Solution Q1 [Paul’s Online Maths Notes]

∫∫RdA =

∫ θ=π2

θ=0

∫ r=1

r=0(r cos(θ))2(r sin(θ))3rdr dθ

=

∫ θ=π2

θ=0r6 cos2(θ) sin3(θ)dr dθ

=

∫ θ=π2

θ=0

1

7cos2(θ) sin2 θ sin(θ)dθ

=1

7

∫ 0

1u2(1− u2) · −du

=1

7

∫ 1

0u2 − u4du

=2

105



Solution Q2 [MATH2011 Q109/MATH2111 Q149]

∫∫RdA =

∫ θ= 3π4

θ=0

∫ r=3−3 cos θ

r=0rdr dθ

=

∫ 3π4

0

1

2(3− 3 cos θ)2dθ

= la-da-daa-da-da-de-di-da-di-day

=81π

16− 9

8(4√

2 + 1)



Examples on Cylindrical co-ordinates

Examples from homework

1 Find the volume of the solid enclosed between the spheresx2 + y2 + z2 = 4 and x2 + y2 + z2 = 4z

2 Find the volume inside the cone z + 2 =√

x2 + y2 betweenthe planes z = 0, z = 1




∫∫∫SdV =

∫ r=√

3

r=0

∫ θ=2π

θ=0

∫ z=2+√

4−r2

z=2−√

4−r2

rdz dθ dr

=

∫ r=√

3

r=0

∫ θ=2π

θ=0

∫ z=2+√

4−r2

z=2−√

4−r2

rdz dθ dr

The outer integral for the bounds on r are because we only considerhow far away from the original in the xy plane we travel, θ rangesfrom 0 to 2π because we can go around in a full circle in the validregion for r , and z ranges from the lower sphere to the upper sphere.




∫∫∫SdV =

∫ r=3

r=2

∫ θ=2π

θ=0

∫ z=r−2

z=0rdz dθ dr



Examples of Spherical co-ordinates


1 Find the volume of the region above the cone z =√x2 + y2

and inside the sphere x2 + y2 + z2 = 2az

2 Use spherical coordinates to find the volume enclosed by thesurface (

√x2 + y2 − 1)2 + z2 = 1




∫∫∫SdV =

∫ θ=2π

θ=0

∫ ϕ=π4

ϕ=0

∫ ρ=2a cosϕ

ρ=0ρ2 sin(ϕ)dρ dϕ dθ



Solution Q2 [MATH2011 Q128/MATH2111 Q156 iii)]

∫∫∫SdV =

∫ θ=2π

θ=0

∫ ϕ=π

ϕ=0

∫ ρ= 1sin(ϕ)

ρ=0ρ2 sin(ϕ)dρ dϕ dθ



Examples on change of variables


1 Let Ω be the region in the first quadrant bounded by thehyperbolas x2 − 2y2 = 1, x2 − 2y2 = 3, xy = 1, xy = 2. Letu = x2 − 2y2, v = xy . Sketch the region Ω in the x − y planeand the region Ω′ in the u, v plane that corresponds to Ω.Hence evaluate

∫∫Ω(x2 − 2y2)x2y2(2x2 + 4y2)dx dy

2 Integrate the function 1xy over the region Ω′ bounded by the 4

circlesx2 + y2 = ax , x2 + y2 = a′x , x2 + y2 = by , x2 + y2 = b′y .



Solution Q1 [MATH2011 Q137/MATH2111 Q160]

We shall consider the substitution u = x2 − 2y2, v = xy . Then theJacobian of this substitution is:

det∂(u, v)

∂(x , y)= det

[2x y−4y x

]= 2x · x − (−4y) · y = 2x2 + 4y2

∫∫Ω

(x2 − 2y2)x2y2(2x2 + 4y2)dx dy =

∫∫Ω′

uv2(2x2 + 4y2)·

1

2x2 + 4y2du dv

=

∫ u=3

u=1

∫ v=2

v=1uv2du dv



Solution Q2 [MATH2111 Q161] I

Suppose we re-arrange the given equations: x2+y2

x = a, a′ andx2+y2

y = b, b′. Then we use the subtitutions u = x2+y2

x , v = x2+y2

y .Thus, we have:

det

[1− y2

x22xy

2yx − x2

y2 + 1

]=(

1− y2

x2

)(1− x2

y2

)−(2y

x

)(2x

y

)= 2− x2

y2− y2

x2− (4)

= −x2

y2− 2− y2

x2

= −(xy

+y

x

)2

= −(x2 + y2

xy

)2



Solution Q2 [MATH2111 Q161] II

Hence we have: −(x2+y2

xy

)2= uv

xy

Thus the integral becomes:∫∫Ω′

1

xydx dy =

∫∫Ω

1

xy

xy

uvdu dv =

∫ v=b′

v=b

∫ u=a′

u=a

1

uvdu dv


Integral Calculus Centre of Mass

Centre of Mass

Consider a solid (either a plane or 3D solid) where everyinfinitesimally small part of the solid has a specific mass densitydenoted ρ(x). This means that the overall mass of the object willbe given by

∫∫Ω ρ(x)dA or

∫∫∫Ω ρ(x)dV depending on how many

dimensions the solid has.

Mass of a solid

The mass of a solid is given by the formula:

M(S) =

∫...

∫Ωρ(x)dR


Integral Calculus Centre of Mass

Centre of Mass (MATH2011 ONLY)

Centre of Mass

The i th coordinate for the centre of mass is given according to theformula:

ci =

∫∫Ω xiρ(x)

M(S)dR

Here, M(S) describes the mass of the solid, and ρ(x) describes thedensity of the solid as a function of x

To compute the centre of mass can be done using whatever rules youwant to, provided that are valid under double and triple integration.


Integral Calculus Leibniz Rule

Leibniz Rule (MATH2111 ONLY)

Commonly known as Differentiation under the integral sign, LeibnizRule helps evaluate weird looking integrals based on relatedfunctions.

Leibniz Rule

d

dx

∫ t=b

t=af (x , t)dt =

∫ t=b

t=a

∂

∂xf (x , t)dt

For a more general integral, we have:

d

dx

∫ b(x)

a(x)f (x , t)dt =

∫ b(x)

a(x)

∂

∂xf (x , t)dt

+ f (x , b(x)) · b′(x)− f (x , a(x)) · a′(x)



Examples

2018 Q1iii)

You are given that: ∫ ∞−∞

cos(3x)

x2 + a2dx =

π

ae−3a

Evaluate the following integral:∫ ∞−∞

cos(3x)

(x2 + 4)2dx



Solution 2018 Q1iii)

Beginning with the given integral, we differentiate both sides withrespect to a

d

da

∫ ∞−∞

cos(3x)

x2 + a2dx =

d

da

π

ae−3a

∫ ∞−∞

∂

∂a

cos(3x)

x2 + a2dx = −−πe

−3x(3x + 1)

x2∫ ∞−∞−2a cos(3x)

(x2 + a2)2dx = −πe

−3a(3a + 1)

a2

Substituting a = 2∫ ∞−∞−4

cos(3x)

(x2 + 4)2dx = −7πe−6

4∫ ∞−∞

cos(3x)

(x2 + 4)2dx =

7πe−6

16



Examples

2014 Q2i)

Suppose that f : [0, 1]× [0, 1]→ R has continuous partialderivatives. Define F : [0, 1]× [0, 1]→ R by:

F (u, v) =

∫ u

0f (v , y)dy

1 If u = u(x), v = v(x), find an expression for dFdx

2 Hence, or otherwise, compute:

d

dx

∫ x

0

sin(xy)

ydy



Solution 2014 Q2i) a)

By Leibniz Rule:

dF

dx=

d

dx

∫ u(x)

0f (v(x), y)dy

=

∫ u(x)

0

∂

∂xf (v(x), y)dy + f (v(x), u(x))

d

dxu(x)− f (v(x), 0)

d

dx(0)

=

∫ u(x)

0v ′(x)

∂f (v(x), y)

∂xdy + f (v(x), u(x))u′(x)



Solution 2014 Q2i) b)

Let u(x) = x , v(x) = x . Then upon substitution into the expressionfrom a), we have:

d

dx

∫ x

0

sin(xy)

ydy =

∫ x

01 · y · cos(xy)

y+

sin(x2)

x· 1

=

∫ x

0cos(xy)dy +

sin(x2)

x

=[sin(xy)

x

]x0

+sin(x2)

x

= 2sin(x2)

x


Second Year Mathematics Revision€¦ · Integral Calculus Change of Variables Jacobians In single...

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