Download Mathematics II: Multiple, Line, and Surface Integrals and more Cheat Sheet Bioengineering in PDF only on Docsity!
Important! This is an abridged lecture notes designed to introduce the key concepts in a more linear way. This does not substitute the contents covered in the lecture slides and the hand-written notes.
1 Multiple integrals & Theorems
2 Line integrals
2.1 Arc length
Given a curve defined in 3D, an interesting question to ask is what is the length of the curve? To calculate this quantity, we can partition the curve into infinitesimal straight segment and then sum them over. This is the strategy we will use here. But before we do so, we need a convenient way to define a curve, which can be done by parametrising the curve with a parameter, t, say. For instance, the unit circle on the xy-plane can be parametrised as follows:
x(t) = cos t , y(t) = sin t (1)
with t goes from 0 to 2π. Note that in this case, the curve starts at (1, 0), runs in a counter-clockwise manner and ends again at (1, 0). Generically, all curves can be parametrised by an auxiliary variable t. Now, let a generic curve L be defined as ~l(t) = (x(t), y(t), z(t)) with t ∈ [a, b]. Partitioning the curve into small increments (d~l(t) = (dx(t), dy(t), dz(t)), the overall length then becomes:
Arc length =
∫ (^) b
t=a
|d~l(t)| (2)
∫ (^) b
t=a
√ dx(t)^2 + dy(t)^2 + dz(t)^2 (3)
∫ (^) b
t=a
√ x ˙^2 + ˙y^2 + ˙z^2 dt (4)
where ˙x ≡ dx/dt, etc.
2.1.1 Other coordinates
For a curve defined using the Cartesian coordinates, the arc length can be calculated by Eq. (4). In other coordinate systems, one can express x, y, z above in the desired coordinates. For instance, in the cylindrical coordinates, x = r cos θ and y = r sin θ. Hence,
x˙ = ˙r cos θ − r θ˙ sin θ , y˙ = ˙r sin θ + r θ˙ cos θ. (5)
Substituting these back into Eq. (4), we have
Arc length =
∫ (^) b
t=a
√ r ˙^2 + r^2 θ˙^2 + ˙z^2 dt. (6)
Note the extra factor r^2 in front of θ˙^2 in the above formula. It comes from the fact unless r is specified dθ itself does not provide any information on the amount of displacement.
2.1.2 Example
Let us calculate the arc length of the curve L defined by x(t) = 2 cos t, y(t) = 2 sin t, t ∈ [0, 2 π] (i.e., a circle of radius 2). Using Eq. (4), we find
Arc length =
∫ (^2) π
t=
√ 4 cos^2 t + 4 sin^2 tdt = 4π. (7)
We can also re-parametrise the same curve as r(t) = 2, θ = t, t ∈ [0, 2 π]. Using now Eq. (6), we have
Arc length =
∫ (^2) π
t=
4dt = 4π , (8)
which of course gives the same result.
2.2 Line integrals of vector fields
Beside calculating the length of a curve, a line integral can also enable us to calculate the work done on a particle travelling along the curve under the influence of a force field f~. As we know from mechanics, the work done is given by f~ · d~l where ~l is the displacement vector. As usual, to calculate the total work done, we sum over these infinitesimal displacement vectors to arrive at
Work done =
∫ (^) b
t=a
f^ ~ · d~l(t) (9)
where we have again assumed that the curve is parametrised by t.
2.2.1 Example
Let us perform the line integral in Eq. (9) for f~ = ˆx + ˆy, and for the curve defined by x(t) = t^2 , y(t) = t (i.e., y =
x), for t ∈ [0, 1]. Eq. (9) then gives ∫ (^1)
0
(ˆx + ˆy) · (2tˆx + ˆy)dt =
∫ (^1)
0
(2t + 1)dt = 2. (10)
3 Surface integrals
3.1 Tangent planes and normals to a surface
There are many ways to define a surface in 3D, the most intuitive way is to let the coordinate z be a function of x and y. In other words, we can view z = f (x, y) as the height at the point (x, y). Given this definition, how do we obtain the normal vector to the surface at point (x 0 , y 0 )? To do so, we first find the tangent plane at (x 0 , y 0 ), which can be obtained using the Taylor’s theorem:
z = f (x 0 , y 0 ) + ∂xf (x 0 , y 0 )(x − x 0 ) + ∂yf (x 0 , y 0 )(y − y 0 ) + h.o.t. (11)
where ∂xf = ∂f /∂x, etc. If we ignore the higher order terms (denoted by h.o.t.) above, the above equation defines a plane, and as we know, the normal to this plane is proportional to
ˆn(x 0 , y 0 ) ∝ ∂xf (x 0 , y 0 )ˆx + ∂yf (x 0 , y 0 )ˆy − z .ˆ (12)
We can then normalised the above normal factor to obtain
nˆ(x 0 , y 0 ) = ± √∂xf^ (x^0 , y^0 )ˆx^ +^ ∂yf^ (x^0 , y^0 )ˆy^ −^ ˆz 1 + (∂xf (x 0 , y 0 ))^2 + (∂yf (x 0 , y 0 ))^2
Note the plus of minus sign above since the normal factor can be pointing in either direction away from the surface.
3.1.1 Example
Given a surface z = f (x, y) = x^3 + y^2 , the normal factor at (x, y) is given by
nˆ = ±
3 x^2 ˆx + 2y yˆ − zˆ √ 1 + 9x^4 + 4y^2
The surface integral is then given by
∫ (^1)
0
∫ (^1)
0
(ˆx + ˆy + ˆz) · (2ˆx + 3ˆy − zˆ)
1 + 4 + 9dxdy = ± 4 , (21)
where dS =
√ 1 + (∂xf )^2 + (∂yf )^2 dxdy =
1 + 4 + 9dxdy where f (x, y) = 2x + 3y.
3.4 Volume integrals
A volume integral involves three integration variables and its most important use is to obtain the vol- ume of an arbitrarily defined object. As an example, let us calculate the volume of a unit hemisphere. In the Cartesian coordinates, the volume is
V =
∫ (^1)
0
∫ √ 1 −x 2
− √ 1 −x^2
∫ √ 1 −x (^2) −y 2
0
dzdydx. (22)
Note that the order of integration is important. For this problem, it is obvious that the Cartesian coordinates are not most suitable, so we will switch to the spherical coordinate system instead. As a result,
V =
∫ (^2) π
0
∫ (^) π/ 2
0
∫ (^1)
0
r^2 sin φdrdφdθ = 2 π 3
4 Theorems
We have seen the motivations for doing multiple integrations, we will now see that there are theorems to help us to reduce the number of integrals to be performed.
4.1 Gradient Theorem
The first theorem enables us to reduce some line integrals to no integrals at all. Specifically, if a vector field f~ is given by the gradient of a scalar, i.e., f~ = ∇~c for some scalar field c. Then, ∫
L
f^ ~ · d~l = c(end point) − c(start point) (24)
where the start point and the end point refer to the start location and end location of the curve L. To use this theorem, one would need to check that a given vector field can be expressed as the gradient of some scalar field, if it is possible, then the vector field is called conservative.
4.1.1 Example
Let f~ = yˆx + xˆy. To see whether f~ is conservative, we try to find a scalar c such that ∇~c = f~. If such a c exists, then
c =
∫ fxdx =
∫ ydx = xy + h(y) (25)
and simultaneously,
c =
∫ fydy =
∫ xdy = xy + g(x) (26)
where h(y) and g(x) are arbitrary functions of y and x, respectively. Comparing the two equations above, we conclude that c can be given by xy + K where K is a constant. Therefore, f~ is conservative.
4.2 Stokes’ Theorem
The second theorem, the Stokes’ theorem, enables us to reduce a double integral to a single integral. The theorem says (^) ∫ ∫
D
(∇ × ~u · nˆ)dS =
∮
L
~u · d~l. (27)
In the above equation, D denotes an arbitrarily defined surface with the boundary given by the curve L. Since L is bounding a surface, it is necessarily a loop, hence the circle in the integral sign on the R.H.S. At first sight, there are two ambiguities in the above equation: first the normal vector field ˆn can be of plus or minus sign since it depends on how we define what is inside and what is outside for a given surface; second, the line integral depends on the direction of integration along the loop. For instance, for a circle on the xy-plane, the integral can be done in clockwise or counter-clockwise manner. These two ambiguities are in fact related as we will use the right hand rule to relate the line integral direction and the direction of the normal vector field. Specifically, we use the index finger to point along the direction of the curve, the middle finger to point towards the interior of the domain of the surface integral, the thumb will then indicate the direction of the normal vector field. As an example, consider the surface being the hemisphere defined for z > 0 centred at the original. The curve is then a unit circle on the xy-plane. If we choose to integrate the curve in a counter clockwise manner, the normal vector field will then be pointing up, i.e., ˆn = ˆr in the spherical coordinate system.
4.3 Divergence Theorem
The last theorem, the Divergence theorem, enables us to reduce a triple integral to a double integral. The theorem says (^) ∫ ∫ ∫
DV
(∇ · ~u)dV =
∫ ∫
DS
~u · nˆdS. (28)
In the above equation, DV denotes an arbitrarily defined volume with the boundary given by the surface DS. Here, the normal vector fields points away from the enclosed volume.
4.3.1 Archimedes’ principle
Let us now use the divergence theorem to prove the Archimedes’ principle, which says that the upward buoyant force that is exerted on a body immersed in a fluid is equal to the weight of the fluid displaced by the body. Given an arbitrarily shaped object bounded by the surface DS , the total force exerted by the fluid pressure outside is
F^ ~tot = −
∫ ∫
DS
p(z)ˆndS (29)
where p(z) = −ρgz + p 0 with ρ being the fluid density, g the gravitational acceleration and p 0 the atmospheric pressure (the pressure at the surface of the water. Focusing now on the z-component, we get
Fz = −
∫ ∫
DS
pnz dS (30)
=
∫ ∫
DS
(−pˆz) · nˆdS. (31)
Using the divergence theorem, we get
Fz =
∫ ∫ ∫
DV
∇ · (−pzˆ)dV (32)
=
∫ ∫ ∫
DS
(−∂z p)dV (33)
=
∫ ∫ ∫
DS
(ρg)dV , (34)
which is exactly the weight of the fluid displaced by the body.
