In this part we’ll cover how we change our variables into others, in the most general case.
General case of variable change
∬Df(x,y)dA
We want to rewrite this in terms of u and v. Let’s say that:
x=g(u,v)y=h(u,v)
We can say that we have a transformation, T, from the uv-plane to the xy-plane, given by our functions g and h.
We need to make some assumptions of T to make our lives easier.
T is a C1-transformation, meaning that g and h have continuous partial derivatives.
T is an injective transformation (meaning it is 1-to-1). This means that we can express u and v in terms of x and y.
With this, we can define the Jacobian matrix.
Definition
The jacobian matrix of a transformation, T, is:
∂(u,v)∂(x,y)=det∂u∂x∂u∂y∂v∂x∂v∂y
Suppose T is a transformation, from the uv-plane to the xy-plane. Assuming that the jacobian for T, is non-zero, then:
∬Df(x,y)dA=∬Sf(x(u,v),y(u,v))∣∂(u,v)∂(x,y)∣dA
Example
Find
∬Dex−yx+ydA
Where D is the trapezoidal region, with vertices, (1,0),(2,0),(0,−2) and (0,−1).
Let u=x+y and v=x−y.
This means that:
x=2u+vy=2u−v
Let’s compute the jacobian:
∂(u,v)∂(x,y)=det∂u∂x∂u∂y∂v∂x∂v∂y
∂(u,v)∂(x,y)=det[212121−21]
∂(u,v)∂(x,y)=−21
Since our transformation is linear, we only need to recompute our vertices. Using our formulas we find that
(1,0)→(1,1)(2,0)→(2,2)(0,−2)→(−2,2)(0,−1)→(−1,1)
If we plot our trapezodial, we find that it is a type II region. The left curve being u=−v and the right curve u=v.
Which means our integral becomes:
21∫12∫−vvevududv
21∫12vevuu=−vu=vdv
21e−e−1∫12vdv
21e−e−12v2v=1v=2
43(e−e−1)
In three variables
It is the same approach. Let x=x(u,v,w),y=y(u,v,w),z=z(u,v,w)
The jacobian is:
∂(u,v,w)∂(x,y,z)=det∂u∂x∂u∂y∂u∂z∂v∂x∂v∂y∂v∂z∂w∂x∂w∂y∂w∂z
Just as we did before:
∭Ef(x,y,z)dV=∭Sf(x(u,v,w),y(u,v,w),z(u,v,w))∣∂(u,v,w)∂(x,y,z)∣dV
Let’s now cover some special cases that we’ll often encounter.
Cylindrical coordinates
Is just like polar coordinates but for three variables:
(x,y,z)→(r,θ,z)
If we have symmetry around the z-axis, it makes sense to use these variables.
We know that:
x=rcos(θ)y=rsin(θ)z=z
Therefore, the jacobian for this becomes:
∂(u,v,w)∂(x,y,z)=detcos(θ)sin(θ)0−rsin(θ)rcos(θ)0001
It makes sense to change to spherical coordinates when our solid is bounded by spheres and/or cones.
Example
Find
∬Be(x2+y2+z2)23dV
Where B is the unit ball.
B={(x,y,z)∣x2+y2+z2≤1}
In spherical coordinates:
S={(ρ,φ,θ)∣0≤ρ≤1,0≤φ≤π,0≤θ≤2π}
Which is easy to integrate over.
∬Be(x2+y2+z2)23dV=∭Se(ρ2)23⋅ρ2sin(φ)dV
∭Seρ3⋅ρ2sin(φ)dV
∫02π∫01∫0πeρ3⋅ρ2sin(φ)dφdρdθ
∫02π∫01eρ3⋅ρ2[−cos(φ)φ=0φ=π]dρdθ
∫02π∫012eρ3⋅ρ2dρdθ
Let
u=ρ3
du=3ρ2dρ
3du=ρ2dρ
∫02π∫0132eududθ
∫02π32euu=0u=1dθ
∫02π32(e−1)dθ
∫02π32(e−1)dθ
32(e−1)θθ=0θ=2π
34π(e−1)
Some applications of multiple integrals
Let’s see what we can achieve with multiple integrals.
Computing mass of solid
Let E be an arbitrary solid, with a density function, ρ(x,y,z)
We say that the mass of the solid is:
∭Eρ(x,y,z)dV
We perform the same proof as for integrals for solids, but with the density function instead. Taking the sum and limit of small points and their densities.
Center of Mass
One very important application is finding the center of mass of a solid.
Given a physical system with mi points, the center of mass has the coordinates, (x0,y0,z0).
Therefore:
x0=∑mi∑mixi
y0=∑mi∑miyi
z0=∑mi∑mizi
For an arbitrary solid, it’s almost sum same:
x0=∭Eρ(x,y,z)dV∭Eρ(x,y,z)⋅xdV
y0=∭Eρ(x,y,z)dV∭Eρ(x,y,z)⋅ydV
z0=∭Eρ(x,y,z)dV∭Eρ(x,y,z)⋅zdV
Again, the proof is the exact same as the previous one.
Example
A rectangular solid of width 2, length 10 and height 2 has the density function, ρ(x,y,z)=y3, where x,y,z are counted from the lower left farthest corner.
Now the solid is place on a table, such that 2/3 of its length, counting from the left, lies on the table.
Does the rectangular solid fall of the table?
So, we need to compute the center of mass of the solid. If we think logically about this, the only coordinate that matters is the y coordinate.
If y0≤32⋅10 it stays, otherwise it falls from the table. So let’s compute y0.
y0=m∭Eρ(x,y,z)⋅ydV
y0=m∭Ey3⋅ydV
y0=m∭Ey4dV
Since we have a rectangular box:
y0=m∫02∫02∫010y4dydxdz