15.8: Substitutions in Multiple Integrals
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This section discusses the translation of a graph from the xy Cartesian plane to the uv Cartesian plane and defines the Jacobian.
Introduction
As observed in other sections regarding polar coordinates, some integration of functions on the xyz-space are more easily integrated by translating them to different coordinate systems. These substitutions can make the integrand and/or the limits of integration easier to work with, as "U" Substitution did for single integrals. In this section, we will translate functions from the x-y-z Cartesian coordinate plane to the u-v-w Cartesian coordinate plane to make some integrations easier to solve.
One key component of this translation is called the Jacobian determinant, or simply the Jacobian, which measures how much the volume at a certain point changes when being transformed from one coordinate system to another.
It is important to note that although we are changing the coordinate system on which we graph our function, the theory behind multiple integrals has not changed. The limits of integration still create the domain under the curve, and the integration will help us find the volume of the figure created by the original function and the domain.
Theoretical discussion with Descriptive Elaboration
For any given function f(x,y), we can define x and y as a function of other variables g(u,v). To do this, we first find u and v as a function of x and y that will allow for an easier integrand. Then solve for x and y in order to translate the function so that x=g(u,v) and y=h(u,v). This translates the are region from R in the x-y plane to D in the u-v plane.
Remember:
I=∬Rf(x,y) dA
So we must find dA:
dA changes from dxdy to |J(u,v)| dudv. Each change in u (Δu) and change in v (Δv) create parallelograms that are small areas ΔA or dA . We can find the area of each of these parallelograms (P) by taking the cross product of the two vectors that create it (Δu and Δv).
Area of P=|→V1×→V2|=J(u,v)
J(u,v)=|∂x∂u∂x∂v∂y∂u∂y∂v|=∂x∂u∂y∂v−∂y∂u∂x∂v=∂(x,y)∂(u,v)
|J(u,v)| represents the area of the parallelogram, and it is the determinant of the Jacobian matrix, shown above. The Jacobian measures how much the transformation is changing from the region R to the region G. Therefore, the translation of the integration of f(x,y) is represented by
∬Rf(x,y) dx dy=∬Gf(g(u,v), h(u,v))|J(u,v)|du dv.
The same can be applied for triple integrals, where the translations are represented by
x=g(u,v,w),y=h(u,v,w),z=k(u,v,w)
J(u,v,w)=∂(x,y,z)∂(u,v,w)=|∂x∂u∂x∂v∂x∂w∂y∂u∂y∂v∂y∂w∂z∂u∂z∂v∂z∂w|=∂x∂u|∂y∂v∂y∂w∂z∂v∂z∂w|−∂x∂v|∂y∂u∂y∂w∂z∂u∂z∂w|+∂x∂w|∂y∂u∂y∂v∂z∂u∂z∂v|=∂x∂u(∂y∂v⋅∂z∂w−∂z∂v⋅∂y∂w)−∂x∂v(∂y∂u⋅∂z∂w−∂z∂u⋅∂y∂w)+∂x∂w(∂y∂u⋅∂z∂v−∂z∂u⋅∂y∂v)
This method for getting the Jacobian is called cofactor expansion.
Although the introduction focused primarily on translating a Cartesian function to a different Cartesian coordinate system, the concept of the Jacobian can also be used to explain how translations into polar coordinates work as well.
For cylindrical coordinates
x=r cos θ, y=r sin θ, z=z
Therefore:
J(u,v,w)=|∂x∂u∂x∂v∂x∂w∂y∂u∂y∂v∂y∂w∂z∂u∂z∂v∂z∂w|=|cos θ−r sin θ0sinr cos θ0001|
∭DF(x,y,z) dx dy dz=∭GH(r,θ,z)|r| dr dθ dz
For spherical coordinates
J(ρ,ϕ,θ)=|∂x∂ρ∂x∂ϕ∂x∂θ∂y∂ρ∂y∂ϕ∂y∂θ∂z∂ρ∂z∂ϕ∂z∂θ|=ρ2 sinϕ
∭DF(x,y,z) dx dy dz=∭GH(ρ,ϕ,θ)|ρ2sin ϕ | dρ dϕ dθ
Hence, dxdydz becomes rdrdθ in cylindrical coordinates and ρ2sin ϕ dρ dϕ dθ in spherical coordinates.
Use the following transformation to evaluate the integral.
u=yx and v=xy
∬RyxdA
Where R is bounded by: 1≤u≤2 and 1≤v≤2
Solution
First find x and y as functions of u and v:
u=yx v=xy
y=xu → x=vy
x=vxu
x2=vu
x=√vu
y=xu
y=(√vu)u
y=√vu
x=g(u,v)=√vu and y=h(u,v)=√vu
Using x=g(u,v) and y=h(u,v), find the integrand in terms of u and v:
yx=√vu√vu=u
And dA changes from dxdy to |J(u,v)| du dv. The Jacobian equals:
J(u,v)=|∂x∂u∂x∂v∂y∂u∂y∂v|=∂x∂u∂y∂v−∂y∂u∂x∂v
∂x∂u=12u−32v12 ∂x∂v=12u−12v−12 ∂y∂u=12u−12v12 ∂y∂v=12u12v−12
J(u,v) =|12u−32v1212u−12v−1212u−12v1212u12v−12|=(−14u−1−14u−1)=12u
Therefore, evaluate:
∫21∫21u (12u) du dv=∫21∫2112 du dv=∫2112u|21 dv=∫211−12 dv=12v|21=1−12=12
Use the following transformation to evaluate the integral.
u=x−12y and v=y
∫120∫y+42y2y3(2x−y)e2x−y2 dx dy
Solution
First solve for x and y:
u=x−12y v=y
u=x−12v y=v
x=u+12v.
Then substitute x and y for g(u,v) and h(u,v) :
The integrand:
y3(2x−y)e2x−y2 → v3[2(u+12v)−v]e[2(u+12v)−v]2
=v3(2u)e2u2
=(2uv3)e4u2
The transformation also changes the bounds of integration
\(x=y+42 → u+12v=v+42=y4 → u+v2=v2 \]
u=42
u=0
u=2
And dxdy changes to |J(u,v)| du dv. The Jacobian equals:
J(u,v)=|∂x∂u∂x∂v∂y∂u∂y∂u|=∂x∂u∂y∂v−∂y∂u∂x∂v
=|1501|=1
Thus,
∬Rf(x,y) dx dy=∬Gf(g(u,v), h(u,v))|J(u,v)|du dv=∫120∫202uv3e4u2 (1) dv du=∫120ue4u2v42|20 du=∫1208ue4u2 du=e4u2|120=e−1
Find the mass of an object bounded by
1≤x≤2, 0≤xy≤1, 0≤z≤2
with a density that can be described by the formula x2y+2xyz by using the transformation u=x, v=xy, w=3z.
Solution
Set up the integral in cartesian coordinates:
∫21∫1x0∫20x2y+2xyz dzdydx.
To apply the substitution, first solve for x and y using the given transformations:
u=xv=xyw=3z
x=uy=vxz=w3
y=vu.
Then make the appropriate substitutions within the integrand:
x2y+2xyz→u2(vu)+2u(vu)(w3) → uv+2vw3.
Next, find the new boundaries to the region we want to integrate:
1≤x≤2 → 1≤u≤2
0≤xy≤1 → 0≤v≤1
0≤z≤2 → 0≤w3≤2 → 0≤w≤6.
To complete the transformation, find the Jacobian:
J(u,v,w)=∂(x,y,z)∂(u,v,w)=|∂x∂u∂x∂v∂x∂w∂y∂u∂y∂v∂y∂w∂z∂u∂z∂v∂z∂w|=|100−vu21u00013|=13u.
Notice the Jacobian of a lower triangular matrix (the values above the diagonal are all zero) is the multiplication of the diagonal entries. You can confirm this with cofactor expansion.
Using all of our calculated transformations, we can compute the new integral:
Mass =∫60∫10∫21(uv+2vw3)13u dudvdw=13∫60∫10∫21v+2vw3u dudvdw=12∫60∫10vu+2vw3ln|u||21 dvdw=13∫60∫10v+2vw3ln2 dvdw=13∫60v22+2wln23(v22)|10 dw=13∫6012+wln23 dw=13[12w+w2ln26]60=13[3+6ln2]=1+2ln2.
Contributors and Attributions
- Sydney Wong (UCD), Alagu Chidambaram (UCD)
Integrated by Justin Marshall.