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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)=|xuxvyuyv|=xuyvyuxv=(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)=|xuxvxwyuyvywzuzvzw|=xu|yvywzvzw|xv|yuywzuzw|+xw|yuyvzuzv|=xu(yvzwzvyw)xv(yuzwzuyw)+xw(yuzvzuyv)

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)=|xuxvxwyuyvywzuzvzw|=|cos θr 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)=|xρxϕxθyρyϕyθzρzϕzθ|=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.

Example 15.8.1

Use the following transformation to evaluate the integral.

u=yx  and  v=xy

RyxdA

Where R is bounded by:  1u2  and  1v2

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=vuvu=u

And dA changes from dxdy to |J(u,v)| du dv. The Jacobian equals:

J(u,v)=|xuxvyuyv|=xuyvyuxv

xu=12u32v12   xv=12u12v12   yu=12u12v12   yv=12u12v12

J(u,v) =|12u32v1212u12v1212u12v1212u12v12|=(14u114u1)=12u

Therefore, evaluate:

2121u (12u) du dv=212112 du dv=2112u|21 dv=21112 dv=12v|21=112=12

Example 15.8.2

Use the following transformation to evaluate the integral.

u=x12y and v=y

120y+42y2y3(2xy)e2xy2 dx dy

Solution

First solve for x and y:

u=x12y v=y

u=x12v y=v

x=u+12v.

Then substitute x and y for g(u,v) and h(u,v) :

The integrand:

y3(2xy)e2xy2  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)=|xuxvyuyu|=xuyvyuxv

=|1501|=1

Thus,

Rf(x,y) dx dy=Gf(g(u,v), h(u,v))|J(u,v)|du dv=120202uv3e4u2 (1) dv du=120ue4u2v42|20 du=1208ue4u2 du=e4u2|120=e1

Example 15.8.3

Find the mass of an object bounded by

1x2,  0xy1,  0z2

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:

211x020x2y+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+2xyzu2(vu)+2u(vu)(w3)  uv+2vw3.

Next, find the new boundaries to the region we want to integrate:

1x2  1u2

0xy1  0v1

0z2  0w32  0w6.

To complete the transformation, find the Jacobian:

J(u,v,w)=(x,y,z)(u,v,w)=|xuxvxwyuyvywzuzvzw|=|100vu21u00013|=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 =601021(uv+2vw3)13u dudvdw=13601021v+2vw3u dudvdw=126010vu+2vw3ln|u||21 dvdw=136010v+2vw3ln2 dvdw=1360v22+2wln23(v22)|10 dw=136012+wln23 dw=13[12w+w2ln26]60=13[3+6ln2]=1+2ln2.

Contributors and Attributions

  • Sydney Wong (UCD), Alagu Chidambaram (UCD)
  • Integrated by Justin Marshall.


15.8: Substitutions in Multiple Integrals is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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