# Calculus

Triple Integrals# Change of Variables in Triple Integrals

Problem 1

Problem 2

As with double integrals, triple integrals can often be easier to evaluate by making the change of variables. This allows to simplify the region of integration or the integrand.

Let a triple integral be given in the Cartesian coordinates \(x, y, z\) in the region \(U:\)

\[\iiint\limits_U {f\left( {x,y,z} \right)dxdydz} .\]
We need to calculate this integral in the new coordinates \(u, v, w.\) The relationship between the old and new coordinates is given by

\[

{x = \varphi \left( {u,v,w} \right),\;\;}\kern-0.3pt

{y = \psi \left( {u,v,w} \right),\;\;}\kern-0.3pt

{z = \chi \left( {u,v,w} \right).}

\]
It is supposed here that the following conditions are satisfied:

- The functions \(\varphi, \psi, \chi\) are continuous together with their partial derivatives;
- There’s a single valued relation between points of the region of integration \(U\) in the \(xyz\)-space and points of the region \(U’\) in the \(uvw\)-space;
- The Jacobian of transformation \(I\left( {u,v,w} \right)\) equal to

\[

{I\left( {u,v,w} \right) }={ \frac{{\partial \left( {x,y,z} \right)}}{{\partial \left( {u,v,w} \right)}} }

= {\left| {\begin{array}{*{20}{c}}

{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}&{\frac{{\partial x}}{{\partial w}}}\\

{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}&{\frac{{\partial y}}{{\partial w}}}\\

{\frac{{\partial z}}{{\partial u}}}&{\frac{{\partial z}}{{\partial v}}}&{\frac{{\partial z}}{{\partial w}}}

\end{array}} \right|,}

\] is non-zero and keeps a constant sign everywhere in the region of integration \(U.\)

Then the formula for change of variables in triple integrals is written as

\[

{\iiint\limits_U {f\left( {x,y,z} \right)dxdydz} \text{ = }}\kern-0.3pt

{\iiint\limits_{U’} {f\left( {\varphi,\psi,\chi} \right) \cdot}\kern0pt{\left|{I\left( {u,v,w} \right)} \right|dudvdw}.}

\]
Here \(\varphi,\psi,\chi\) are functions of the variables \({u,v,w}\) and \(\left| {I\left( {u,v,w} \right)} \right|\) means the absolute value of the Jacobian.

Triple integrals are often easier to evaluate in the cylindrical or spherical coordinates. The corresponding examples are considered on the pages

Some examples of using other transformations of coordinates are given below.

## Solved Problems

Click on problem description to see solution.

### ✓ Example 1

Find the volume of the region \(U\) defined by the inequalities

\[

{0 \le z \le 2,\;\;\;}\kern0pt

{0 \le y + z \le 5,\;\;\;}\kern0pt

{0 \le x + y + z \le 10.}

\]

### ✓ Example 2

Find the volume of the parallelepiped defined by the inequalities

\[{0 \le 2x – 3y + z \le 5,\;\;\;}\kern0pt

{1 \le x + 2y \le 4,\;\;\;}\kern0pt

{ – 3 \le x – z \le 6.}\]

### Example 1.

Find the volume of the region \(U\) defined by the inequalities

\[

{0 \le z \le 2,\;\;\;}\kern0pt

{0 \le y + z \le 5,\;\;\;}\kern0pt

{0 \le x + y + z \le 10.}

\]

*Solution.*

Obviously, this region is a parallelepiped. It’s convinient to change variables in such a way to transform the parallelepiped into a rectangular box. In this case the triple integral becomes the product of three integrals of one variable.

Make the following replacement:

\[

{u = x + y + z,\;\;\;}\kern-0.3pt

{v = y + z,\;\;\;}\kern-0.3pt

{w = z.}

\]
The region of integration \(U’\) in the new variables \(u, v, w\) is defined by the inequalities

\[

{0 \le u \le 10,\;\;\;}\kern0pt

{0 \le v \le 5,\;\;\;}\kern0pt

{0 \le w \le 2.}

\]
The volume of the solid is

\[

{V = \iiint\limits_U {dxdydz} }

= {\iiint\limits_{U’} {\left| {I\left( {u,v,w} \right)} \right|dudvdw} .}

\]
Calculate the Jacobian of this transformation. In order to not express the old variables \(x, y, z\) through the new ones \(u, v, w,\) we find first the Jacobian of the inverse transformation:

\[

{\frac{{\partial \left( {u,v,w} \right)}}{{\partial \left( {x,y,z} \right)}} }

= {\left| {\begin{array}{*{20}{c}}

{\frac{{\partial u}}{{\partial x}}}&{\frac{{\partial u}}{{\partial y}}}&{\frac{{\partial u}}{{\partial z}}}\\

{\frac{{\partial v}}{{\partial x}}}&{\frac{{\partial v}}{{\partial y}}}&{\frac{{\partial v}}{{\partial z}}}\\

{\frac{{\partial w}}{{\partial x}}}&{\frac{{\partial w}}{{\partial y}}}&{\frac{{\partial w}}{{\partial z}}}

\end{array}} \right| }

= {\left| {\begin{array}{*{20}{c}}

1&1&1\\

0&1&1\\

0&0&1

\end{array}} \right| }

= {1 \cdot \left| {\begin{array}{*{20}{c}}

1&1\\

0&1

\end{array}} \right| }

={ 1 – 0 = 1.}

\]
Then

\[

{\left| {I\left( {u,v,w} \right)} \right| }

= {\left| {\frac{{\partial \left( {x,y,z} \right)}}{{\partial \left( {u,v,w} \right)}}} \right| }

= {\left| {{{\left( {\frac{{\partial \left( {u,v,w} \right)}}{{\partial \left( {x,y,z} \right)}}} \right)}^{ – 1}}} \right| }={ 1.}

\]
Hence, the volume of the solid is

\[

{V }={ \iiint\limits_{U’} {\left| {I\left( {u,v,w} \right)} \right|dudvdw} }

= {\iiint\limits_{U’} {dudvdw} }

= {\int\limits_0^{10} {du} \int\limits_0^5 {dv} \int\limits_0^2 {dw} }

= {10 \cdot 5 \cdot 2 }={ 100.}

\]

Problem 1

Problem 2