Change of Variables in Triple Integrals
Solved Problems
Example 1.
Find the volume of the region \(U\) defined by the inequalities
\[0 \le z \le 2,\ 0 \le y + z \le 5, 0 \le x + y + z \le 10.\]
Solution.
Obviously, this region is a parallelepiped. It's convenient 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:
\[u = x + y + z, v = y + z, w = z.\]
The region of integration \(U'\) in the new variables \(u, v, w\) is defined by the inequalities
\[0 \le u \le 10, 0 \le v \le 5, 0 \le w \le 2.\]
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
\[V = \iiint\limits_{U'} {\left| {I\left( {u,v,w} \right)} \right|dudvdw} = \iiint\limits_{U'} {dudvdw} = \int\limits_0^{10} {du} \int\limits_0^5 {dv} \int\limits_0^2 {dw} = 10 \cdot 5 \cdot 2 = 100.\]
Example 2.
Find the volume of the parallelepiped defined by the inequalities
\[0 \le 2x - 3y + z \le 5, 1 \le x + 2y \le 4, - 3 \le x - z \le 6.\]
Solution.
We introduce the new variables
\[u = 2x - 3y + z, v = x + 2y, w = x - z.\]
Calculate the Jacobian of the inverse transformation:
\[
\frac{{\partial \left( {u,v,w} \right)}}{{\partial \left( {x,y,z} \right)}}
= \left| {\begin{array}{*{20}{c}}
{\frac{{\partial u}}{{\partial x}}}&{\frac{{\partial u}}{{\partial y}}}&{\frac{{\partial u}}{{\partial z}}}\\
{\frac{{\partial v}}{{\partial x}}}&{\frac{{\partial v}}{{\partial y}}}&{\frac{{\partial v}}{{\partial z}}}\\
{\frac{{\partial w}}{{\partial x}}}&{\frac{{\partial w}}{{\partial y}}}&{\frac{{\partial w}}{{\partial z}}}
\end{array}} \right|
= \left| {\begin{array}{*{20}{c}}
2&{ - 3}&1\\
1&2&0\\
1&0&{ - 1}
\end{array}} \right|.\]
By expanding the determinant along the third row, we find its value to be
\[
\frac{{\partial \left( {u,v,w} \right)}}{{\partial \left( {x,y,z} \right)}}
= \left| {\begin{array}{*{20}{c}}
2&{ - 3}&1\\
1&2&0\\
1&0&{ - 1}
\end{array}} \right|
= 1 \cdot \left| {\begin{array}{*{20}{c}}
{ - 3}&1\\
2&0
\end{array}} \right| - 1 \cdot \left| {\begin{array}{*{20}{c}}
2&{ - 3}\\
1&2
\end{array}} \right|
= - 2 - 7 = - 9.\]
Then the absolute value of the Jacobian of the direct transformation is
\[\left| {I\left( {u,v,w} \right)} \right| = \left| {\frac{{\partial \left( {x,y,z} \right)}}{{\partial \left( {u,v,w} \right)}}} \right| = \left| {{{\left( {\frac{{\partial \left( {u,v,w} \right)}}{{\partial \left( {x,y,z} \right)}}} \right)}^{ - 1}}} \right| = \left| {\frac{1}{{ - 9}}} \right| = \frac{1}{9}.\]
Now it is easy to calculate the volume of the solid:
\[V = \iiint\limits_{U'} {\left| {I\left( {u,v,w} \right)} \right|dudvdw} = \iiint\limits_{U'} {\frac{1}{9}dudvdw} = \frac{1}{9}\int\limits_0^5 {du} \int\limits_1^4 {dv} \int\limits_{ - 3}^6 {dw} = \frac{1}{9} \cdot 5 \cdot 3 \cdot 9 = 15.\]
