How to Solve Double Integral Transformation (Triangular Region of Integration)
You may be given a transformation to use to solve a double integral, and sometimes the region of integration is a shape, such as a parallelogram or ellipse.
Before continuing, consider looking at an example of a double integral that only has a rectangular region of integration as practice before moving onto more complicated region of integrations.
We have worked-out examples of double integrals over rectangular regions here and here.
Let’s solve an example where the region of integration is a triangle.
The general steps are to construct lines using the given points of the triangle, make your own limits of integration, then solve the double integral normally.
Example Problem: Use the transformation
\[x = 2u + v\]
\[y = u + 2v\]
to solve the double integral \(\int_{}^{} {\int_R^{} {(x – 3y)dA} } \), where R is the triangular region with vertices (0,0), (2,1), and (1,2).
Since this is a multivariable transformation of variables (you have two equations and two variables in the transformation), we need to find the Jacobian, which is just a determinant of all the partial derivatives.
\[J = \left| {\begin{array}{*{20}{c}}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}\end{array}} \right|\]
\[J = \left| {\begin{array}{*{20}{c}}2&1\\1&2\end{array}} \right|\]
\[J = (2)(2) – (1)(1)\]
\[J = 3\]
This means that:
\[dxdy = 3dudv = 3dvdu\]
For the region of integration, we’re told the shape is a triangle and we’re given the vertices in x and y coordinates.
Plot the triangle shape, and find the three equations that make up the triangle.
Just use y = mx + b, use slope m = rise/run and get those equations, don’t make algebraic mistakes.
The three equations that make up the triangle will be:
\[y = 2x\]
\[y = – x + 3\]
\[y = \frac{x}{2}\]
What you want to do next is substitute in the given transformation (x = 2u + v) and (y = u + 2v) into each of the three lines we acquired just now.
When you do this, you’ll get new integral boundaries for the u and v variables.
Remember, we can’t use the same boundaries of the triangle, we need to get new boundaries in terms of u and v first.
First:
\[\begin{array}{l}y = 2x\\u + 2v = 2(2u + v)\end{array}\]
\[u = 0\]
Second:
\[\begin{array}{l}x = 2y\\2u + v = 2(u + v)\\v = 0\end{array}\]
Third:
\[\begin{array}{l}x + y = 3\\(2u + v) + (u + 2v) = 3\\v = 1 – u\end{array}\]
So we got new boundaries, but we only have three new boundaries but we need four for a double integral. Just plot these u and v boundaries and you can figure out the last boundary.
Plot a new triangle in the u and v coordinates.
We found u = 0, that’s just a horizontal line on the x-axis (now the u-axis), and v = 0 which is a vertical line on the y-axis (now the v-axis), and plot the line v = 1 – u, which is a line that encloses a triangle.
See the sketch below we made:
There are two ways to do the boundary, but I would do the boundary as : v is between 0 and 1-u, and u is between 0 and 1.
The alternative is v is between 0 and 1, and u is between 0 and 1-v.
Don’t over-complicate it, just remember these general rules for determining the integral boundaries. You only need to use the actual diagonal line equation for one of the boundaries.
We will continue with the former boundaries we found.
Let “I” be the double integral.
Substitute in the transformation into the integral as well as the Jacobian value.
\[I = \int_0^1 {\int_0^{1 – u} {\left[ {(2u + v) – 3(u + 2v)} \right]} } 3dvdu\]
\[I = 3\int_0^1 {\int_0^{1 – u} {\left[ {(2u + v) – 3(u + 2v)} \right]} } dvdu\]
\[I = 3\int_0^1 {\int_0^{1 – u} {( – u – 5v)dvdu} } \]
\[I = 3\int_0^1 {\left[ { – u(1 – u) – \frac{5}{2}{{(1 – u)}^2}} \right] – } \left[ { – u(0) – \frac{5}{2}{{(0)}^2}} \right]du\]
\[I = 3\int_0^1 { – u(1 – u) – \frac{5}{2}{{(1 – u)}^2}du} \]
\[I = 3\int_0^1 {\left( { – \frac{5}{2} – \frac{3}{2}{u^2} + 4u} \right)} du\]
\[I = 3\left( { – \frac{5}{2}(1) – \frac{1}{2}{{(1)}^3} + 2{{(1)}^2}} \right)\]
\[I = – 3\]