### Double Integral Transformation (Triangular Region) Example 3

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. Let's solve an example where the region of integration is a triangle.

Use the transformation \(x = 2u + v\) and \(y = u + 2v\) 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, 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 u and v first.

First:

\[y = 2x\]

\[u + 2v = 2(2u + v)\]

\[u + 2v = 2(2u + v)\]

Second:

\[x = 2y\]

\[2u + v = 2(u + v)\]

\[v = 0\]

Third:

\[x + y = 3\]

\[(2u + v) + (u + 2v) = 3\]

\[v = 1 - u\]

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:

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. I will continue with the former. Let "I" be the double integral. Substitute in the transformation into the integral as well as the Jacobian value.

\[I = \int_0^1 {\int_0^{1 - u} {\left[ {(2u + v) - 3(u + 2v)} \right]} } 3dvdu\]

\[I = 3\int_0^1 {\int_0^{1 - u} {\left[ {(2u + v) - 3(u + 2v)} \right]} } dvdu\]

\[I = 3\int_0^1 {\int_0^{1 - u} {( - u - 5v)dvdu} } \]

\[I = 3\int_0^1 {\left[ { - u(1 - u) - \frac{5}{2}{{(1 - u)}^2}} \right] - } \left[ { - u(0) - \frac{5}{2}{{(0)}^2}} \right]du\]

\[I = 3\int_0^1 { - u(1 - u) - \frac{5}{2}{{(1 - u)}^2}du} \]

\[I = 3\int_0^1 {\left( { - \frac{5}{2} - \frac{3}{2}{u^2} + 4u} \right)} du\]

\[I = 3\left( { - \frac{5}{2}(1) - \frac{1}{2}{{(1)}^3} + 2{{(1)}^2}} \right)\]

\[I = - 3\]

Please see other double integral transformation examples below:

Double Integral Transformation (Parallelogram Region) Example 1

Double Integral Transformation (Ellipse Region) Example 2

Alternatively return to the double integral transformation hub page:

Double Integrals (Transformations)