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How to Solve A Double Integral Using Transformation (Parallelogram Region of Integration)
Now we have a double integral, but the region of integration is a different shape, a parallelogram with four points we’re given!
You need to construct lines and make your own limits of integration, but once you do that it turns into a normal double integral problem.
You’ll generally be provided the transformation to use.
Example problem: Find \(\int\limits_{}^{} {\int\limits_R^{} {(4x + 8y)dA} } \) where R is the parallelogram with vertices (-1,3), (1,-3), (3,-1) and (1,5), given the transformations \(x = \frac{{(u + v)}}{4}\) and \(y = \frac{{(v – 3u)}}{4}\)
The first thing to do is make equations for the lines that connect the vertices of the parallelogram, so just find those slopes and intercepts.
You will get the following 4 equations: \[\begin{array}{l}y = 8 – 3x\\y = x + 4\\y = – 3x\\y = x – 4\end{array}\]
Then, substitute the transformation equations into each of the four equations above.
For example, we will do this for the first equation above. \[\frac{{v – 3u}}{4} = 8 – \frac{{3(u + v)}}{4}\]
The u variable cancels out and we obtained that v = 8.
You need to do this for the other three line equations and you’ll acquire u = -4, v = 0, and u = 4.
These serve as the boundaries for the double integral we will be creating.
The next step is to find the Jacobian. I will let “J” represent this.
\[J = \left| {\frac{{\partial x}}{{\partial u}} \cdot \frac{{\partial y}}{{\partial v}} – \frac{{\partial x}}{{\partial v}} \cdot \frac{{\partial y}}{{\partial u}}} \right|\]
\[J = \left( {\frac{1}{4}} \right) \cdot \left( {\frac{1}{4}} \right) – \left( {\frac{1}{4}} \right) \cdot \left( { – \frac{3}{4}} \right)\]
\[J = \frac{1}{4}\]
And we just substitute all this information in to make a double integral to solve.
We can take the Jacobian factor out since it’s not a variable, it’s a constant.
Let’s call the integral “I”. Now the problem turns into a regular double integral problem.
\[\begin{array}{l}I = \frac{1}{4}\int_0^8 {\int_{ – 4}^4 {\left( {3v – 5u} \right)dudv} } \\I = \frac{1}{4}\int_0^8 {\left( 3 \right.(4)v – \frac{5}{2}} \left. {{{(4)}^2}} \right) – \left( {3( – 4)v – \frac{5}{2}{{( – 4)}^2}} \right)dv\\I = \frac{1}{4}\int_0^8 {24vdv} \\I = \frac{1}{4} \cdot \frac{{24}}{2}\left( {{8^2} – {0^2}} \right)\\I = 192\end{array}\]
Try an example where you find a double integral over a different region, such as a triangle region!
Try an example where you find a double integral over a different region, such as an ellipse region!