Home»Math Guides»Double Iterated Integrals (over ellipse region of integration)
Double Integral Transformation Example (With Region of Integration over an Ellipse Region)
You may be given a transformation to use to solve a double integral, and sometimes the region of integration is over 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.
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We have worked-out examples of double integrals over rectangular regions here and here.
When you solved double integrals with only numbers in the integral boundaries, that meant you were actually integrating over a region of integration that was a rectangle because the numbers represented the equations for four straight lines, which formed boundaries of a rectangle when plotted together.
When the integral boundaries instead have equations, the boundary formed becomes more complicated. In this practice problem we’ll go over, the region of integration is an ellipse, which is just a stretched or shrunk equation of a circle.
You will generally be given the transformation to use in the question! If you’re not, you probably just need to use polar coordinates as a transformation. The transformation to use isn’t always apparent and instructors realize that.
If you’re unfamiliar with the ellipse shape, it’s just a stretched or compressed circle.
Example problem: Use the transformation (x = 2u) and (y = 3v) to solve the integral \(\int_{}^{} {\int_R^{} {{x^2}dA} } \) where R is the ellipse \(9{x^2} + 4{y^2} = 36\)
Immediately substitute x and y into the equation for the ellipse.
\[9{x^2} + 4{y^2} = 36\]
\[9{(2u)^2} + 4{(3v)^2} = 36\]
\[9(4{u^2}) + 4(9{v^2}) = 36\]
\[36{u^2} + 36{v^2} = 36\]
\[{u^2} + {v^2} = 1\]
By using the transformation we got a unit circle, just a regular circle with its center in the origin and radius of the root of 1, which is just one.
Next we should find the Jacobian because this transformation has multiple variables.
\[J = \left| {\begin{array}{*{20}{c}}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}\end{array}} \right|\]
\[\begin{array}{l}J = \left| {\begin{array}{*{20}{c}}2&0\\0&3\end{array}} \right|\\J = (2)(3) – (0)(0)\\J = 6\end{array}\]
What does the Jacobian here tell us and why did we need to find it? It is a matrix of partial derivatives, and tells us that \(dxdy = 6dudv\)
So now we can make our double integral.
Remember that we’re integrating over a unit circle for the u and v variables.
How can we write that out? Well, you can say v goes from -1 to 1, and u goes from
\[ – \sqrt {1 – {v^2}} \]
to
\[\sqrt {1 – {v^2}} \]
Substituting this information in,
\[I = \int_{ – 1}^1 {\int_{ – \sqrt {1 – {v^2}} }^{\sqrt {1 – {v^2}} } {4{u^2} \cdot 6dudv} } \]
And when integrating over a circle, use polar coordinates! See polar coordinate examples for more details, but we’ll show all the steps here.
\[u = r\cos (\theta )\]
\[v = r\sin (\theta )\]
\[dudv = rdrd\theta \]
In polar coordinates, the radius goes from 0 to 1. Theta goes from 0 to \(2\pi \) since it’s an unrestricted full circle.
Now our integral will look like:
\[I = \int_0^{2\pi } {\int_0^1 {24{{(r\cos (\theta ))}^2}rdrd\theta } } \]
\[I = \int_0^{2\pi } {\int_0^1 {24{r^3}{{\cos }^2}(\theta )drd\theta } } \]
\[I = \int_0^{2\pi } {\frac{{24}}{4}} {(1)^4}{\cos ^2}(\theta ) – \frac{{24}}{4}{(0)^4}{\cos ^2}(\theta )d\theta \]
\[I = \int_0^{2\pi } {6{{\cos }^2}(\theta )d\theta } \]
Now you can use trigonometric identities, such as double angle formulas to simplify our expression.
\[I = 3\int_0^{2\pi } {(1 + \cos (2\theta ))d\theta } \]
\[I = 3\left[ {(2\pi ) + \frac{{\sin (2(2\pi ))}}{2} – 0 – \frac{{\sin (2(0))}}{2}} \right]\]
\[I = 6\pi \]