### Double Integral Transformation (Ellipse Region) Example 2

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.

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.

$J = \left| {\begin{array}{*{20}{c}}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}\end{array}} \right|$

$J = \left| {\begin{array}{*{20}{c}}2&0\\0&3\end{array}} \right|$

$J = (2)(3) - (0)(0)$

$J = 6$

What does the Jacobian here tell us? It tells us $$dxdy = 6dudv$$.

That's good. 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 $$\sqrt {1 - {v^2}}$$.

$I = \int_{ - 1}^1 {\int_{ - \sqrt {1 - {v^2}} }^{\sqrt {1 - {v^2}} } {4{u^2} \cdot 6dudv} }$

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.

$I = 3\int_0^{2\pi } {(1 + \cos (2\theta ))d\theta }$

$I = 3\left[ {(2\pi ) + \frac{{\sin (2(2\pi ))}}{2} - 0 - \frac{{\sin (2(0))}}{2}} \right]$

$I = 6\pi$

Please see other double integral transformation examples below:

Double Integral Transformation (Parallelogram Region) Example 1

Double Integral Transformation (Triangle Region) Example 3 