Let $D$ be an ellipse rotated counterclockwise by $\frac{\pi}{3}$ radians, as pictured below. Before rotation, the horizontal radius was 2 and vertical radius was 6.
Convert $\iint\limits_{D} \, \bigl(8x + 4y\bigr)\, dA$ into an integral over a polar integral over a unit circle.
My work is as follows: $$\frac{1}{4} \left(\frac{x}{2} + \frac{\sqrt{3} y}{2}\right)^{2} + \frac{1}{36} \left(\frac{\sqrt{3} x}{2} - \frac{y}{2}\right)^{2}=1$$ $$\frac{x^2}{12} + \frac{1}{8} \sqrt{3} x y - \frac{x y}{24 \sqrt{3}} + \frac{7 y^2}{36} = 1$$ $$\frac{1}{36} (3 x^2 + 4 \sqrt{3} x y + 7 y^2) = 1$$ $$\frac{x^2}{12} + \frac{x y}{3 \sqrt{3}} + \frac{7 y^2}{36} = 1$$
$$x = 2\sqrt{3}u , \, y = \frac{6}{\sqrt{7}}v$$
$$u^{2} + \frac{4 u v}{\sqrt{7}} + v^{2} = 1$$
This is not a form I am used to. I am not sure actually how to obtain the $x$ and $y$ equations that do the magic and this was just a blind attempt.
I am stuck at this point. Any guidance is appreciated.
To follow along, draw three copies of the plane. Label one set of axes $x,y$, another set of axes $u,v$ and the third set of axes $\xi, \eta$. In the $(x,y)$ plane, draw the rotated ellipse $D$, as in the figure. In the $(u,v)$ plane, draw the same ellipse, but with semi-major axes along the coordinate axes; i.e the equation of the boundary ellipse should be \begin{align} \left( \dfrac{u}{6}\right)^2 + \left( \dfrac{v}{2}\right)^2 = 1 \end{align} Let's call this ellipse $E$. Finally, draw a circle of radius $1$ in the $(\xi, \eta)$ plane; i.e the equation of the circle is $\xi^2 + \eta^2 = 1$. We shall call the unit ball $B$.
Currently you have the integral $\int_D (\text{stuff involving $x,y$})\, dx \, dy$. Your task is to express everything in terms of an integral over the unit circle, $B$; i.e you want to write \begin{align} \int_D(\text{stuff involving $x,y$})\, dx \, dy &= \int_E(\text{stuff involving $u,v$})\, du \, dv = \int_B(\text{stuff involving $\xi,\eta$})\, d\xi \, d\eta \end{align} So, you just have to figure out what the "stuff" is. To do this, you have to figure out how the $(x,y)$, $(u,v)$ and $(\xi, \eta)$ coordinates are related to each other. We do this step by step.
First, what's the relation between $(x,y)$ and $(u,v)$? Well, to get from the funny ellipse $D$ to the standard ellipse $E$, we have to rotate points in the $(x,y)$ plane anti-clockwise by an angle $\dfrac{\pi}{3}$. This means \begin{align} \begin{pmatrix} u \\ v\end{pmatrix} &= \begin{pmatrix} \cos(\pi/3) & - \sin (\pi/3)\\ \sin(\pi/3) & \cos(\pi/3) \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix} \end{align} Or equivalently, \begin{align} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} \cos(-\pi/3) & - \sin (-\pi/3)\\ \sin(-\pi/3) & \cos(-\pi/3) \end{pmatrix} \cdot \begin{pmatrix} u \\ v\end{pmatrix} \\ &= \dfrac{1}{2} \begin{pmatrix} u + \sqrt{3} v \\ - \sqrt{3} u + v\end{pmatrix} \end{align} Recall the general formula for the change of variables: $\int_D f(x,y) \, dx \, dy = \int_E f(x(u,v), y(u,v)) \cdot \left| \frac{\partial (x,y)}{\partial(u,v)} \right| \, du \, dv$. So, you have to calculate the absolute value of the determinant of the Jacobian matrix. You can directly verify that in this case, the result will be 1 (rotation matrices always have determinant 1). Therefore, \begin{align} \int_D 8x + 4y \, dx \, dy &= 4 \int_D (2x + y) \, dx \, dy \\ &= 4 \int_E \left[2 \left( \dfrac{u + \sqrt{3}v}{2}\right) + \left( \dfrac{- \sqrt{3}u +v}{2}\right) \right] \cdot |1|\, du \, dv \\ &= 2 \int_E (2 - \sqrt{3})u + (\sqrt{3} + 1)v \, du \, dv \end{align}
Finally, what's the realtionship between $(u,v)$ and $(\xi, \eta)$? Well, to deform an ellipse of horizontal semi-axis length $6$ and veritcal semi-axis length $2$ into a unit circle, we must have $u = 6 \xi$ and $v = 2 \eta$. The determinant of this transformation is $6 \times 2 = 12$. Hence, \begin{align} \int_D 8x + 4y \, dx \, dy &= 2 \int_E (2 - \sqrt{3})u + (\sqrt{3} + 1)v \, du \, dv \\ &= 2 \int_B \left[(2 - \sqrt{3})(6 \xi) + (\sqrt{3} + 1)(2 \eta) \right] \cdot |12|\, d\xi \, d\eta \\ &= 24 \int_B \left[ 6(2 -\sqrt 3)\xi + 2(\sqrt{3} + 1) \eta \right] \, d \xi \, d \eta \end{align} This is the integral expressed over the unit ball/disk $B:= \{(\xi, \eta) \in \Bbb{R}^2 | \, \, \xi^2 + \eta^2 = 1\}$.
Of course, I intentionally gave several intermediate steps, but with some experience, it is possible to directly compute everything in one line.