Computing an ellipse's area by substitution I want to do it by substitution, please don't put an answer that isn't about substitution.
Say I go from something like this $\frac{x²}{c²}+\frac{y²}{d²}=k²$ to $\frac{x²}{a²}+\frac{y²}{b²}=1$.
Let $x=a\frac{u}{b}$, the equation becomes then that of a circle: $\frac{u²}{b²}+\frac{y²}{b²}=1$. The area is then $\pi\left(\frac{u²}{b²}+\frac{y²}{b²}\right)=\pi\left(\frac{x²}{a²}+\frac{y²}{b²}\right)$ which is wrong, where is the problem?
 A: The problem with what you tried is that the $(x,y)$ and $(u,y)$ dimensions are not the same. In $(u,y)$ dimensions you have
$$u^2 + y^2 = b^2 \tag{1}\label{eq1A}$$
This is a standard equation for a circle centered at the origin, with radius $|b|$ and area $b^2\pi$.
For using $(x,y)$ dimensions, you have the transformation equation of
$$x = \left(\frac{a}{b}\right)u \tag{2}\label{eq2A}$$
This applies a horizontal stretch or compression factor of $\left|\frac{a}{b}\right|$. Thus, the resulting new area in $(x,y)$ co-ordinates would be the original area in $(u,y)$ co-ordinates multiplied by this factor (e.g., as explained in the answer to Stretching a Hexagon). As such, the area of the ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \tag{3}\label{eq3A}$$
would be
$$\left(b^2\pi\right)\left|\frac{a}{b}\right| = |ab|\pi \tag{4}\label{eq4A}$$
This is the standard formula for the area of an ellipse.
Note you can also do this in the other direction. Suppose you let $A_e(a,b)$ be the area of an ellipse with an equation given in \eqref{eq3A}. Then you can apply the transformation equation
$$u = \left(\frac{b}{a}\right)x \tag{5}\label{eq5A}$$
to get \eqref{eq1A}, which is an equation of a circle with area $\pi b^2$. As the transformation applies a horizontal stretch or compression factor of $\left|\frac{b}{a}\right|$, the resulting area of the circle would be the area of the ellipse multiplied by this, i.e., you get
$$\pi b^2 = \left|\frac{b}{a}\right|A_e(a,b) \tag{6}\label{eq6A}$$
A: Consider how the circle becomes an elongated ellipse. Although not strictly pre-calculus, area finding for variable ordinates is by integration... each vertical strip area increases from $y_1 dx$  to $ y_2 dx$ where $y_2>y_1,$ that is multiplied by a constant factor.
Area is computed by integration, and $y_2>y_1$ being a constant can be taken out of the integral sign.
There is a constant stretch factor along y- direction
$$ = \frac{y_2}{y_1} =\frac{b}{a}, \,b>a. $$
Area $ \pi a^2 = \pi a\cdot a$ becomes $ \pi \,\dfrac{b}{a} a\cdot a =\pi b a$ 
