Finding the Area of a Figure Given by an Equation

[WMDhamnekar]

Question:- Find the area of the figure given by the cartesian equation below:

$$\frac{(x+y)^2}{16}+\frac{(x-y)^2}{9}=1$$

Solution given:-

Let $x+y= 4\cos{\alpha},x-y=3\sin{\alpha}$ Then $x=\frac{4\cos{\alpha}+3\sin{\alpha}}{2}$ $\Rightarrow dx=\frac{3\cos{\alpha}-4\sin{\alpha}}{2}d\alpha$

$y=\frac{4\cos{\alpha}-3\sin{\alpha}}{2}$. So, $ydx=\left(3-\frac{25}{8}*\sin{2\alpha}\right)*d\alpha$ What is this ydx?

Hence the required area is

$$\displaystyle\int_0^{2\pi}\left(3-\frac{25}{8}*\sin{2\alpha}\right)*d\alpha$$ Why does the author select ydx as integrand for computation of area of ellipse? What is the logic behind that?

$=6\pi=18.849$

If any member of Math help Board knows the explanations for my queries, may reply to this question.

 

[I like Serena]

$y$ is the y-coordinate of a point on the edge of the figure. $dx$ is an infinitely small increment of the x-coordinate.
The product $y\,dx$ represents a rectangle of height $y$ and width $dx$. Its area is $y\,dx$.
If we add all such rectangles together, we get the area of the figure.

It's actually more or less the definition of an integral.
See here how that interpretation works.