We have an elliptic area defined by $$A := \{(x, y) \in \mathbb{R}^2 \mid (\tfrac{x}{a})^2+(\tfrac{y}{b})^2 \leq 1 \}$$ and a height function $$h \colon \mathbb{R}^2 \to \mathbb{R}, (x, y) \mapsto h(x, y) = c\left[1-(\tfrac{x}{a})^2-(\tfrac{y}{b})^2\right]$$ by transforming the area constraint we get $$y = \pm b\sqrt{1-(\tfrac{x}{a})^2}$$ which gives us the bounds for integrating over the y-axis. By setting $y = 0$ we get the bounds for the x-axis which is $\pm a$.
Finally we can write down our integral $$\int_Ah(x,y)dxdy = \int_{x_1=-a}^{x_2=a} \int_{y_1=-b\sqrt{1-(\tfrac{x}{a})^2}}^{y_2=b\sqrt{1-(\tfrac{x}{a})^2}}c\left[1-(\tfrac{x}{a})^2-(\tfrac{y}{b})^2\right]dydx = \\\ 4c\int_{0}^{x_2} \int_{0}^{y_2}\left[1-(\tfrac{x}{a})^2-(\tfrac{y}{b})^2\right]dydx$$ now comes the hard part.
1. Approach
This was my own attempt which does get stuck at some point
$$\dfrac{8}{3}cb\int_0^a\left[1-(\tfrac{x}{a})^2\right]^\frac{3}{2}dx \stackrel{x = a\sin(z)}{=} \dfrac{8}{3}\dfrac{cb}{a}\int_{z_1=0}^{z_2=a\sin(a)} \left[1-\sin(z)^2\right]^\frac{3}{2}\dfrac{1}{\cos(z)}dz = \dfrac{8}{3}\dfrac{cb}{a}\int_{z_1}^{z_2} \cos(z)^2dz = \dfrac{8}{3}\dfrac{cb}{a}\left[z+\cos(z)\sin(z)\right]\vert^{z_2}_{z_1} = \dfrac{8}{3}\dfrac{cb}{a}\left[a\sin(a)+\cos(a\sin(a))\sin(a\sin(a))\right]$$
2. Approach
This is taken from the solutions.
$$\dfrac{8}{3}cb\int_0^a\left[1-(\tfrac{x}{a})^2\right]^\frac{3}{2}dx \stackrel{x = az}{=} \dfrac{8}{3}cba\int_0^1\left[1-z^2\right]^\frac{3}{2}dz$$
where $\int_0^2\left[1-z^2\right]^\frac{3}{2}dz$ was given with $\dfrac{3}{16}\pi$.
Questions
Why does the first approach come to another solution? What rule did I break?
What happened with the substitution $x = az$ in the second approach? How does the a in the upper bound turn to 1 and why is $dz \neq \dfrac{1}{a}$.
For APPROACH 1:
There were two errors in Approach 1.
First, $dx= a\,\cos z$ and not $\frac1{a\cos z}$.
Second, the upper limit transforms from $x=a$ to $z=\pi/2$.
Then, the integral of interest becomes
$$\frac83 abc\int_0^{\pi/2}\cos^4 z\,dz=\frac83 abc\times \frac{3\pi}{16}=\frac{\pi}{2}abc$$
For APPROACH 2:
Here, the substitution $x=az\implies dx=adz$ and when the upper limit on $x$ is $a$, the corresponding upper limit on $z$ is, therefore, $1$ since $a \times 1=a$.