Problem: Find the area of the part of the hyperbola $az = z^2-y^2$ that intersects the cylinder $z^2+y^2=a^2$.
My attempt: I thought we could do this by projecting the surface onto the $yz$-plane and taking the surface integral of the function $x=g(y,z)=\sqrt{z^2-y^2}.$ Letting $S$ be the surface and $E$ be the projection onto the $yz$-plane. We wish to project the cylinder $az=z^2-y^2$ onto the $yz$-plane. This is done by first calculating $\mathrm{d}S$ in $\iint_{\mathcal{S}}dS$ which gives the surface area. We have that $$dS = \sqrt{1+\left(\frac{\partial x}{\partial y}\right)^{2}+\left(\frac{\partial x}{\partial z}\right)^{2}} \ \mathrm{d}z\mathrm{d}y=\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y$$
Letting $E$ represent the area of the projection of the cylinder onto the $yz$-plane in the first octant we get: $$\iint_{\mathcal{S}}\mathrm{d}S=4\iint_{E}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y$$ The particular limits of $z$ and $y$ are $0\leq z \leq \sqrt{2ay} \quad \text{and} \quad 0\leq y \leq 2a$ So: \begin{align*}4\iint_{E}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y & = 4\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y \\ &= 4 \int_{0}^{2a}\sqrt{2ay}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}y\\ &= 4(4a^2)=16a^2\end{align*}