So I am asked to find the surface area of the part of the curve $$x=5y+z^2$$ that lies between the planes $y=0,y=z, z=0,$ and $z=2$.
I parametized my curve as $r(y,z)=\left<5y+z^2,y,z\right>$, found the absolute value of cross product $\left|r_{x} \times r_{y} \right|$ to be $$\left|r_{x} \times r_{y} \right|=\sqrt{4z^2-25}$$ and integrated $$\int_{0}^{2}\int_{0}^{z} \sqrt{4z^2-25}\space \mathrm d y \mathrm d z$$ with a bit of help from Wolfram Alpha, and got $\frac{1}{12}(41\sqrt{41}-125)$ as an answer. Could someone please explain why this is incorrect?
There are several problems with your computation.
First, you write $r_x \times r_y$ when computing the differential area. I think you mean $r_y \times r_z$, based on your parameterization.
Your computation of the magnitude of the cross product doesn't make sense: $4z^2 - 25$ is strictly negative on your region, and so its square root is complex.
Your computation of the integral is apparently incorrect as well. WolframAlpha shows that $$\int_0^2 \int_0^z \sqrt{4z^2 - 25} \, dy \, dz = \frac{49i}{6}$$ which is a complex number because of the complex integrand.
For a starting point in a better direction, notice that
$$r_y \times r_z = \langle 5, 1, 0 \rangle \times \langle 2z, 0, 1\rangle = \langle1, -5, -2z\rangle$$
so that $|r_y \times r_z| = \sqrt{4z^2 + 26}.$