What is the area of a function $z = \frac{1}{2}y+4$ over a triagular region given by points $A(0,-1)$, $B(3,-1)$ and $C(3,2)$?
I've tried two different approaches but came to a different results, where is my mistake?
$(1)$ Since the function is flat, I parametrized the region on the function, call it $\phi$:$$x = s$$ $$y=t$$ $$z=\frac{1}{2}t+4$$ where $s \in (0,3)$ and $t \in (-1, s-1)$.
I came to an integral $$\int_{0}^{3} \int_{-1}^{s-1}1 \left \Vert\frac{\partial \phi}{\partial s} × \frac{\partial \phi}{\partial t} \right\Vert \,dt \,ds= \int_{0}^{3} \int_{-1}^{s-1}1 \frac{\sqrt{5}}{2} \,dt \,ds= \frac{9\sqrt{5}}{4}$$
$(2)$ I used surface integral over the function: $$\int_{0}^{3} \int_{-1}^{x-1}(\frac{1}{2}y+4) \sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y}})^2 \,dy \,dx= \int_{0}^{3} \int_{-1}^{x-1}(\frac{1}{2}y+4) \frac{\sqrt{5}}{2} \,dy \,dx=9\sqrt{5}.$$ What is my mistake and why?
If $T$ is the following triangle
we have that the area of the surface defined by $f(x,y)=4+\frac{y}{2}$ over $T$ is given by $$ \iint_{T} \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}\,dx\,dy =\frac{\sqrt{5}}{2}\mu(T)=\frac{9\sqrt{5}}{4}.$$