Find the surface area of the plane $$x+\frac{1}{\sqrt{2}}y+\frac 14 z=1$$ limited by the coordinate system planes
My findings :
I suppose we should express the scalar $z=f(x,y) \rightarrow z=4-4x-2\sqrt{2}y$ and then $\mathbf{r}(s,t)$ is $$\mathbf{r}(s,t)=s\mathbf{i}+t\mathbf{j}+(4-4s-2\sqrt{2}t)\mathbf{k}$$ and normal is $$\mathbf{n}(s,t)=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -4 \\ 0 & 1 & -2\sqrt{2} \end{vmatrix}=4\mathbf{i}+2\sqrt{2}\mathbf{j}+\mathbf{k}$$ Good so far? I guess I should find $(u,s)\in \text{"limits"}$ and use the right integral. Any hints/help? $$\int \int_S gdS = \int \int_D g(\mathbf{r}(s,t)) \sqrt{(\frac{\partial{f}}{\partial{s}})^2+(\frac {\partial{f}}{\partial{t}})^2+1}dsdt$$
The area of a surface is $$\iint\limits_{\Sigma} dS,$$ where $\Sigma$ is the surface and $dS$ is the surface area element. It is computed by finding the length of the normal vector to the surface. In your case we have the graph of a function, therefore
$$dS = \sqrt{ 1 + \left( \frac{\partial f}{\partial s} \right)^2 + \left( \frac{\partial f}{\partial t} \right)^2 } \, dA,$$
where $dA$ is the area element in a projected region, assuming parameters $s,t$ as you did. Writing
$$z = 4 - 4x - 2 \sqrt{2} y = 4 - 4s - 2 \sqrt{2}t = f(s,t)$$
I am taking $x=s$ and $y=t$ (arbitrarily). The projected region will be when $z=0$, yielding $4s + 2\sqrt{2} t = 4$ or $2s + \sqrt{2} t = 2$. Since $s,t \geq 0$ we have $s \in [0,1]$ and $t \in [0, \sqrt{2}]$.
The length of the normal vector is
$$\sqrt{ 1 + \left( \frac{\partial f}{\partial s} \right)^2 + \left( \frac{\partial f}{\partial t} \right)^2 } = \sqrt{ 1 + (-4)^2 + (-2 \sqrt{2})^2 } = \sqrt{1 + 16 + 8} = 5.$$
Our surface area will be
$$\iint\limits_{\Sigma} dS = \iint\limits_{R} 5 \,dA = 5 \iint\limits_{R} \, dA,$$
where $R$ is the region projected in the $xy$ plane. Setting up the integral gives
$$5 \iint\limits_{R} \, dA = 5 \int_0^{\sqrt{2}} \hspace{-5pt} \int_0^1 dx \, dy.$$
You can compute the integrals or realize this is a right triangle with sides 1 and $\sqrt{2}$ and therefore the area is $1 \cdot \frac{\sqrt{2}}{2}$ and the surface area is
$$\iint\limits_{\Sigma} dS = \frac{5 \sqrt{2}}{2}.$$