So, first, we need to find vector function for the surface.Find the area of the portion of $x + 2y + 4z = 10$ in the first octant.

A few point that lie on the plane are $\langle 2,1,1 \rangle$, $\langle 0,5,0 \rangle$, $\langle 2,4,0 \rangle$.

Now we can find two vectors parallel to the plane by subtracting a few points:

$\langle 0,5,0 \rangle - \langle 2,4,0 \rangle = \langle -2,1,0 \rangle$

$\langle 0,5,0 \rangle - \langle 2,1,1 \rangle = \langle -2,4,-1 \rangle$

So our vector function is: $\vec r(u,v)= \langle 2,1,1\rangle +u\langle -2,1,0\rangle + v\langle -2,4,-1\rangle=\langle 2-2u-2v,1+u+4v,1-v \rangle$.

To find area we'll also need partial derivatives of $\vec r$:

$\vec r_u = \langle -2,1,0\rangle$ and $\vec r_v = \langle -2,4,-1 \rangle$

So area of the surface is

\begin{align*}

\int_{a}^{b} \int_{c}^{d} ||\vec r_u \times \vec r_v||\,du\,dv&=\int_{a}^{b} \int_{c}^{d} ||\langle -2,1,0\rangle \times \langle -2,4,-1 \rangle||\,du\,dv\\

&=\int_{a}^{b} \int_{c}^{d} ||\langle -1,-2,-6 \rangle||\,du\,dv\\

&=\int_{a}^{b} \int_{c}^{d} \sqrt{(-1)^2 + (-2)^2 + (-6)^2}\,du\,dv\\

&=\int_{a}^{b} \int_{c}^{d} 41\,du\,dv

\end{align*}

Well, it seems to be the procedure in the book, but I'm not sure this was correct? Also I'm not sure how to find the boundaries for the integrals? Can someone help?