I was trying to find the surface integral of the surface defined by equation of plane passing through the points $(0,0,2)$, $(0,1,0)$ and $(2,1,0)$ over the vector field $\vec{F}=x\vec{i}+y^3\vec{j}+\vec{k}$
My try: The equation of the plane is $2y+z=2$ whose unit outward normal being:
$$\hat{n}=\frac{2\vec{j}+\vec{k}}{\sqrt{5}}$$
So the surface integral is now:
$$\int\int_{S}\vec{F}.\hat{n}dS=\int\int_{S}\frac{2y^3+1}{\sqrt{5}}dS$$
Now how to start the procedure of projection? the book has projected the surface on $x-z$ plane, but i am not clear the concept behind this projection
You could parameterize the plane. This means write it as a mapping $M:\mathbb{R}^2\rightarrow\mathbb{R}^3$ in terms of two variables $u$ and $v$. You want to find two vectors $\vec{a}$ and $\vec{b}$ and a point $P$ so that the plane is described by $M(u,v) = \vec{a}u + \vec{b}v + P$. The point $P$ can be any point on the plane. So let $P=(0,1,0)$. The vectors $a$ and $b$ must be parallel to the plane and linearly independent. We can find suitable choices for $a$ and $b$ by subtracting pairs of known points on the plane. So let $a = (0,-1,2)$ and let $b=(2,0,0)$. This makes it so that the $(u,v)$ coordinates of the points are $(0,0)$ for $P=(0,1,0)$, $(1,0)$ for $Q=(0,0,2)$ and $(0,1)$ for $R=(2,1,0)$. In other words, the $(u,v)$ domain is the unit triangle. We can now write the surface integral over the triangle $PQR$ as $$ \iint_S \frac{2y^3+1}{\sqrt{5}} \,dS = \int_0^1 \int_0^{1-u} \frac{2(a_yu + b_yv + P_y)^3+1}{\sqrt{5}} |M_u \times M_v| \,dv\,du. $$ Since $M_u=a$ and $M_v=b$, $|M_u \times M_v| = 2\sqrt{5}$ and the surface integral becomes $$ \int_0^1 \int_0^{1-u} \left(4(1-u)^3 + 2\right) \,dv\,du. $$
Projecting the triangle onto the $xz$ plane is just another way of parameterizing the plane. The parameterization in this case is $x=v$, $z=u$, and $y=f(x,z) = f(v,u) = 1-z/2 = 1-u/2$. For this parameterization, $M(u,v) = (v, f(v,u), u)$, $M_u = (0, f_z, 1)$, and $M_v=(1, f_x, 0)$. The surface normal is $M_u \times M_v = (-f_x, 1, -f_z) = (0, 1, 1/2)$. Note that this is consistent with the direction of the unit normal computed in the question, $(0, 2/\sqrt{5}, 1/\sqrt{5})$. If we had used $x=u$ and $z=v$, the normal would have been reversed. Now we can define the surface integral using this second parameterization: $$ \begin{align} \iint_S \frac{2y^3+1}{\sqrt{5}} \,dS &= \int_0^2 \int_0^{2-v} \frac{2(1-u/2)^3 + 1}{\sqrt{5}} \frac{\sqrt{5}}{2} \,du\,dv \\ &= \int_0^2 \int_0^{2-v} \left((1-u/2)^3 + \frac{1}{2}\right) \,du\,dv. \end{align} $$ The limits of integration of the inner integral are 0 and $2-x$ because the projection of the line segment connecting $(0,0,2)$ and $(2,1,0)$ onto the $xz$ plane is the line $z=2-x$.