Find the area of the part of the surface $z = x + y^2$ that lies above the triangle with vertices $(0, 0)$, $(1, 1)$ and $(0, 1)$. Find the area of the part of the surface $z = x + y^2$ that
lies above the triangle with vertices $(0, 0)$, $(1, 1)$ and $(0, 1)$.

My Solution
$z = x + y^2$
We begin by parameterising the surface in terms of $u$ and $v$:
Let $x = u$ and $v = y$.
$\therefore z = u + v^2$
We now have a parameterisation of the surface:
$(x, y, z) = S(u, v) = (u, v, u+v^2)$
We must now find the equations of the triangle:
From $(0, 0)$ to $(1, 1)$, $y = x$.
From $(0, 0)$ to $(0, 1)$, $x = 0$.
From $(1, 1)$ to $(0, 1)$, $y = 1$.
$\therefore$ The triangle is bounded by $y = 1$, $x = 0$, and $y = x$.
$y$ and $x$ intersect at $y = 0$ and $x = 1$
$\therefore 1 \ge x \ge 0, 1 \ge y \ge 0$ where $x = u$ and $y = v$.
$\therefore S = \{ S(u, v) | 1 \ge u \ge 0, 1 \ge v \ge 0 \}$ is the region we are integrating over in the $(u, v)$ plane.
If we graph this, we can see that the linear transformation mapped from a triangle in the xy-plane to a square in the uv-plane.
$S_u(u, v) = (1, 0, 1)$
$S_v(u, v) = (0, 1, 2v)$
$S_u(u, v) \times S_v(u, v) = (-1, -2v, 1) = N(u, v)$
$|N(u, v)| = \sqrt{2 + 4v^2}$
Therefore, we can calculate the area using the double integral $\int_{u = 0}^1 \int_{v = 0}^1 \sqrt{2 + 4v^2} \: dvdu$

Textbook Solution
The textbook solution states that the integral should be $\int_{v = 0}^1 \int_{u = 0}^v \sqrt{2 + 4v^2} \: dudv$ and the region as $S = \{ S(u, v) | 1 \ge v \ge 0, v \ge u \ge 0 \}$.

I would greatly appreciate it if people could please take the time to explain why my solution is incorrect and why the textbook's is the correct one. I would like to know what concept(s) I'm misunderstanding, so that I may study them further.
 A: First off, in general, you can just calculate surface area over a region using
$$\int\int\sqrt{f_x^2+f_y^2+1}dxdy$$
You can probably prove that using what you already have used in your solution.
Onward to the explanation: Everything is correct except your bounds. Your bounds would simply vary from $0$ to $1$ for all $x$ and $y$ and thus form a square, not the triangle you are looking for. The textbook offers a nice explanation, but I will go into more detail to show that $$S = \{ S(u, v) | 1 \ge v \ge 0, v \ge u \ge 0 \}$$
So for the first integral bounds, it is probably easier to view the bounding lines as the inequalities 
$$0 \leq x \leq 1$$
$$0 \leq y \leq 1$$
and finally, the one which you missed, the line $y=x$ is the equivalence case, so
$$y\leq x$$
is our final inequality. If this makes sense, then you should see why we must write the textbook's line that $0 \leq y \leq x$ Which translates directly into the integral.
Hope this helps!
A: You've done all the hard work parametrising the surface in terms of $x$ and $y$ correctly. The key is that the parametrisation "lifts" the triangle to the 2-dimensional surface embedded in $\mathbb{R}^3$ and so our domain of integration should be the triangle itself.
If you write the equation of the line between points $(0,0)$ and $(1,1)$, you've gotten $y=x$ (or $v=u$). This line bounds the region of integration from below and so our domain of integration is constrained by the fact that $v \geq u \geq 0$, where $0 \leq v \leq 1$. So, as in the case where one simply integrates a two-variable function over a triangular region, one integrates with respect to $u$ from $0$ to $v$ (as $u$ is bounded to the left by $u=0$ and bounded to the right by $u=v$) and then with respect to the $v$ variable from $0$ to $1$.
