Integral in $3D$ 
Let $R = [0,1]\times [0,1]$ and $f(x,y) = x.$ Note that $f$ is uniformly continuous on $R$. Use the definition of an integral to show that $\displaystyle\int_R f(v)dv = 0.5.$

Drawing this out, it is easy to see that this integral is the volume of a triangular prism with base area $0.5$ and height $1$, which is $0.5$. However, I'm more concerned about how to prove why this is indeed $0.5$.
By the definition of integration, $\displaystyle\int_R f(v)dv = L\displaystyle\int_R f(v)dv = \sup \{L(f, G): \text{ G grid on $R$}\} = U\displaystyle\int_R f(v)dv = \inf\{U(f, G) : \text{G grid on $R$}\}.$ For any grid $G = \{R_1,\cdots, R_p\},$ we have that $L(f, G) = \sum_{R_i\cap R\neq \emptyset} m_i|R_i| = \sum_{R_i} m_i|R_i|$ and $U(f, G) = \sum_{R_i} M_i|R_i|,$ where $M_i = \mathrm{sup}\{f(x) : x \in R_i\}$ and $m_i = \mathrm{inf}\{f(x) : x \in R_i\}.$ To show that $\sup\{L(f, G) : \text{ G grid on $R$}\} = 0.5,$ it suffices to show that for any $\epsilon > 0,$ we can find a grid $G$ so that $0.5 - \epsilon < L(f, G)$ and that for all grids $G,L(f, G)\leq 0.5$. Since $\sup\{L(f, G) : \text{ G grid on $R$}\} = \inf\{U(f, G) : \text{ G grid on $R$}\}$, it suffices to show the latter case, that is, for any $\epsilon > 0, \exists$ a grid $G = \{R_1,\cdots, R_k\}$ so that $0.5 - \epsilon < \displaystyle\sum_{R_i}m_i|R_i|$ and for all grids $G,L(f, G)\leq 0.5$. Let $G$ be a grid of $R$. Then $G$ is a collection of rectangles $R_k$ of the form $[x_{k-1}^1, x_k^1]\times [x_{k-1}^2\times x_k^2],$ where $P_j(G)=\{x_k^j : 1\leq k\leq l, x_{k-1}^j < x_k^j, x_0^j = x_j, x_l^j = b_j\}$ is a partition of $[a_j, b_j]$, and $R = [a_1,b_1]\times [a_2,b_2].$ We have that each $[x_{k-1}^j, x_k^j]\subseteq [0,1].$ Also, $m_k = \inf \{f(x) : x \in R_k\} = x_{k-1}^1.$ Hence $\displaystyle\sum_{R_k}m_k|R_k| = \displaystyle\sum_{R_k}x_{k-1}^1(x_k^1 - x_{k-1}^1)(x_k^2 - x_{k-1}^2).$ However, I'm not quite sure how to show that  $\displaystyle\sum_{R_k}x_{k-1}^1(x_k^1 - x_{k-1}^1)(x_k^2 - x_{k-1}^2) \leq 0.5$ and that we may find a set of $R_k's$ so that $\displaystyle\sum_{R_k}x_{k-1}^1(x_k^1 - x_{k-1}^1)(x_k^2 - x_{k-1}^2)>0.5-\epsilon.$
I know that $\displaystyle\sum_{R_k}x_{k-1}^1(x_k^1 - x_{k-1}^1)(x_k^2 - x_{k-1}^2) \leq \displaystyle\sum_{R_k} x_{k-1}^1$ as $[x_{k-1}^j, x_k^j]\subseteq [0,1]$ for $1\leq j\leq 2$
 A: For a general grid (partition) $G$   of $[0,1]\times[0,1]$ we have subrectangles of the form $[x_{j-1},x_j] \times [y_{k-1},y_k]$ for $j = 1, \ldots, n$ and $k = 1, \ldots,m.$
The upper sum is given by
$$U(f,G) = \sum_{j=1}^n\sum_{k=1}^m x_{j}(x_j-x_{j-1})(y_k - y_{k-1}) = \sum_{j=1}^nx_{j}(x_j-x_{j-1})\sum_{k=1}^m (y_k - y_{k-1})  =  \sum_{j=1}^nx_{j}(x_j-x_{j-1}), $$
and, similarly, the lower sum is
$$L(f,G) =  \sum_{j=1}^nx_{j-1}(x_j-x_{j-1}), $$
Since $x_{j-1} \leqslant \frac{1}{2}(x_{j-1} + x_j)\leqslant x_j$, we have
$$L(f,G)\leqslant \underbrace{\sum_{j=1}^n \frac{1}{2}(x_{j-1} + x_j)(x_j - x_{j-1})}_{ = \frac{1}{2}\sum_{j=1}^n (x_j^2 - x_{j-1}^2) = \frac{1}{2}} \leqslant U(f,G)$$
We also have
$$U(f,G) - L(f,G) = \sum_{j=1}^n (x_j - x_{j-1})^2 \leqslant \max_{j=1,\ldots,n}(x_j - x_{j-1})$$
Thus,
$$\tag{1}L(f,G) \leqslant \frac{1}{2} \leqslant U(f,G) \leqslant L(f,G) + \max_{j=1,\ldots,n}(x_j - x_{j-1})$$
Choosing a sufficiently fine grid $G$ where $\max_{j=1,\ldots,n}(x_j - x_{j-1})< \epsilon$, we get
$$\tag{2}\frac{1}{2} - \epsilon \leqslant L(f,G) \leqslant \frac{1}{2}$$

(1) and (2) together show that for any $\epsilon > 0$ there is a grid $G$ such that
$$\frac{1}{2} - \epsilon \leqslant L(f,G) \leqslant \frac{1}{2} \leqslant U(f,G) \leqslant \frac{1}{2} + \epsilon$$
This proves Riemann integrability and since the integral lies between lower and upper sums we must have
$$\int_R f = \frac{1}{2}$$
A: Ummm.... here's your graph of $f(x,y) = x$.  It sure seems like you're working really hard on a simple question:

