Form the lower sum underline S and upper sum S
Form the lower sum $\underline{S}$ and upper sum $\overline{S}$ for the function $z(x,y)=x^2+y^2$ on the region $1\leqslant x\leqslant 2,\ 1\leqslant y\leqslant 3$, dividing this region into rectangles by the lines $$ x=1+\frac{i}{n}, \qquad y=1+\frac{2j}{n} \qquad (i,j=0,1,\dots,n). $$ Find the limit value of these sums as $n\to\infty$.
Explanation of Method
Since $z(x,y)=x^2+y^2$ is strictly increasing in both $x$ and $y$ on the given region, on every subrectangle the minimum sits at the lower-left corner and the maximum at the upper-right corner. We build $\underline{S}$ and $\overline{S}$ explicitly with the given partition, simplify using power-sum formulas, and take $n\to\infty$.
- Compute $\Delta x,\Delta y,\Delta A$ from the given partition.
- Locate the min/max of $z$ on each subrectangle using monotonicity.
- Form $\underline S$ and $\overline S$ and simplify with $\sum k$, $\sum k^2$.
- Take the limit as $n\to\infty$ and confirm both sums agree.
- Verify with an iterated integral.
Step 1 · Understanding the partition
The region is $R=[1,2]\times[1,3]$, partitioned by $x_i=1+\frac{i}{n}$ ($i=0,\dots,n$) and $y_j=1+\frac{2j}{n}$ ($j=0,\dots,n$). The subinterval widths are constant:
$$ \Delta x = x_{i+1}-x_i=\frac1n, \qquad \Delta y = y_{j+1}-y_j=\frac2n, \qquad \Delta A=\Delta x\,\Delta y=\frac{2}{n^2}. $$
Step 2 · Analyze the function
$$ \frac{\partial z}{\partial x}=2x>0, \qquad \frac{\partial z}{\partial y}=2y>0 \quad\text{on the region.} $$
Since $z$ is strictly increasing in both variables, on each subrectangle $[x_i,x_{i+1}]\times[y_j,y_{j+1}]$ the minimum occurs at $(x_i,y_j)$ and the maximum at $(x_{i+1},y_{j+1})$.
Step 3 · Form the lower sum $\underline{S}$
$$ \underline{S}=\sum_{i=0}^{n-1}\sum_{j=0}^{n-1} z(x_i,y_j)\,\Delta A = \frac{2}{n^2}\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\left[\left(1+\frac{i}{n}\right)^2+\left(1+\frac{2j}{n}\right)^2\right]. $$
Expanding, $x_i^2+y_j^2 = 2+\frac{2i+4j}{n}+\frac{i^2+4j^2}{n^2}$.
Step 4 · Separate the sums for the lower sum
Using $\sum_{k=0}^{n-1} k=\frac{(n-1)n}{2}$ and $\sum_{k=0}^{n-1} k^2=\frac{(n-1)n(2n-1)}{6}$:
$$ \sum_{i=0}^{n-1}\sum_{j=0}^{n-1} 2 = 2n^2, \qquad \sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\frac{2i+4j}{n}=3n(n-1), \qquad \sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\frac{i^2+4j^2}{n^2}=\frac{5(n-1)(2n-1)}{6}. $$
Step 5 · Combine for the lower sum
$$ \underline{S}=\frac{2}{n^2}\left[2n^2+3n(n-1)+\frac{5(n-1)(2n-1)}{6}\right]=\frac{2}{n^2}\cdot\frac{40n^2-33n+5}{6}=\frac{40n^2-33n+5}{3n^2}. $$
Step 6 · Form the upper sum $\overline{S}$
Shifting indices to the opposite corner and using $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$, $\sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6}$:
$$ \overline{S}=\frac{2}{n^2}\left[2n^2+3n(n+1)+\frac{5(n+1)(2n+1)}{6}\right]=\frac{40n^2+33n+5}{3n^2}. $$
Step 7 · Limit of the lower sum
$$ \lim_{n\to\infty}\underline{S}=\lim_{n\to\infty}\frac{40-\frac{33}{n}+\frac{5}{n^2}}{3}=\frac{40}{3}. $$
Step 8 · Limit of the upper sum
$$ \lim_{n\to\infty}\overline{S}=\lim_{n\to\infty}\frac{40+\frac{33}{n}+\frac{5}{n^2}}{3}=\frac{40}{3}. $$
Step 9 · Conclusion about integrability
Since $\lim_{n\to\infty}\underline{S}=\lim_{n\to\infty}\overline{S}=\frac{40}{3}$, the function $z(x,y)=x^2+y^2$ is Riemann integrable on $[1,2]\times[1,3]$ and
$$ \iint_{[1,2]\times[1,3]} (x^2+y^2)\,dA = \frac{40}{3}. $$
Step 10 · Verification using iterated integrals
$$ \int_1^2\int_1^3 (x^2+y^2)\,dy\,dx = \int_1^2\left(2x^2+\frac{26}{3}\right)dx = \frac{68}{3}-\frac{28}{3}=\frac{40}{3}. $$
Perfect match. ✓
Summary Table
| Quantity | Formula |
|---|---|
| Lower sum $\underline{S}$ | $\dfrac{40n^2-33n+5}{3n^2}$ |
| Upper sum $\overline{S}$ | $\dfrac{40n^2+33n+5}{3n^2}$ |
| $\lim_{n\to\infty}\underline S=\lim_{n\to\infty}\overline S$ | $\dfrac{40}{3}$ |
Additional Notes
- Since $z(x,y)=x^2+y^2$ is strictly increasing in both variables on the region, the minimum/maximum on each subrectangle sit at opposite corners — no calculus needed to find them.
- The common limit $\frac{40}{3}\approx 13.333$ is the volume under the paraboloid $z=x^2+y^2$ over $[1,2]\times[1,3]$.
Result
The solution is: