Answer
Using a computer algebra system, we obtain:
1. for $N=25$: ${S_{50,25}} \simeq 14.6315$
2. for $N=50$: ${S_{100,50}} \simeq 14.4857$
3. for $N=100$: ${S_{200,100}} \simeq 14.4127$
Work Step by Step
The Riemann sum ${S_{2N,N}}$ to estimate $\mathop \smallint \limits_0^4 \mathop \smallint \limits_0^2 \ln \left( {1 + {x^2} + {y^2}} \right){\rm{d}}y{\rm{d}}x$ is given by
${S_{2N,N}} = \mathop \sum \limits_{i = 1}^{2N} \mathop \sum \limits_{j = 1}^N f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j}$,
where $f\left( {x,y} \right) = \ln \left( {1 + {x^2} + {y^2}} \right)$.
The domain is ${\cal R} = \left[ {0,4} \right] \times \left[ {0,2} \right]$. We evaluate ${S_{2N,N}}$ using upper right-hand vertices as sample points.
Case 1. For $N=25$
Using the regular partition, dimensions of the subrectangles are
$\Delta x = \frac{{4 - 0}}{{2\cdot25}} = \frac{2}{{25}}$, ${\ \ \ \ }$ $\Delta y = \frac{{2 - 0}}{{25}} = \frac{2}{{25}}$
The Riemann sum becomes
${S_{50,25}} = \mathop \sum \limits_{i = 1}^{50} \mathop \sum \limits_{j = 1}^{25} f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j}$
${S_{50,25}} = \frac{4}{{625}}\left( {f(\frac{2}{{25}},\frac{2}{{25}}\} + f(\frac{4}{{25}},\frac{2}{{25}}\} + \cdot\cdot\cdot + f\left( {\frac{{98}}{{25}},2} \right) + f\left( {4,2} \right)} \right)$
Using a computer algebra system, we obtain ${S_{50,25}} \simeq 14.6315$.
Case 2. For $N=50$
Using the regular partition, dimensions of the subrectangles are
$\Delta x = \frac{{4 - 0}}{{2\cdot50}} = \frac{1}{{25}}$, ${\ \ \ \ }$ $\Delta y = \frac{{2 - 0}}{{50}} = \frac{1}{{25}}$
The Riemann sum becomes
${S_{100,50}} = \mathop \sum \limits_{i = 1}^{100} \mathop \sum \limits_{j = 1}^{50} f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j}$
${S_{100,50}} = \frac{1}{{625}}\left( {f\left( {\frac{1}{{25}},\frac{1}{{25}}} \right) + f(\frac{2}{{25}},\frac{1}{{25}}\} + \cdot\cdot\cdot + f\left( {\frac{{99}}{{25}},2} \right) + f\left( {4,2} \right)} \right)$
Using a computer algebra system, we obtain ${S_{100,50}} \simeq 14.4857$.
Case 3. For $N=100$
Using the regular partition, dimensions of the subrectangles are
$\Delta x = \frac{{4 - 0}}{{2\cdot100}} = \frac{1}{{50}}$, ${\ \ \ \ }$ $\Delta y = \frac{{2 - 0}}{{100}} = \frac{1}{{50}}$
The Riemann sum becomes
${S_{200,100}} = \mathop \sum \limits_{i = 1}^{200} \mathop \sum \limits_{j = 1}^{100} f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j}$
${S_{200,100}} = \frac{1}{{2500}}\left( {f\left( {\frac{1}{{50}},\frac{1}{{50}}} \right) + f(\frac{1}{{25}},\frac{1}{{50}}\} + \cdot\cdot\cdot + f\left( {\frac{{199}}{{50}},2} \right) + f\left( {4,2} \right)} \right)$
Using a computer algebra system, we obtain ${S_{200,100}} \simeq 14.4127$.
