Answer
(a) ${S_{4,4}} \simeq 0.947644$
(b)
${S_{10,10}} \simeq 0.946334$
${S_{50,50}} \simeq 0.946093$
${S_{100,100}} \simeq 0.946086$
Work Step by Step
(a) We have the double integral: $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \cos \left( {xy} \right){\rm{d}}x{\rm{d}}y$.
Since we use the regular partition to compute the Riemann sum ${S_{4,4}}$, we have
$\Delta x = \frac{{1 - 0}}{4} = \frac{1}{4}$, ${\ \ \ \ \ }$ $\Delta y = \frac{{1 - 0}}{4} = \frac{1}{4}$
Method: use midpoints as sample points.
The Riemann sum ${S_{4,4}}$ to estimate $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \cos \left( {xy} \right){\rm{d}}x{\rm{d}}y$ is given by
${S_{4,4}} = \mathop \sum \limits_{i = 1}^4 \mathop \sum \limits_{j = 1}^4 f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j} = \frac{1}{{16}}\mathop \sum \limits_{i = 1}^4 \mathop \sum \limits_{j = 1}^4 f\left( {{P_{ij}}} \right)$
$ = \frac{1}{{16}}(f\left( {\frac{1}{8},\frac{1}{8}} \right) + f\left( {\frac{3}{8},\frac{1}{8}} \right) + f\left( {\frac{5}{8},\frac{1}{8}} \right) + f\left( {\frac{7}{8},\frac{1}{8}} \right)$
${\ \ \ }$ $ + f\left( {\frac{1}{8},\frac{3}{8}} \right) + f\left( {\frac{3}{8},\frac{3}{8}} \right) + f\left( {\frac{5}{8},\frac{3}{8}} \right) + f\left( {\frac{7}{8},\frac{3}{8}} \right)$
${\ \ \ }$ $ + f\left( {\frac{1}{8},\frac{5}{8}} \right) + f\left( {\frac{3}{8},\frac{5}{8}} \right) + f\left( {\frac{5}{8},\frac{5}{8}} \right) + f\left( {\frac{7}{8},\frac{5}{8}} \right)$
${\ \ \ }$ $ + f\left( {\frac{1}{8},\frac{7}{8}} \right) + f\left( {\frac{3}{8},\frac{7}{8}} \right) + f\left( {\frac{5}{8},\frac{7}{8}} \right) + f\left( {\frac{7}{8},\frac{7}{8}} \right))$
$ = \frac{1}{{16}}(\cos \left( {\frac{1}{{64}}} \right) + \cos \left( {\frac{3}{{64}}} \right) + \cos \left( {\frac{5}{{64}}} \right) + \cos \left( {\frac{7}{{64}}} \right)$
${\ \ \ }$ $ + \cos \left( {\frac{3}{{64}}} \right) + \cos \left( {\frac{9}{{64}}} \right) + \cos \left( {\frac{{15}}{{64}}} \right) + \cos \left( {\frac{{21}}{{64}}} \right)$
${\ \ \ }$ $ + \cos \left( {\frac{5}{{64}}} \right) + \cos \left( {\frac{{15}}{{64}}} \right) + \cos \left( {\frac{{25}}{{64}}} \right) + \cos \left( {\frac{{35}}{{64}}} \right)$
${\ \ \ }$ $ + \cos \left( {\frac{7}{{64}}} \right) + \cos \left( {\frac{{21}}{{64}}} \right) + \cos \left( {\frac{{35}}{{64}}} \right) + \cos \left( {\frac{{49}}{{64}}} \right))$
${S_{4,4}} \simeq 0.947644$
(b)
1. Case $N=10$
$\Delta x = \frac{{1 - 0}}{{10}} = \frac{1}{{10}}$, ${\ \ \ \ \ }$ $\Delta y = \frac{{1 - 0}}{{10}} = \frac{1}{{10}}$
The Riemann sum ${S_{10,10}}$ to estimate $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \cos \left( {xy} \right){\rm{d}}x{\rm{d}}y$ is given by
${S_{10,10}} = \mathop \sum \limits_{i = 1}^{10} \mathop \sum \limits_{j = 1}^{10} f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j} = \frac{1}{{100}}\mathop \sum \limits_{i = 1}^{10} \mathop \sum \limits_{j = 1}^{10} f\left( {{P_{ij}}} \right)$
Using a computer algebra system, we obtain
${S_{10,10}} \simeq 0.946334$
2. Case $N=50$
$\Delta x = \frac{{1 - 0}}{{50}} = \frac{1}{{50}}$, ${\ \ \ \ \ }$ $\Delta y = \frac{{1 - 0}}{{50}} = \frac{1}{{50}}$
The Riemann sum ${S_{50,50}}$ to estimate $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \cos \left( {xy} \right){\rm{d}}x{\rm{d}}y$ is given by
${S_{50,50}} = \mathop \sum \limits_{i = 1}^{50} \mathop \sum \limits_{j = 1}^{50} f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j} = \frac{1}{{2500}}\mathop \sum \limits_{i = 1}^{50} \mathop \sum \limits_{j = 1}^{50} f\left( {{P_{ij}}} \right)$
Using a computer algebra system, we obtain
${S_{50,50}} \simeq 0.946093$
3. Case $N=100$
$\Delta x = \frac{{1 - 0}}{{100}} = \frac{1}{{100}}$, ${\ \ \ \ \ }$ $\Delta y = \frac{{1 - 0}}{{100}} = \frac{1}{{100}}$
The Riemann sum ${S_{100,100}}$ to estimate $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \cos \left( {xy} \right){\rm{d}}x{\rm{d}}y$ is given by
${S_{100,100}} = \mathop \sum \limits_{i = 1}^{100} \mathop \sum \limits_{j = 1}^{100} f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j} = \frac{1}{{10000}}\mathop \sum \limits_{i = 1}^{100} \mathop \sum \limits_{j = 1}^{100} f\left( {{P_{ij}}} \right)$
Using a computer algebra system, we obtain
${S_{100,100}} \simeq 0.946086$
