Answer
1. Case $N=10$
${S_{10,10,10}} \simeq 0.561054$
2. Case $N=20$
${S_{20,20,20}} \simeq 0.572576$
3. Case $N=30$
${S_{30,30,30}} \simeq 0.57649$
$I \simeq 0.584441$
For $N=60$, ${S_{60,60,60}}$ approximates $I$ to two decimal places.
Work Step by Step
We have $f\left( {x,y,z} \right) = {{\rm{e}}^{{x^2} - y - z}}$.
1. Case $N=10$
The Riemann sum approximation becomes
${S_{10,10,10}} = \frac{1}{{{{10}^3}}}\mathop \sum \limits_{i = 1}^{10} \mathop \sum \limits_{j = 1}^{10} \mathop \sum \limits_{k = 1}^{10} f\left( {\frac{i}{{10}},\frac{j}{{10}},\frac{k}{{10}}} \right)$
Using a computer algebra system, we calculate ${S_{10,10,10}}$, the result is
${S_{10,10,10}} \simeq 0.561054$
2. Case $N=20$
${S_{20,20,20}} = \frac{1}{{{{20}^3}}}\mathop \sum \limits_{i = 1}^{20} \mathop \sum \limits_{j = 1}^{20} \mathop \sum \limits_{k = 1}^{20} f\left( {\frac{i}{{20}},\frac{j}{{20}},\frac{k}{{20}}} \right)$
${S_{20,20,20}} \simeq 0.572576$
3. Case $N=30$
${S_{30,30,30}} = \frac{1}{{{{30}^3}}}\mathop \sum \limits_{i = 1}^{30} \mathop \sum \limits_{j = 1}^{30} \mathop \sum \limits_{k = 1}^{30} f\left( {\frac{i}{{30}},\frac{j}{{30}},\frac{k}{{30}}} \right)$
${S_{30,30,30}} \simeq 0.57649$
Using a computer algebra system, we evaluate $I$:
$I = \mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 {{\rm{e}}^{{x^2} - y - z}}{\rm{d}}V = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{z = 0}^1 {{\rm{e}}^{{x^2} - y - z}}{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$I \simeq 0.584441$
To find an $N$ such that ${S_{N,N,N}}$ approximates $I$ to two decimal places, we increase the value of $N$ and found that for $N=60$, ${S_{60,60,60}} \simeq 0.580444$.
Thus, for $N=60$, ${S_{60,60,60}}$ approximates $I$ to two decimal places.
