Answer
(a) ${S_{2,3}} = 240$
(b) ${S_{2,3}} = 510$
$\mathop \smallint \limits_{y = 1}^4 \mathop \smallint \limits_{x = 2}^6 {x^2}y{\rm{d}}x{\rm{d}}y = 520$
Work Step by Step
We have the double integral: $\mathop \smallint \limits_{y = 1}^4 \mathop \smallint \limits_{x = 2}^6 {x^2}y{\rm{d}}x{\rm{d}}y$.
Since we use the regular partition to compute the Riemann sum ${S_{2,3}}$, we have
$\Delta x = \frac{{6 - 2}}{2} = 2$, ${\ \ \ \ \ }$ $\Delta y = \frac{{4 - 1}}{3} = 1$
(a) Use sample points: lower-left vertex
The Riemann sum ${S_{2,3}}$ to estimate $\mathop \smallint \limits_{y = 1}^4 \mathop \smallint \limits_{x = 2}^6 {x^2}y{\rm{d}}x{\rm{d}}y$ is given by
${S_{2,3}} = \mathop \sum \limits_{i = 1}^2 \mathop \sum \limits_{j = 1}^3 f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j} = 2\mathop \sum \limits_{i = 1}^2 \mathop \sum \limits_{j = 1}^3 \left( {{P_{ij}}} \right)$
${S_{2,3}} = 2\left( {f\left( {2,1} \right) + f\left( {4,1} \right) + f\left( {2,2} \right) + f\left( {4,2} \right) + f\left( {2,3} \right) + f\left( {4,3} \right)} \right)$
$ = 2\left( {4 + 16 + 8 + 32 + 12 + 48} \right)$
${S_{2,3}} = 240$
(b) Use sample points: midpoint of rectangle
The Riemann sum ${S_{2,3}}$ to estimate $\mathop \smallint \limits_{y = 1}^4 \mathop \smallint \limits_{x = 2}^6 {x^2}y{\rm{d}}x{\rm{d}}y$ is given by
${S_{2,3}} = \mathop \sum \limits_{i = 1}^2 \mathop \sum \limits_{j = 1}^3 f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j} = 2\mathop \sum \limits_{i = 1}^2 \mathop \sum \limits_{j = 1}^3 f\left( {{P_{ij}}} \right)$
${S_{2,3}} = 2\left( {f\left( {3,\frac{3}{2}} \right) + f\left( {5,\frac{3}{2}} \right) + f\left( {3,\frac{5}{2}} \right) + f\left( {5,\frac{5}{2}} \right) + f\left( {3,\frac{7}{2}} \right) + f\left( {5,\frac{7}{2}} \right)} \right)$
${S_{2,3}} = 2\left( {\frac{{27}}{2} + \frac{{75}}{2} + \frac{{45}}{2} + \frac{{125}}{2} + \frac{{63}}{2} + \frac{{175}}{2}} \right)$
${S_{2,3}} = 510$
Now, we evaluate the exact value of the double integral:
$\mathop \smallint \limits_{y = 1}^4 \mathop \smallint \limits_{x = 2}^6 {x^2}y{\rm{d}}x{\rm{d}}y = \left( {\mathop \smallint \limits_{y = 1}^4 y{\rm{d}}y} \right)\left( {\mathop \smallint \limits_{x = 2}^6 {x^2}{\rm{d}}x} \right)$
$ = \frac{1}{6}\left( {{y^2}|_1^4} \right)\left( {{x^3}|_2^6} \right) = \frac{1}{6}\left( {15} \right)\left( {216 - 8} \right) = 520$
So, $\mathop \smallint \limits_{y = 1}^4 \mathop \smallint \limits_{x = 2}^6 {x^2}y{\rm{d}}x{\rm{d}}y = 520$.
