Answer
$${\text{Area}} = \frac{{44}}{3}$$
Work Step by Step
$$\eqalign{
& g\left( y \right) = 4{y^2} - {y^3},{\text{ }}1 \leqslant y \leqslant 3 \cr
& g\left( 1 \right) = 3{\text{ and }}g\left( 3 \right) = 9,{\text{ }} \cr
& g\left( y \right){\text{ is continuous and there are no negatives on the interval }} \\
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {g\left( {{c_i}} \right)} \Delta y,{\text{ }}\Delta y = \frac{{3 - 1}}{n} = \frac{2}{n} \cr
& {c_i} = a + i\Delta y \to 1 + \frac{{2i}}{n} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {4{{\left( {1 + \frac{{2i}}{n}} \right)}^2} - {{\left( {1 + \frac{{2i}}{n}} \right)}^3}} \right]} \left( {\frac{2}{n}} \right) \cr
& {\text{Expanding }}4{\left( {1 + \frac{{2i}}{n}} \right)^2} - {\left( {1 + \frac{{2i}}{n}} \right)^3}{\text{ by hand and replacing}} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( { - \frac{{8{i^3}}}{{{n^3}}} + \frac{{4{i^2}}}{{{n^2}}} + \frac{{10i}}{n} + 3} \right)} \left( {\frac{2}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( { - \frac{{16{i^3}}}{{{n^4}}} + \frac{{8{i^2}}}{{{n^3}}} + \frac{{20i}}{{{n^2}}} + \frac{6}{n}} \right)} \cr
& {\text{Area}} = - \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{16{i^3}}}{{{n^4}}}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{8{i^2}}}{{{n^3}}}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{20i}}{{{n^2}}}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{6}{n}} \cr
& {\text{Using the Summation Formulas from THEOREM 4}}{\text{.2}} \cr
& {\text{and simplifyng by hand or a CAS}} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{4}{{{n^2}}} + \frac{8}{n} + 4} \right) + \mathop {\lim }\limits_{n \to \infty } \left( {\frac{4}{{3{n^2}}} + \frac{4}{n} + \frac{8}{3}} \right) \cr
& + \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{10}}{n} + 10} \right) + 6 \cr
& {\text{Evaluate the limit when }}n \to \infty \cr
& {\text{Area}} = - \left( {0 + 0 + 4} \right) + \left( {0 + 0 + \frac{8}{3}} \right) + \left( {0 + 10} \right) + 6 \cr
& {\text{Area}} = \frac{{44}}{3} \cr} $$
