Solution 4: First solve the equation for $x$ getting $ x=y^{2/3} $. Here is a carefully labeled sketch of the graph with a radius $r$ marked together with $y$ on the $y$-axis.
Thus the total Area of this Surface of Revolution is
$$ Surface \ Area = 2 \pi \int_{1}^{8} (radius) \sqrt{ 1 + \Big({dx \over dy}\Big)^2 } \ dy $$
$$ = 2 \pi \int_{1}^{8} (y^{2/3}) \sqrt{ 1 + ((2/3)y^{-1/3})^2 } \ dy $$
$$ = 2 \pi \int_{1}^{8} y^{2/3} \ \sqrt{ 1 + (4/9)y^{-2/3} } \ dy $$
$$ = 2 \pi \int_{1}^{8} y^{2/3} \ \sqrt{ 1 + {4 \over 9y^{2/3} } } \ dy $$
$$ = 2 \pi \int_{1}^{8} y^{2/3} \ \sqrt{ {9y^{2/3}\over 9y^{2/3} } + {4 \over 9y^{2/3} } } \ dy $$
$$ = 2 \pi \int_{1}^{8} y^{2/3} \ \sqrt{ 9y^{2/3} +4 \over 9y^{2/3} } \ dy $$
$$ = 2 \pi \int_{1}^{8} y^{2/3} \ { \sqrt{ 9y^{2/3} +4 } \over \sqrt{ 9y^{2/3} } } \ dy $$
$$ = 2 \pi \int_{1}^{8} y^{2/3} \ { \sqrt{ 9y^{2/3} +4 } \over 3 y^{1/3} } \ dy $$
$$ = {2 \pi \over 3} \int_{1}^{8} y^{2/3} \ { \sqrt{ 9y^{2/3} +4 } \over y^{1/3} } \ dy $$
$\Big($ Let's do a $u$-substitution. Let $ u= 9y^{2/3} +4 \ $ so that $ \ y^{2/3}= {1 \over 9}(u-4) \ $ and $ \ du = 9(2/3)y^{-1/3} dy = {6 \over y^{1/3} } dy \ \ \longrightarrow \ \ {1 \over y^{1/3} } dy = {1 \over 6} du.\ \ $ Since $ \ y:1 \rightarrow 8 \ $ and $ \ u= 9y^{2/3} +4 \ $, it follows that $ \ u:13 \rightarrow 40. \ $ Now make the substitutions. $\Big)$
$$ = {2 \pi \over 3} \int_{13}^{40} ({1 \over 9}(u-4)) \cdot \sqrt{ u } \cdot {1 \over 6} du $$
$$ = {2 \pi \over 3} {1 \over 9} {1 \over 6} \int_{13}^{40} (u-4) \sqrt{ u } \ du $$
$$ = { \pi \over 81} \int_{13}^{40} (u^{3/2}-4u^{1/2}) \ du $$
$$ = { \pi \over 81} \Big( \ {2 \over 5}u^{5/2} - 4 \Big({2 \over 3}\Big)u^{3/2} \Big) \Big|_{13}^{40} $$
$$ = { \pi \over 81} \Big( \ \Big( {2 \over 5}(40)^{5/2} - {8 \over 3}(40)^{3/2} \Big)
- \Big( {2 \over 5}(13)^{5/2} - {8 \over 3}(13)^{3/2} \Big) \Big) $$
$$ = { \pi \over 81} \Big( \ \Big( {2 \over 5}(4 \cdot 10)^{5/2} - {8 \over 3}(4 \cdot 10)^{3/2} \Big)
- \Big( {2 \over 5}(13)^{5/2} - {8 \over 3}(13)^{3/2} \Big) \Big) $$
$$ = { \pi \over 81} \Big( \ \Big( {2 \over 5}(32)(10)^{5/2} - {8 \over 3}(8)( 10)^{3/2} \Big)
- \Big( {2 \over 5}(13)^{5/2} - {8 \over 3}(13)^{3/2} \Big) \Big) $$
$$ = { \pi \over 81} \Big( {64 \over 5}(10)^{5/2} - {64 \over 3} (10)^{3/2}
- {2 \over 5}(13)^{5/2} + {8 \over 3}(13)^{3/2} \Big) $$
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