Solution 3: Here is a carefully labeled sketch of the graph with a radius $r$ marked together with $x$ on the $x$-axis.
Thus the total Area of this Surface of Revolution is
$$ Surface \ Area = 2 \pi \int_{0}^{1} (radius) \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx $$
$$ = 2 \pi \int_{0}^{1} ((4/3)x^{3/4}) \sqrt{ 1 + ((4/3)(3/4)x^{-1/4})^2 } \ dx $$
$$ = 2 \pi {4 \over 3} \int_{0}^{1} x^{3/4} \ \sqrt{ 1 + x^{-1/2} } \ dx $$
$$ = {8 \pi \over 3} \int_{0}^{1} x^{3/4} \ \sqrt{ 1 + {1 \over x^{1/2} } } \ dx $$
$$ = {8 \pi \over 3} \int_{0}^{1} x^{3/4} \ \sqrt{ {x^{1/2} \over x^{1/2}}+ {1 \over x^{1/2}} } \ dx $$
$$ = {8 \pi \over 3} \int_{0}^{1} x^{3/4} \ \sqrt{ x^{1/2} + 1 \over x^{1/2} } \ dx $$
$$ = {8 \pi \over 3} \int_{0}^{1} x^{3/4} \ { \sqrt{ x^{1/2} +1 } \over \sqrt{ x^{1/2} } } \ dx $$
$$ = {8 \pi \over 3} \int_{0}^{1} x^{3/4} \ { \sqrt{ x^{1/2} +1 } \over x^{1/4} } \ dx $$
$$ = {8 \pi \over 3} \int_{0}^{1} x^{1/2} \ \sqrt{ x^{1/2} +1 } \ dx $$
$\Big($ Let's do a $u$-substitution. Let $ u= x^{1/2}+1 \ $ so that $ \ x^{1/2}= u-1 \ $ and $ \
du = (1/2)x^{-1/2} dx = {1 \over 2 x^{1/2} } dx = {1 \over 2 (u-1) } dx \ \ \longrightarrow \ \ dx=2(u-1) du.\ \ $ Since $ \ x:0 \rightarrow 1 \ $ and $ \ u= x^{1/2} +1 \ $, it follows that $ \ u:1 \rightarrow 2. \ $ Now make the substitutions. $\Big)$
$$ = {8 \pi \over 3} \int_{1}^{2} (u-1) \ \sqrt{ u } \cdot \ 2(u-1)du $$
$$ = {8 \pi \over 3} (2) \int_{1}^{2} (u-1)^2 \ u^{1/2} \ du $$
$$ = {16 \pi \over 3} \int_{1}^{2} (u^2-2u+1) \ u^{1/2} \ du $$
$$ = {16 \pi \over 3} \int_{1}^{2} (u^{5/2}-2u^{3/2}+ u^{1/2} ) \ du $$
$$ = {16 \pi \over 3} \Big( \ ( {2 \over 7} u^{7/2}-{4 \over 5} u^{5/2} + {2 \over 3} u^{3/2} ) \Big|_{1}^{2} \ \Big) $$
$$ = {16 \pi \over 3} \Big( \ \Big( {2 \over 7} (2)^{7/2}-{4 \over 5} (2)^{5/2} + {2 \over 3} (2)^{3/2} \Big)
- \Big( {2 \over 7} (1)^{7/2}-{4 \over 5} (1)^{5/2} + {2 \over 3} (1)^{3/2} \Big) \ \Big) $$
$$ = {16 \pi \over 3} \Big( \ \Big( {2^{9/2} \over 7} -{2^{9/2} \over 5} + {2^{5/2} \over 3} \Big)
- \Big( {2 \over 7}-{4 \over 5} + {2 \over 3} \Big) \ \Big) $$
$$ = {16 \pi \over 3} \Big( \ \Big( 5{2^{9/2} \over 35} -7{2^{9/2} \over 35} + {2^{5/2} \over 3} \Big)
- \Big( {30 \over 105}-{84 \over 105} + {70 \over 105} \Big) \ \Big) $$
$$ = {16 \pi \over 3} \Big( \ \Big( {2^{5/2} \over 3} - 2{2^{9/2} \over 35} \Big)
- \Big( {16 \over 105} \Big) \ \Big) $$
$$ = {16 \pi \over 3} \Big( \ {2^{5/2} \over 3} - {2^{11/2} \over 35}
- {16 \over 105} \ \Big) $$
$$ = {2^4 \pi \over 3} \Big( \ {2^{5/2} \over 3} - {2^{11/2} \over 35}
- {2^4 \over 105} \ \Big) $$
$$ = \pi \Big( \ {2^{13/2} \over 9} - {2^{19/2} \over 105}
- {256 \over 315} \ \Big) $$
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