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Area Solution 10
SOLUTION 10: $ \ \ $ If $ x = \displaystyle{ { y^4 \over 8 } + { 1 \over 4y^2 } } = (1/8)y^4+(1/4)y^{-2} $ for $ 1 \le y \le 2 $, then
$$ \displaystyle{ { dx \over dy} = (1/8)(4)y^{3} + (1/4)(-2)y^{-3} } $$
$$ = \displaystyle{ {y^3 \over 2} - { 1 \over 2y^3 } } $$
so that
$$ ARC = \displaystyle{ \int_{1}^{2} \sqrt{ 1 + \Big({dx \over dy}\Big)^2 } \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} \sqrt{ 1 + \Big({y^3 \over 2} - { 1 \over 2y^3 } \Big)^2 } \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} \sqrt{ 1 + \Big({y^6 \over 4} - { 1 \over 2 } + { 1 \over 4y^6 } \Big) } \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} \sqrt{ {y^6 \over 4} + { 1 \over 2 } + { 1 \over 4y^6 } } \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} \sqrt{ {y^6 \over 4}{y^6 \over y^6} + { 1 \over 2 }{ 2y^6 \over 2y^6 } + { 1 \over 4y^6 } } \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} \sqrt{ { { y^{12} + 2y^6 + 1 } \over 4y^6} } \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} { \sqrt{ y^{12} + 2y^6 + 1 } \over \sqrt{ 4y^6} } \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} { \sqrt{ (y^{6} + 1)^2 } \over \sqrt{ (2y^3)^2} } \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} { { y^{6} + 1 } \over 2y^3 } \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} { \Big( { y^{6} \over 2y^3 } + { 1 \over 2y^3} \Big) } \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} { \Big( (1/2)y^{3} + (1/2)y^{-3} \Big) } \ dy } $$
$$ = \displaystyle{ { \Big( (1/2){y^{4} \over 4} + (1/2) {y^{-2} \over -2 } } \Big) \ \Big\vert_{1}^{2} } $$
$$ = \displaystyle{ { \Big( {y^{4} \over 8} - { 1 \over 4y^2 } } \Big) \ \Big\vert_{1}^{2} } $$
$$ = \displaystyle{ { \Big( { (2)^{4} \over 8 } - { 1 \over 4(2)^2 } \Big) }
- { \Big( { (1)^{4} \over 8 } - { 1 \over 4(1)^2 } \Big) } } $$
$$ = \displaystyle{ {16 \over 8} - { 1 \over 16 } - { 1 \over 8 } + { 1 \over 4 } } $$
$$ = \displaystyle{ {32 \over 16} - { 1 \over 16 } - { 2 \over 16 } + { 4 \over 16 } } $$
$$ = \displaystyle{ 33 \over 16 } $$
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