:browse confirm e 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 $$ARC = \displaystyle{ \int_{1}^{2} \sqrt{ 1 + \Big({dx \over dy}\Big)^2 } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} \sqrt{ 1 + \Big({y^3 \over 2} - { 1 \over 2y^3 } \Big)^2 } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} \sqrt{ 1 + \Big({y^6 \over 4} - { 1 \over 2 } + { 1 \over 4y^6 } \Big) } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} \sqrt{ {y^6 \over 4} + { 1 \over 2 } + { 1 \over 4y^6 } } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} \sqrt{ {y^6 \over 4}{y^6 \over y^6} + { 1 \over 2 }{ 2y^6 \over 2y^6 } + { 1 \over 4y^6 } } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} \sqrt{ { { y^{12} + 2y^6 + 1 } \over 4y^6} } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} { \sqrt{ y^{12} + 2y^6 + 1 } \over \sqrt{ 4y^6} } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} { \sqrt{ (y^{6} + 1)^2 } \over \sqrt{ (2y^3)^2} } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} { { y^{6} + 1 } \over 2y^3 } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} { \Big( { y^{6} \over 2y^3 } + { 1 \over 2y^3} \Big) } \ dy }$$ $$= \displaystyle{ \int_{1}^{2} { \Big( (1/2)y^{3} + (1/2)y^{-3} \Big) } \ dy }$$ $$= \displaystyle{ { \Big( (1/2){y^{4} \over 4} + (1/2) {y^{-2} \over -2 } } \Big) \ \Big\vert_{1}^{2} }$$ $$= \displaystyle{ { \Big( {y^{4} \over 8} - { 1 \over 4y^2 } } \Big) \ \Big\vert_{1}^{2} }$$ $$= \displaystyle{ { \Big( { (2)^{4} \over 8 } - { 1 \over 4(2)^2 } \Big) } - { \Big( { (1)^{4} \over 8 } - { 1 \over 4(1)^2 } \Big) } }$$ $$= \displaystyle{ {16 \over 8} - { 1 \over 16 } - { 1 \over 8 } + { 1 \over 4 } }$$ $$= \displaystyle{ {32 \over 16} - { 1 \over 16 } - { 2 \over 16 } + { 4 \over 16 } }$$ $$= \displaystyle{ 33 \over 16 }$$