:browse confirm e Area Solution 11

SOLUTION 11: $\ \$ If $x = \displaystyle{ { y^5 \over 5 } + { 1 \over 12y^3 } } = (1/5)y^5+(1/12)y^{-3}$ for $1/2 \le y \le 1$, then $$\displaystyle{ { dx \over dy} = (1/5)(5)y^{4} + (1/12)(-3)y^{-4} }$$ $$= \displaystyle{ {y^4} - { 1 \over 4y^4 } }$$ so that $$ARC = \displaystyle{ \int_{1/2}^{1} \sqrt{ 1 + \Big({dx \over dy}\Big)^2 } \ dy }$$ $$= \displaystyle{ \int_{1/2}^{1} \sqrt{ 1 + \Big({y^4} - { 1 \over 4y^4 } \Big)^2 } \ dy }$$ $$= \displaystyle{ \int_{1/2}^{1} \sqrt{ 1 + \Big({y^8} - { 1 \over 2 } + { 1 \over 16y^8 } \Big) } \ dy }$$ $$= \displaystyle{ \int_{1/2}^{1} \sqrt{ {y^8} + { 1 \over 2 } + { 1 \over 16y^8 } } \ dy }$$ $$= \displaystyle{ \int_{1/2}^{1} \sqrt{ {y^8}{16y^8 \over 16y^8} + { 1 \over 2 }{ 8y^8 \over 8y^8 } + { 1 \over 16y^8 } } \ dy }$$ $$= \displaystyle{ \int_{1/2}^{1} \sqrt{ { { 16y^{16} + 8y^8 + 1 } \over 16y^8 } } \ dy }$$ $$= \displaystyle{ \int_{1/2}^{1} { \sqrt{ 16y^{16} + 8y^8 + 1 } \over \sqrt{ 16y^8 } } \ dy }$$ $$= \displaystyle{ \int_{1/2}^{1} { \sqrt{ (4y^{8} + 1)^2 } \over \sqrt{ (4y^4)^2} } \ dy }$$ $$= \displaystyle{ \int_{1/2}^{1} { { 4y^{8} + 1 } \over 4y^4 } \ dy }$$ $$= \displaystyle{ \int_{1/2}^{1} { \Big( { 4y^{8} \over 4y^4 } + { 1 \over 4y^4} \Big) } \ dy }$$ $$= \displaystyle{ \int_{1/2}^{1} { \Big( y^{4} + (1/4)y^{-4} \Big) } \ dy }$$ $$= \displaystyle{ { \Big( {y^{5} \over 5} + (1/4) {y^{-3} \over -3 } } \Big) \ \Big\vert_{1/2}^{1} }$$ $$= \displaystyle{ { \Big( {y^{5} \over 5} - { 1 \over 12y^3 } } \Big) \ \Big\vert_{1/2}^{1} }$$ $$= \displaystyle{ { \Big( { (1)^{5} \over 5 } - { 1 \over 12(1)^3 } \Big) } - { \Big( { (1/2)^{5} \over 5 } - { 1 \over 12(1/2)^3 } \Big) } }$$ $$= \displaystyle{ {1 \over 5} - { 1 \over 12 } - { 1 \over 160 } + { 2 \over 3 } }$$ $$= \displaystyle{ {96 \over 480} - { 40 \over 480 } - { 3 \over 480 } + { 320 \over 480 } }$$ $$= \displaystyle{ 373 \over 480 }$$