Find the exact area of the surface obtained by rotating about the \( x \)-axis the curve: \( x=\frac{1}{3}\left(y^{2}+2\right)^{\frac{3}{2}}, 1 \leq y \leq 2 \).

\( \frac { \sqrt[ 3 ] { ( ( x _ { 0 } + \Delta x ) - 1 ) ^ { 2 } + \sqrt[ 3 ] { ( x _ { 0 } - 1 ) ^ { 2 } } } } { \Delta x \cdot ( \sqrt[ 3 ] { x _ { 0 } - 1 } + \sqrt[ 3 ] { x _ { 0 } - 1 } ) } \)

\( \int _{}^{} d \tan (x \)

\( \frac { \sqrt[ 3 ] { ( ( x _ { 0 } + \Delta x ) - 1 ) ^ { 2 } + \sqrt[ 3 ] { ( x _ { 0 } - 1 ) ^ { 2 } } } } { \Delta x \cdot ( \sqrt[ 3 ] { x _ { 0 } - 1 } + \sqrt[ 3 ] { x _ { 0 } - 1 } ) } \)

\( \int \int _ { s } f ( x , y , z ) d s = \int _ { 0 } ^ { \pi } \int _ { 0 } ^ { 2 \pi } f ( s ( \phi , \theta ) ) \| \frac { \partial s } { \partial \phi } \times \frac { \partial s } { \partial \theta } \| d \theta d \phi \)

The enclosed area of the graph of the function \( Y=\left[x^{2}\right] \) and \( x \)-ares \( x=-2 \) to \( x=2 \) which is? \( 10-4(\sqrt{2}+\sqrt{2}) \) \( 10-5(\sqrt{5}+\sqrt{2}) \) \( 10-2(\sqrt{3}+\sqrt{2}) \) \( 10-8(\sqrt{2}+\sqrt{3}) \)