The Area of the Surface Generated by Revolving the Curve $x ( t ) = 1 - t ^ { 2 } , y = 3 + 2 t$

Multiple Choice

The area of the surface generated by revolving the curve $x ( t ) = 1 - t ^ { 2 } , y = 3 + 2 t$ with $t \in [ 0,1 ]$ about the x-axis is

A) $4 \pi ( \sqrt { 2 } + 1 ) \left( 1 + \sinh ^ { - 1 } 1 \right)$
B) $\pi ( \sqrt { 2 } + 1 ) \left( 1 + \sinh ^ { - 1 } 1 \right)$
C) $2 \pi ( \sqrt { 2 } - 1 ) \left( 1 + \sinh ^ { - 1 } 1 \right)$
D) $4 \pi ( \sqrt { 2 } - 1 ) \left( 1 + \sinh ^ { - 1 } 1 \right)$
E) $8 \pi ( \sqrt { 2 } - 1 ) \left( 1 + \sinh ^ { - 1 } 1 \right)$