The area of the surface generated by revolving the curve $x ( t ) = 8 \sin t , y = 2 t - \sin ( 2 t )$ with $t \in \left[ 0 , \frac { \pi } { 2 } \right]$ about the y-axis is

A) $\frac { 32 \pi } { 3 }$
B) $\frac { 64 \pi } { 3 }$
C) $\frac { 128 \pi } { 3 }$
D) $\frac { 256 \pi } { 3 }$
E) $\frac { 512 \pi } { 3 }$

Question

The area of the surface generated by revolving the curve $x ( t ) = 8 \sin t , y = 2 t - \sin ( 2 t )$ with $t \in \left[ 0 , \frac { \pi } { 2 } \right]$ about the y-axis is

A) $\frac { 32 \pi } { 3 }$
B) $\frac { 64 \pi } { 3 }$
C) $\frac { 128 \pi } { 3 }$
D) $\frac { 256 \pi } { 3 }$
E) $\frac { 512 \pi } { 3 }$

Quizplus · Accepted Answer

The Answer of The area of the surface generated by revolving the curve $x ( t ) = 8 \sin t , y = 2 t - \sin ( 2 t )$ with $t \in \left[ 0 , \frac { \pi } { 2 } \right]$ about the y-axis is

A) $\frac { 32 \pi } { 3 }$
B) $\frac { 64 \pi } { 3 }$
C) $\frac { 128 \pi } { 3 }$
D) $\frac { 256 \pi } { 3 }$
E) $\frac { 512 \pi } { 3 }$

The Area of the Surface Generated by Revolving the Curve x(t)=8sin⁡t,y=2t−sin⁡(2t)x ( t ) = 8 \sin t , y = 2 t - \sin ( 2 t )x(t)=8sint,y=2t−sin(2t)

The Area of the Surface Generated by Revolving the Curve $x ( t ) = 8 \sin t , y = 2 t - \sin ( 2 t )$