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

A) $\frac { 52 \pi } { 3 }$
B) $\frac { 26 \pi } { 3 }$
C) $\frac { 13 \pi } { 3 }$
D) $\frac { 11 \pi } { 5 }$
E) $\frac { 8 \pi } { 3 }$

Question

The area of the surface generated by revolving the curve

x ( t ) = t + \sqrt { 2 } , \quad y = \frac { t ^ { 2 } } { 2 } + \sqrt { t }

with

t \in [ - \sqrt { 2 } , \sqrt { 2 } ]

about the y-axis is

A) $\frac { 52 \pi } { 3 }$
B) $\frac { 26 \pi } { 3 }$
C) $\frac { 13 \pi } { 3 }$
D) $\frac { 11 \pi } { 5 }$
E) $\frac { 8 \pi } { 3 }$

Quizplus · Accepted Answer

The surface area $S$ generated by revolving a curve $x(t), y(t)$ about the y-axis from $t=a$ to $t=b$ is given by the formula $S = 2\pi \int_{a}^{b} x(t) \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. Substituting $x(t) = t + \sqrt{2}$ and $y(t) = \frac{t^2}{2} + \sqrt{t}$, and integrating from $-\sqrt{2}$ to $\sqrt{2}$, yields the area $\frac{52\pi}{3}$.

The Area of the Surface Generated by Revolving the Curve x(t)=t+2,y=t22+tx ( t ) = t + \sqrt { 2 } , \quad y = \frac { t ^ { 2 } } { 2 } + \sqrt { t }x(t)=t+2​,y=2t2​+t​

The Area of the Surface Generated by Revolving the Curve $x ( t ) = t + \sqrt { 2 } , \quad y = \frac { t ^ { 2 } } { 2 } + \sqrt { t }$