The area of the surface generated by revolving the curve $x ( t ) = \cos t , \quad y ( t ) = 2 + \sin t$ with $t \in [ 0,2 \pi ]$ about the x-axis is

A) $4 \pi ^ { 2 }$
B) $8 \pi ^ { 2 }$
C) $12 \pi ^ { 2 }$
D) $16 \pi ^ { 2 }$
E) $2 \pi ^ { 2 }$

Question

The area of the surface generated by revolving the curve $x ( t ) = \cos t , \quad y ( t ) = 2 + \sin t$ with $t \in [ 0,2 \pi ]$ about the x-axis is

A) $4 \pi ^ { 2 }$
B) $8 \pi ^ { 2 }$
C) $12 \pi ^ { 2 }$
D) $16 \pi ^ { 2 }$
E) $2 \pi ^ { 2 }$

Quizplus · Accepted Answer

The surface area $S$ generated by revolving a curve around the x-axis is given by the formula $S = 2\pi \int_{a}^{b} y(t) \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. For the given curve $x(t) = \cos t$ and $y(t) = 2 + \sin t$, we have $\frac{dx}{dt} = -\sin t$ and $\frac{dy}{dt} = \cos t$. Substituting these into the formula gives $S = 2\pi \int_{0}^{2\pi} (2 + \sin t) \sqrt{(-\sin t)^2 + (\cos t)^2} dt = 2\pi \int_{0}^{2\pi} (2 + \sin t) dt$, which simplifies to $8\pi^2$, making option B correct.

The Area of the Surface Generated by Revolving the Curve x(t)=cos⁡t,y(t)=2+sin⁡tx ( t ) = \cos t , \quad y ( t ) = 2 + \sin tx(t)=cost,y(t)=2+sint

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