The area of the surface generated by revolving the curve with $x ( t ) = 4 - 3 \cos t,$ about the x-axis is

A) $\pi \ln \left( \frac { \pi + 4 } { 4 } \right)$
B) $6 \pi \ln \left( \frac { \pi - 2 } { 4 } \right)$
C) $6 \pi \ln \left( \frac { \pi + 2 } { 4 } \right)$
D) $6 \pi \ln \left( \frac { \pi + 2 } { 2 } \right)$
E) $6 \pi \ln \left( \frac { \pi - 2 } { 2 } \right)$

Question

The area of the surface generated by revolving the curve with $x ( t ) = 4 - 3 \cos t,$ about the x-axis is

A) $\pi \ln \left( \frac { \pi + 4 } { 4 } \right)$
B) $6 \pi \ln \left( \frac { \pi - 2 } { 4 } \right)$
C) $6 \pi \ln \left( \frac { \pi + 2 } { 4 } \right)$
D) $6 \pi \ln \left( \frac { \pi + 2 } { 2 } \right)$
E) $6 \pi \ln \left( \frac { \pi - 2 } { 2 } \right)$

Quizplus · Accepted Answer

The surface area $S$ of a curve generated by revolving around the x-axis can be found using the formula $S = 2\pi \int y \sqrt{1 + (\frac{dy}{dx})^2} dx$, where $y$ is a function of $x$. In this case, the curve is parametric, so we need to express $y$ and $x$ in terms of the parameter $t$, and then find $\frac{dy}{dx}$. However, the given curve does not directly provide a $y(t)$, suggesting a potential misunderstanding in the question's formulation as it only provides $x(t)$ in terms of $t$. Without a $y(t)$ or further information, directly calculating the surface area as described is not feasible. The correct answer is inferred based on typical outcomes of such problems, not direct calculation from the provided information.

The Area of the Surface Generated by Revolving the Curve x(t)=4−3cos⁡t,x ( t ) = 4 - 3 \cos t,x(t)=4−3cost,

The Area of the Surface Generated by Revolving the Curve $x ( t ) = 4 - 3 \cos t,$