Find the area of the surface of revolution $\mathbf { r } ( u , v ) = 11 \sin u \cos v \mathbf { i } + 11 u \mathbf { j } + 11 \sin u \sin v \mathbf { k }$, where $0 \leq u \leq \pi$ and $0 \leq v \leq 2 \pi$.
A) $121 \pi [ \sqrt { 2 } + \ln | \sqrt { 2 } - 1 | ]$
B) $22 \pi [ \sqrt { 2 } + \ln | \sqrt { 2 } + 1 | ]$
C)$242 \pi \left[ \sqrt { 2 } + \ln \left| \frac { \sqrt { 2 } - 1 } { \sqrt { 2 } + 1 } \right| \right]$
D) $\pi [ \sqrt { 2 } - \ln | \sqrt { 2 } + 1 | ]$
E)$242 \pi \left[ \sqrt { 2 } + \ln \left| \frac { \sqrt { 2 } + 1 } { \sqrt { 2 } - 1 } \right| \right]$

Question

Quizplus · Accepted Answer

The area of the surface of revolution generated by revolving a curve around an axis can be calculated using the formula for the surface area of a parametric surface. The given parametric equations describe a surface in 3D space. The correct answer involves calculating the integral that represents the surface area, taking into account the given bounds for $u$ and $v$. The formula involves derivatives of the parametric equations with respect to $u$ and $v$, and the correct choice (E) reflects the correct application of this formula, including the integration over the specified bounds for $u$ and $v$. The presence of $242\pi$ in the answer indicates the integration and multiplication by the constants derived from the parametric equations and their derivatives. The expression inside the brackets is related to the specific form of the parametric equations and their derivatives, indicating the geometric properties of the surface being revolved.

Find the Area of the Surface of Revolution r(u,v)=11sin⁡ucos⁡vi+11uj+11sin⁡usin⁡vk\mathbf { r } ( u , v ) = 11 \sin u \cos v \mathbf { i } + 11 u \mathbf { j } + 11 \sin u \sin v \mathbf { k }r(u,v)=11sinucosvi+11uj+11sinusinvk

Find the Area of the Surface of Revolution $\mathbf { r } ( u , v ) = 11 \sin u \cos v \mathbf { i } + 11 u \mathbf { j } + 11 \sin u \sin v \mathbf { k }$