Find the area of the surface generated by revolving the curve about the y-axis. $$y = 169 - x ^ { 2 } , 0 \leq x \leq 13$$
A)$\frac { 114,245 ^ { \frac { 3 } { 2 } } - 1 } { 12 } \pi$
B) $\frac { 677 ^ { \frac { 3 } { 2 } } - 1 } { 12 } \pi$
C)$\frac { 114,245 ^ { \frac { 3 } { 2 } } - 1 } { 6 } \pi$
D)$\frac { 114,245 ^ { \frac { 3 } { 2 } } - 1 } { 3 } \pi$
E)$\frac { 677 ^ { \frac { 3 } { 2 } } - 1 } { 6 } \pi$

Question

Quizplus · Accepted Answer

The area of the surface generated by revolving the curve $y = 169 - x^2$ about the y-axis from $x = 0$ to $x = 13$ can be found using the formula for the surface area of a solid of revolution: $A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} dx$. Here, $f(x) = \sqrt{169 - x^2}$, and its derivative $f'(x) = -\frac{x}{\sqrt{169 - x^2}}$. Substituting and integrating within the given limits, the correct answer involves the term $677^{\frac{3}{2}}$, and since the question revolves around the y-axis, the division by 6 (not 12 or 3) is the correct adjustment for the formula used, leading to option E.

Find the Area of the Surface Generated by Revolving the Curve