Find the area of the surface generated by revolving the curve about the indicated axis.
-$y = \sqrt { x } , 3 / 2 \leq x \leq 9 / 2 ; x$-axis
A) $\left( \frac { 19 \sqrt { 19 } } { 6 } - \frac { 7 \sqrt { 7 } } { 6 } \right) \pi$
B) $\frac { 7 \sqrt { 7 } } { 2 } \pi$
C) $\left( \frac { 7 \sqrt { 7 } } { 2 } - \frac { \sqrt { 3 } } { 6 } \right) \pi$
D) $\frac { 7 \sqrt { 7 } } { 6 } \pi$

Question

Quizplus · Accepted Answer

The area of the surface generated by revolving the curve $y = \sqrt{x}$ about the x-axis from $x = \frac{3}{2}$ to $x = \frac{9}{2}$ can be found using the formula for the surface area of a solid of revolution: $A = 2\pi \int_{a}^{b} y \sqrt{1 + (dy/dx)^2} dx$. Calculating this integral with the given limits and function yields the answer in choice A.

Find the Area of the Surface Generated by Revolving the Curve