Solve the problem by integration.
-Find the volume generated by revolving the first-quadrant area bounded by $y = \frac { x } { ( x + 2 ) ^ { 2 } }$ and $x = 2$ about the X-axis.
A) $\frac { 1 } { 96 } \pi$
B) $\frac { 1 } { 48 } \pi$
C) $\frac { 1 } { 24 } \pi$
D) $\frac { 1 } { 12 } \pi$

Question

Quizplus · Accepted Answer

The volume of the solid generated by revolving the area bounded by $y = \frac{x}{(x+2)^2}$ and $x=2$ about the x-axis can be found using the disk method. The formula for the volume is $V = \pi \int_{a}^{b} [f(x)]^2 dx$, where $f(x) = \frac{x}{(x+2)^2}$ in this case, and the limits of integration are from $a = 0$ to $b = 2$. The integral becomes $V = \pi \int_{0}^{2} \left(\frac{x}{(x+2)^2}\right)^2 dx$. Simplifying and integrating yields $V = \frac{1}{48}\pi$, which corresponds to choice B.

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