Find the volume of the described solid.
-The solid lies between planes perpendicular to the $x$-axis at $x = \pi / 6$ to $x = \pi / 2$. The cross sections perpendicular to the $x$-axis are circular disks with diameters running from the curve $y = \cot x$ to the curve $y = \csc x$.
A) $\frac { ( \sqrt { 3 } - 1 ) \pi } { 2 } - \frac { \pi ^ { 2 } } { 12 }$
B) $( \sqrt { 3 } - 1 ) \pi + \frac { \pi ^ { 2 } } { 6 }$
C) $( 2 \sqrt { 3 } - 2 ) \pi - \frac { \pi ^ { 2 } } { 3 }$
D) $\frac { ( \sqrt { 3 } + 1 ) \pi } { 2 } - \frac { \pi ^ { 2 } } { 6 }$

Question

Quizplus · Accepted Answer

The volume of the solid can be calculated using the formula for the volume of a solid of revolution with circular cross-sections, $V = \int_{a}^{b} A(x) dx$, where $A(x)$ is the area of the cross-section perpendicular to the x-axis at a point $x$, and $a$ and $b$ are the limits of integration. The diameter of each disk is given by the difference between $\csc x$ and $\cot x$, so the radius is $\frac{\csc x - \cot x}{2}$. The area of each disk is then $A(x) = \pi \left(\frac{\csc x - \cot x}{2}\right)^2$. Integrating this expression from $x = \pi / 6$ to $x = \pi / 2$ gives the volume of the solid. The correct calculation leads to the answer in choice A.

Find the Volume of the Described Solid xxx -Axis At x=π/6x = \pi / 6x=π/6

Find the Volume of the Described Solid $x$ -Axis At $x = \pi / 6$