Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis.
-$$y = 3 \csc x , y = 3 \sqrt { 2 } , \frac { \pi } { 4 } \leq x \leq \frac { 3 \pi } { 4 }$$
A) $3 \pi ^ { 2 } - 9 \pi$
B) $9 \pi ^ { 2 } + 18 \pi$
C) $9 \pi ^ { 2 } - 18 \pi$
D) $\pi ^ { 2 } + 6 \pi$

Question

Quizplus · Accepted Answer

The volume of the solid generated by revolving the region about the x-axis can be found using the method of disks or washers. The given region is bounded by $y = 3 \csc x$, $y = 3 \sqrt{2}$, and $x = \frac{\pi}{4}$ to $x = \frac{3\pi}{4}$. The volume $V$ is given by the integral $$V = \pi \int_{a}^{b} [R(x)]^2 dx$$, where $R(x)$ is the distance from the curve to the axis of rotation (in this case, the x-axis), and $a$ and $b$ are the limits of integration. Here, $R(x) = 3 \csc x$ and $R(x) = 3\sqrt{2}$ are the outer and inner radii, respectively, for the region of interest. The volume generated by revolving $y = 3 \csc x$ about the x-axis from $\frac{\pi}{4}$ to $\frac{3\pi}{4}$ minus the volume generated by revolving $y = 3\sqrt{2}$ over the same interval gives the total volume. The correct calculation involves setting up and evaluating these integrals properly, taking into account the geometry of the situation. The correct answer reflects the volume after performing these calculations, which involves integrating the square of the function (or functions) that define the outer and inner boundaries of the solid, and subtracting the inner volume from the outer volume if necessary.

Find the Volume of the Solid Generated by Revolving the Region