Use the method of disks or washers, or the method of cylindrical shells to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the indicated
Axis.Sketch the region and a representative rectangle. $y = 2 \sqrt { x } , y = x - 3 , y = 0$, the $x$-axis
A) $135 \pi$
B) $90 \pi$
C) $45 \pi$
D) $180 \pi$

Question

Use the method of disks or washers, or the method of cylindrical shells to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the indicated
Axis.Sketch the region and a representative rectangle. $y = 2 \sqrt { x } , y = x - 3 , y = 0$, the $x$-axis
A) $135 \pi$ 
B) $90 \pi$ 
C) $45 \pi$ 
D) $180 \pi$

Quizplus · Accepted Answer

First, we need to sketch the region:

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The region is bounded by three curves: $y=2\sqrt{x}$, $y=x-3$, and $y=0$. We will use the method of cylindrical shells to find the volume of the solid obtained by revolving this region about the $x$-axis.

Consider an infinitesimally thin horizontal shell formed by revolving a vertical strip of width $dx$ about the $x$-axis. The strip is located at a distance $x$ from the $y$-axis and its height varies from $0$ (on the $x$-axis) to $\min\{2\sqrt{x},x-3\}$, which is the height of the topmost point of the strip.

Using the formula for the volume of a cylindrical shell, we have:

$$
dV = 2\pi x \cdot \left(\min\{2\sqrt{x},x-3\}ight) dx
$$

To find the total volume, we integrate from $x=0$ to $x=9$, where the curves $y=2\sqrt{x}$ and $y=x-3$ intersect:

$$
V = \int_0^9 2\pi x \cdot \left(\min\{2\sqrt{x},x-3\}ight) dx
$$

This integral can be evaluated by splitting the integral at $x=4$ and using the fact that $\min\{2\sqrt{x},x-3\}=2\sqrt{x}$ for $0\leq x<4$ and $\min\{2\sqrt{x},x-3\}=x-3$ for $4\leq x\leq 9$:

$$
\begin{aligned}
V &= \int_0^4 2\pi x\cdot 2\sqrt{x} dx + \int_4^9 2\pi x\cdot (x-3) dx \
&= 4\pi\int_0^4 x^{3/2} dx + 2\pi\int_4^9 (x^2-3x) dx \
&= \frac{32}{5}\pi + \frac{125}{3}\pi - \frac{54}{2}\pi \
&= 18\pi + \frac{32}{5}\pi \
&= \frac{90}{5}\pi \
&= \boxed{18\pi}
\end{aligned}
$$

Therefore, the answer is (B) $18\pi$.

Use the Method of Disks or Washers, or the Method y=2x,y=x−3,y=0y = 2 \sqrt { x } , y = x - 3 , y = 0y=2x​,y=x−3,y=0

Use the Method of Disks or Washers, or the Method $y = 2 \sqrt { x } , y = x - 3 , y = 0$