Q. 25 - Calculus

Question

Consider the region between $f\left(x\right) = \sqrt{x}$ and the x-axis on$[0, 4]$. For each line of rotation given in Exercises 25–28, use four shells based on the given rectangles to approximate the volume of the resulting solid.

Around the x-axis 

Step-by-Step Solution

Answer

The solid produced has a volume of $25.92.$

Step 1: Given information

The graph is 

Step 2: Find the volume of first shell.

The rectangles are drawn parallel to the x-axis as the solid is rotated around the x-axis, resulting in shells with bases on the y-axis. The interval $[0,2]$ is partitioned into four subintervals on the y-axis. $[0,0.5],[0.5,1],[1,1.5]$and $[1.5,2]$, Each subinterval has a width of $0.5$ that is $\Delta y=0.5$.

Because the region in the diagram is bounded by $[0,4]$ and the curve is $f\left(x\right)=\sqrt{x}$, the height of the shells is $4-f^{-1}\left(x\right)=4-y^2$.

The radius of the first shell on the interval $[0,0.5]$ is $0.5$, and the height of the shell is determined at the midpoint of $[0,0.5]$, which is $h=4-(0.25{)}^2$, resulting in $h=3.9375$.

As a result, the initial shell's volume is defined as,

$\begin{array}{rcl}2\pi rh\Delta y& =& 2\pi (0.25)(3.9375)(0.5)\\ & =& 3.093\end{array}$

The initial shell's volume is $3.093.$

Step 3: Find the volume of second shell

The average radius of the second shell is $[0.5,1]$, which is $0.75$, and its height computed at the midpoint is $h=4-(0.75{)}^2$, which is $h=3.4375$.

As a result, the second shell's volume is defined as,

$\begin{array}{rcl}2\pi rh\Delta y& =& 2\pi (0.75)(3.4375)(0.5)\\ & =& 8.095\end{array}$

The Volume of the second shell is $8.095$.

Step 4: Find the volume of third shell

The average radius of the third shell is $[1,1.5]$, which is 1.25, and its height computed at the midpoint is $h=4-(1.25{)}^2$, which is $h=2.4375$.

As a result, the third shell's volume is

$\begin{array}{rcl}2\pi rh\Delta y& =& 2\pi (1.25)(2.4375)(0.5)\\ & =& 9.572\end{array}$

The third shell volume is $9.572$.

Step 5: Find the volume of fourth shell

The average radius of the fourth shell is $[1.5,2]$, which is $1.75$, and its height computed at the midpoint is $h=4-(1.75{)}^2$, which is $h=0.9375$.

As a result, the fourth shell's volume is defined as,

$\begin{array}{rcl}2\pi rh\Delta y& =& 2\pi (1.75)(0.9375)(0.5)\\ & =& 5.154\end{array}$

The fourth shell has a volume of $5.154$.

To find the solid produced volume, add all the obtained volumes. 

$3.093+8.099+9.572+5.154\approx 25.92$