Q. 28 - 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 27–30, use four disks or washers based on the given rectangles to approximate the volume of the resulting solid.

Step-by-Step Solution

Answer

The required volume is $\frac{431}{16}\mathrm{\pi }$

Step 1. Given Information

The given figure is    

Step 2. Calculation

To determine the volume of solid of revolution, rotated around y-axis, express the curve as inverse function.

$$\begin{gathered} f\left(x\right)=\sqrt{x} \\ y=\sqrt{x} \\ {y}^2=x \\ x=y^2 \\ g\left(y\right)=y^2 \end{gathered}$$

For the x-interval of $[0,4]$, the corresponding interval of y-variable will be $[0,2]$

The width of each washer is determined as,

$$\begin{gathered} ∆y=\frac{b-a}{n} \\ =\frac{2-0}{4} \\ =\frac{1}{2} \end{gathered}$$

The external radius is $x=4$ and inner radius is given as $g\left(y_k\right)$. 

The starting value of disk is 0. It can be termed as $y_0$.

Determine these end points as follows,

$$\begin{gathered} y_k=y_0+k∆y \\ =0+k\left(\frac{1}{2}\right) \\ =\frac{k}{2} \end{gathered}$$

Step 3. Calculation

Total volume of solid is determined as follows, 

$$\begin{gathered} V=\sum _{k=1}^n\mathrm{\pi }\left({\left(\mathrm{R}\left(\mathrm{y}\right)\right)}^2-{\left(\mathrm{r}\left(\mathrm{y}\right)\right)}^2\right)∆\mathrm{y} \\ =\mathrm{\pi }\sum _{\mathrm{k}=1}^{\mathrm{n}}\left(4^2-{\left(\mathrm{g}\left({\mathrm{y}}_{\mathrm{k}}\right)\right)}^2\right)\left(\frac{1}{2}\right) \\ =\frac{\mathrm{\pi }}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\left({\left(4\right)}^2-{\left({{\mathrm{y}}_{\mathrm{k}}}^2\right)}^2\right) \\ =\frac{\mathrm{\pi }}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\left(16-{\left(\frac{\mathrm{k}}{2}\right)}^4\right) \\ =\frac{\mathrm{\pi }}{2}\left(\left(16-{\left(\frac{1}{2}\right)}^4\right)+\left(16-{\left(\frac{2}{2}\right)}^4\right)+\left(16-{\left(\frac{3}{2}\right)}^4\right)+\left(16-{\left(\frac{4}{2}\right)}^4\right)\right) \\ =\frac{\mathrm{\pi }}{2}\left(\frac{255}{16}+15+\frac{175}{16}+12\right) \\ =\frac{431}{16}\mathrm{\pi } \end{gathered}$$