Question

(a)To determine the difficulty to use slicing to find the volume, \(V\) of solid \(S\).

(b)To sketch the typical approximating shell.

(c)To find the circumference, height and volume using the method of shell.

Short answer

The circumference, height and volume of shell as follows, \(C = 2\pi x,\;\;\;H = \sin \left( {{x^2}} \right)\), and \(V = 2\pi \).

Step-by-step solution

Given

The function is \(y = \sin \left( {{x^2}} \right)\)

The region lies between 0 to \(\sqrt \pi \).

The Concept ofvolume by the use of the method of shell

The volume by the use of the method of shell is\(V = \int_a^b 2 \pi x[f(x)]dx\)

Evaluate the volume

Show the equation as below:

\(y = \sin \left( {{x^2}} \right)\) ….. (1)

Plot a graph for the equation \(y = \sin \left( {{x^2}} \right)\)by the use of the calculation as follows:

Calculate \(y\) value by the use ofEquation (1).

Substitute 0 for \(x\) in Equation (1).

\(dy = \sin \left( {{0^2}} \right){\rm{ }} = 0\)

Hence, the co-ordinate of(x, y) is \((0,0)\).

Calculate \(y\) value by the use ofEquation (1).

Substitute \(\sqrt \pi \) for \(x\) in Equation (1).

\(\begin{array}{c}dy = \sin {(\sqrt \pi )^2}\\ = \sin \pi \\ = 0\end{array}\)

Hence, the co-ordinate of(x, y) is \((\sqrt \pi ,0)\).

Draw the region as shown in Figure

Draw the shell

Refer Figure 2

The radius of shell is \(x\).

Calculate the circumference of shell:

\(C = 2\pi r\) ………. (2)

Substitute \(x\) for \(r\) in Equation (2).

\(C = 2\pi x\)

Calculate the height of shell:

\(H = y\) ..……. (3)

Substitute \(\sin \left( {{x^2}} \right)\) for \(y\) in Equation (3).

\(H = \sin \left( {{x^2}} \right)\)

Calculate the volume by the use ofthe method ofshell:

\(V = \int_a^b 2 \pi x[f(x)]dx\) ..……. (4)

Substitute 0 for \(a,\sqrt \pi \) for \(b\), and \(\sin \left( {{x^2}} \right)\) for \([f(x)]\) in Equation (4).

\[V = \int_0^{\sqrt \pi } 2 \pi x\left[ {\sin \left( {{x^2}} \right)} \right]dx\]

\[V = 2\pi \int_0^{\sqrt \pi } x \sin \left( {{x^2}} \right)dx\] ..……. (5)

Consider \(u = {x^2}\) ..……. (6)

Differentiate both sides of the Equation (6).

\(du = 2xdx\)

Calculate the lower limit value of \(u\)by the use ofEquation (6).

Substitute 0 for \(x\) in Equation (6).

\(du{\rm{ }} = {0^2} = 0\)

Calculate the upper limit value of \(u\)by the use ofEquation (6).

Substitute \(\sqrt \pi b\) for \(x\) in Equation (6).

\(\begin{array}{c}du = {(\sqrt \pi )^2}\\ = \pi \end{array}\)

Apply lower and upper limits for \(u\) in Equation (5).

Substitute \(u\) for \({x^2}\) and \(du\) for xdx in Equation (5).

.\(\begin{array}{c}V = \int_0^\pi \pi \sin udu\\ = \pi \int_0^\pi {\sin } udu\end{array}\)

Integrate Equation (7).

\(\begin{array}{c}V = \pi [ - \cos u]_0^\pi \\ = \pi [ - \cos (\pi ) - ( - \cos 0)]\\ = \pi [1 + 1]\\ = 2\pi \end{array}\)

Therefore, the circumference, height and volume of shell as follows, \[\underline {C = 2\pi x} ,\;\;H = \sin {(x)^2}\], and \(\underline {V = 2\pi } \).

Shell method is preferable to slicing.