Question

How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y = \dfrac{1}{{{x^4}}},y = 0,x = 1,x = 4$ revolved about the $x = - 4$ ?

Answer 1

Hint : Start by using the formula $\int\limits_a^b {2\pi rhdx}$ where $a,b$ are the x-bounds. Substitute the values in place of the terms to make the equation easier to solve. Then we will differentiate the term. Now we will substitute these terms in the original expression and integrate.

Complete step by step solution:
We will start by substituting terms in the formula for the shell method which is given by
$\int\limits_a^b {2\pi rhdx}$ . Here, the values of $a,b$ are $1,4$ . So, $a = 1$ and $b = 4$ .
Also, $r$ is the distance from a certain x-value in the interval $[1,4]$ and the axis of rotation, which is $x = - 4$ .
$r = x - ( - 4) \\\ r = x + 4 \;$
$h$ is the height of the cylinder at a certain x-value in the interval $[1,4]$ which is $\dfrac{1}{{{x^4}}} - 0 = \dfrac{1}{{{x^4}}}$
This is because $\dfrac{1}{{{x^4}}}$ is always greater than zero and $h$ must be positive.
Now if we substitute all the terms, we get
$V = \int\limits_1^4 {\left( {2\pi (x + 4)\left( {\dfrac{1}{{{x^4}}}} \right)} \right)dx}$
$V = 2\pi \int\limits_1^4 {\left( {(x + 4)\left( {\dfrac{1}{{{x^4}}}} \right)} \right)dx}$
$V = 2\pi \int\limits_1^4 {\left( {\dfrac{{x + 4}}{{{x^4}}}} \right)dx}$
$V = 2\pi \int\limits_1^4 {\left( {\dfrac{1}{{{x^3}}} + \dfrac{4}{{{x^4}}}} \right)dx}$
On integration we get,
$V = 2\pi \left[ { - \dfrac{1}{{2{x^2}}} - \dfrac{4}{{3{x^3}}}} \right]_1^4$
Simplifying we get,
V = 2\pi \left[ {\left\\{ { - \dfrac{1}{{2 \times {4^2}}} - \dfrac{4}{{3 \times {4^3}}}} \right\\} - \left\\{ { - \dfrac{1}{{2 \times {1^2}}} - \dfrac{4}{{3 \times {1^3}}}} \right\\}} \right] \\\ \\\ \Rightarrow V = 2\pi \left[ {\left\\{ { - \dfrac{1}{{32}} - \dfrac{1}{{48}}} \right\\} - \left\\{ { - \dfrac{1}{2} - \dfrac{4}{3}} \right\\}} \right] \\\ \\\ On\,simplification\,we\,get, \\\ \\\ \Rightarrow V = 2\pi \left[ {\left\\{ {\dfrac{{ - 3 - 2}}{{96}}} \right\\} - \left\\{ {\dfrac{{ - 3 - 8}}{6}} \right\\}} \right] \\\ \\\ \Rightarrow V = 2\pi \left[ {\dfrac{{ - 5}}{{96}} - \left\\{ {\dfrac{{ - 11}}{6}} \right\\}} \right] \\\ \\\ \Rightarrow V = 2\pi \left[ {\dfrac{{ - 5 + 176}}{{96}}} \right] \\\ \\\ \Rightarrow V = 2\pi \left[ {\dfrac{{171}}{{96}}} \right] \\\ \\\ \Rightarrow V = 2\pi \left[ {\dfrac{{57}}{{32}}} \right] \\\ \\\ And\,Hence\,we\,\,have \\\ \\\ \Rightarrow V = \dfrac{{57\pi }}{{16}} \\\
Therefore required volume is $\dfrac{{57\pi }}{{16}}$ cubic unit.
So, the correct answer is “ $\dfrac{{57\pi }}{{16}}$ cubic unit”.

Note : A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behaviour of the original system under diverse conditions. The power rule allows us to find the indefinite integrals of a variety of functions like polynomials, functions involving roots, and even some rational functions.