Question

What will be the volume of the solid of revolution formed by rotating the finite region bounded by the graphs of $y = \sqrt x $ and $y = {x^3}$ about the X-axis?
A) $\dfrac{{5\pi }}{3}$
B) $\dfrac{{5\pi }}{{14}}$
C) $\dfrac{{5\pi }}{7}$
D) $\dfrac{{7\pi }}{5}$

Answer 1

Hint: Two continuous functions are given. We can find the volume of the solid bounded by the graphs by integration method. For that we have to find which function has greater value and which has smaller value in the interval.

Formula used:
If the functions \[f(x)\] and $g(x)$ are continuous and non-negative on the interval $[a,b]$ and $g(x) \leqslant f(x)$, volume of the solid bounded by the graphs of $f(x)$ and $g(x)$ about X-axis is given by $V = \pi \int\limits_a^b {({{[f(x)]}^2}} - {[g(x)]^2})dx$

Complete step-by-step answer:
Here we are given two functions.
Let $f(x) = \sqrt x = {x^{\dfrac{1}{2}}},g(x) = {x^3}$
Consider the interval $[0,1]$.
We can see that $g(x) \leqslant f(x)$ in $[0,1]$.
And clearly both $f,g$ are continuous and take non-negative values in the interval.
If the functions \[f(x)\] and $g(x)$ are continuous and non-negative on the interval $[a,b]$ and $g(x) \leqslant f(x)$, volume of the solid bounded by the graphs of $f(x)$ and $g(x)$ about X-axis is given by $V = \pi \int\limits_a^b {({{[f(x)]}^2}} - {[g(x)]^2})dx$
So we can use this equation to find the volume.
$ \Rightarrow V = \pi \int\limits_a^b {({{[f(x)]}^2}} - {[g(x)]^2})dx$
Substituting the values we get,
$ \Rightarrow V = \pi \int\limits_0^1 {({{[\sqrt x ]}^2}} - {[{x^3}]^2})dx$
Simplifying we get,
$ \Rightarrow V = \pi \int\limits_0^1 {(x} - {x^6})dx$
Integrating we get,
$ \Rightarrow V = \pi (\int\limits_0^1 x dx - \int\limits_0^1 {{x^6}} dx)$
$ \Rightarrow V = \pi ([\dfrac{{{x^2}}}{2}]_0^1 - [\dfrac{{{x^7}}}{7}]_0^1)$
Simplifying the above equation we get,
$V = \pi ([\dfrac{1}{2} - 0] - [\dfrac{1}{7} - 0]) = [\dfrac{1}{2} - \dfrac{1}{7}]\pi = \dfrac{{(7 - 2)\pi }}{{14}} = \dfrac{{5\pi }}{{14}}$
So, the required volume is $\dfrac{{5\pi }}{{14}}$.
$\therefore $ The answer is option B.

Additional information:
Among the two functions $f(x),g(x)$, if $g(x) \leqslant f(x)$, then $g(x)$ is called the inner radius and $f(x)$ is called the outer radius.

Note: Here since the revolution is done about X-axis we integrate with respect to $x$. If the axis of revolution was the Y-axis, we had to integrate with respect to $y$. Also it is important to satisfy the conditions mentioned in the result. That is, both the functions must be continuous, non-negative and one less than or equal to another in the interval.