Answer:
The volume is:
[tex]\displaystyle\frac{37\pi}{10} [/tex]
Step-by-step explanation:
See the sketch of the region in the attached graph.
We set the integral using washer method:
[tex]\displaystyle\int_a^b\pi r^2dx[/tex]
Notice here the radius of the washer is the difference of the given curves:
[tex]x-\sqrt{x}[/tex]
So the integral becomes:
[tex]\displaystyle\int_1^4\pi(x-\sqrt{x})^2dx[/tex]
We solve it:
Factor [tex]\pi[/tex] out and distribute the exponent (you can use FOIL):
[tex]\displaystyle\pi\int_1^4x^2-2x\sqrt{x}+x\,dx[/tex]
Notice: [tex]x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}[/tex]
So the integral becomes:
[tex]\displaystyle\pi\int_1^4x^2-2x^{3/2}+x\,dx[/tex]
Then using the basic rule to evaluate the integral:
[tex]\displaystyle\pi\left[\frac{x^3}{3}-\frac{2x^{5/2}}{5/2}+\frac{x^2}{2}\right|_1^4[/tex]
Simplifying a bit:
[tex]\displaystyle\pi\left[\frac{x^3}{3}-\frac{4x^{5/2}}{5}+\frac{x^2}{2}\right|_1^4[/tex]
Then plugging the limits of the integral:
[tex]\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(4)^{5/2}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right][/tex]
Taking the root (rational exponents):
[tex]\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(2)^{5}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right][/tex]
Then doing those arithmetic computations we get:
[tex]\displaystyle\frac{37\pi}{10}[/tex]
