What is the volume of the solid produced by revolving $f (x) = \sqrt{1 + x^{2}}$ $f(x)=sqrt(1+x^2)$ around the x-axis?

Question

4108 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-17T02:10:24

$V = π {[x + \frac{1}{3} x^{3}]}_{a}^{3}$ $V = pi[x+1/3x^3]_a^b$

The volume of revolution about $O x$ $Ox$ is given by;

$V = \int_{a}^{b} π y^{2} d x$ $V = int_a^b piy^2dx$

so with $f (x) = \sqrt{1 + x^{2}}$ $f(x)=sqrt(1+x^2)$ , then:

$V = \int_{a}^{b} π {(\sqrt{1 + x^{2}})}^{2} d x$ $V = int_a^b pi(sqrt(1+x^2))^2dx$
$:. V = piint_a^b 1+x^2 dx$
$:. V = pi[x+1/3x^3]_a^b$

As you have not specified the $x$ -coordinate bounds this is as far as we can proceed

Answer 2 · 2016-11-17T02:25:05

Nov 17, 2016

See answer below:
enter image source here

What is the volume of the solid produced by revolving f(x)=sqrt(1+x^2)f(x)=√1+x2f(x)=sqrt(1+x^2) around the x-axis?