Torricelli's Trumpet?

f(x)= 1/x around the x axis. Find the volume of Torricellis Trumpet when X [1,10].


I dont know where to begin for would prefer someone to help point me in the right direction?





Is it Pi (Integral from 1 to 10) 1/x? take the integral of 1/x (being (2/3) X) and substitue? = Pi [(2/3) X] (1 to 10)





2/3 pi - 20/3 pi = -6 pi? This is all guess work because i'm not too sure of myself with finding volume.

Torricelli's Trumpet?
It's pi r^2, so:





pi((1/x)^2)dx





pi*dx / (x^2)





[-pi / x](1 to 10)


-pi/10 + pi





volume = 9pi/10
Reply:Take a "typical rectangle"and rotate it to form a disc.


It has radius f(x) and height dx


So its volume is πr²h = πf(x)² dx.


Now sum these up from 1 to 10,


so your volume is


π ∫ (1..10) dx /x² = -π* 1/x(1..10) = 9π/10.


BTW the integral of 1/x is NOT 2/3 x it is log|x|.
Reply:when the function is revolved around the x-axis it creates a circle at any given point x whose radius is f(x). The area of this circle is f(x), and so the area of the circle is Pi*r^2 or Pi*f(x)^2. If you integrate that from x=1 to x=10, you have the integral(Pi*x^-2) from 1 to 10 which equals





-Pi/x evaluated from 1 to 10





=Pi/10 + Pi --%26gt; Pi-Pi/10 = 0.9 Pi