Torricelli's Trumpet?

Question

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.

thank you guys, I new I was wrong again just now because volume can not be negative...

I have finals in two weeks and I feel as if I have not been getting anything as of recent. I am going to be studying nonstop for the rest of these two weeks, sleeping and eating when I can.

1/x= x^-1, i was thinking of something else that involved a different intergration. I am just so messed up lol.

Answer 1

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

Answer 2

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 --> Pi-Pi/10 = 0.9 Pi

Answer 3

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|.