Volume when rotating around the x-axis (1 Viewer)

[jexca]

Hi,
I've got this question:
The region below the curve y=e^x which is above the x-axis and between the lines x=0 and x=1 is rotated about the x-axis.
a) sketch the region (done)
b) Determine the volume of the resulting solid of revolution.


I can't rememeber how to figure out the volume? I know its pi times the integral of something.. can't work out what the something is
Thankyou!

 

[beentherdunthat]

Hi,
I've got this question:
The region below the curve y=e^x which is above the x-axis and between the lines x=0 and x=1 is rotated about the x-axis.
a) sketch the region (done)
b) Determine the volume of the resulting solid of revolution.


I can't rememeber how to figure out the volume? I know its pi times the integral of something.. can't work out what the something is
Thankyou!

Integral of y squared in terms of x
and multiply by pi...

Is that okay, or do you want me to elaborate ?

 

[jb_nc]

Volume = pi * the intergrand from a to b of y^2 dx on the closed interval [a, b] where b > a

 

[jb_nc]

http://en.wikipedia.org/wiki/Disk_integration

 

[zingerburger]

If it rotates around the x-axis, make y the subject and take the function and square it.

If it rotates around the y-axis, make x the subject and take the function and square it.

 

[jexca]

Thankyou

 

[Forbidden.]

If it rotates around the x-axis, make y the subject and take the function and square it.

If it rotates around the y-axis, make x the subject and take the function and square it.