Volume of Solids of Revolution

Econometricia
Messages
33
Reaction score
0
Find the volume of the solid st,
1. y=cos x , y= 0 in [0,pi] ; Rotated around x=1




2. I am slightly confused, I see that the area will double around twice so I can just use the left half of the curve. I am just not sure how to do so.
 
Physics news on Phys.org
Are you sure that you copied this problem correctly. The line x = 1 does not divide the region between y = cosx and y = 0 into two equal halves, so rotating it around the line x = 1 makes a complicated volume of revolution. Could it be that you're supposed to rotate the region around the line y = 1?
 
Yes, not 2 equal halves but the rotation around the x=1 axis would cover the right hand side of the curve in rotation and double over so can't we ignore that piece?
 
No, I don't think so. The part of the region under y = cosx on [0, 1] would cover the part on [1, pi/2], so I suppose you could ignore that part. The part of the curve on [pi/2, pi] is below the x-axis. The part above [0, 1] is above the x-axis. This is a very unusual problem.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top