Expert: Scotto - 1/30/2010

Question
Hello Scotto, sorry for your time.

I have three volume problems, but my issue is ALWAYS the first step: Do I use disk, washer, or shell? Once someone manages to tell me which one, I can always work from there.

I would just basically like it if you could explain which method to use for each part, and more importantly, why. Solving out the entire problem isn't necessary, but you may if you see fit.


1. Let R be the region in the xy-plane between the graphs of e^x and e^(-x) from x=0 to x=2.

a. Find the volume of the solid generated when revolved around the x-axis.

b. Find the volume of the solid generated when revolved around the y-axis.





2. Let r be the region in the first quadrant enclosed by the y-axis and the graphs of y=sinx and y=cosx.

b. Find the volume of the solid generated when revolved around the x-axis.

(a and c were omitted, b/c they didn't involve volume ^_^)





3. Let R be the region enclosed by the graphs of y=e^x, y=(x-1)^2, and the line x=1.


b. Find the volume of the solid generated when revolved around the x-axis.

c. Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated when R is revolved around the y-axis.




Thanks for any help! :)

Answer
The answer is all three are used.  The outer one is a dsik and the inner one is a disk.
Taking the difference of the two results in a shell.  This has a thickness as well,
so when the volume is looked at from one end to the other, it is like find the volume
of a thick shell.

1. To revolve ex - e^-x around the x axis from 1 to 2, take r1 = e^x and r2 = e^-x.
The area of each disk would be πr², where r = r1 or r = 2.
Take the difference of two disks to get the area, so it is π(r1² - r2²).
Put in r1 and r2, square, which muliplies the exponet by 2, and integrate.

If it is to be revolved around the y axis, then take the ln() of both sides of both equations
to find what the equaion for y is.  If x goes from 0 to 2, y goes from 1 to e² on the first equation and 1 down to e^-² on the second equation.


2.  It is known that sin(x) = cos(x) at x = π/2.  Since the sin() is the lower section and cos()
is the upper secition, there are two integrals with respect to y.  The 1st would be on
π*arcsin²(y) from 0 to √2/2 and the 2nd would be on π*arccos²(y) from √2/2 to 1.


3. y = e^x is above (x-1)² in the interval in question.
To find the volume, that intersection point needs to be found.  Call it a.
To find the volume, find the volume from a up to 1 when both curves are rotated around the axis.
Use the disk method.  Subtract the 2nd from the 1st to get the volume.

c. The equation is ∫πR²(y) dy.