what does y = e^.5 x look like?
at x = 0, y =e^0 = 1
at x = .1, y .= e^.05 = 1.05
at x = .2 y = 1.11
at x = ln(3) = 1.098 = 1.1
y = e^.5 ln (3) = e^ln (sqrt 3) = 1.73
well, y goes from 1 at x = 0
to
y = 1.73 at x =1.1l
so it looks like washers
but the sensible way might be to integrate
y = pi (e^.5x)^2dx from x = 0 to x
= 1.1
and
from that subtract the cylinder
pi (1)^2 dx from 0 to 1.1
The region enclosed by the graph e^(x/2), y=1, and x=ln(3) is revolved around the x-axis. Find the volume of the solid generated.
I don't understand if we have to use the washer method or the disk method for this one because when I drew it out on a graph it looked very confusing.
2 answers
As above, using discs of thickness dx,
v = ∫[0,ln3] π(R^2-r^2) dx
where R=y and r=1
v = ∫[0,ln3] π((e^(x/2)^2-1^2) dx
= ∫[0,ln3] π(e^x-1) dx
= π(2-ln3)
or, using shells of thickness dy,
v = ∫[1,√3] 2πrh dy
where r=y and h=ln3-x
v = ∫[1,√3] 2πy(ln3-2lny) dy
= π(2-ln3)
v = ∫[0,ln3] π(R^2-r^2) dx
where R=y and r=1
v = ∫[0,ln3] π((e^(x/2)^2-1^2) dx
= ∫[0,ln3] π(e^x-1) dx
= π(2-ln3)
or, using shells of thickness dy,
v = ∫[1,√3] 2πrh dy
where r=y and h=ln3-x
v = ∫[1,√3] 2πy(ln3-2lny) dy
= π(2-ln3)