Make a sketch.
I also believe there is a left boundary of x= 0, or else the shape would be open-ended to the left.
Do it in two parts:
1. consider the whole solid generated when y = e^x is rotated about y = -1
2. "hollow-out" the cylinder generated when y = 1 is rotated about y = -1
the solid:
radius = e^x - (-1) = e^x + 1
volume = π ʃ (e^x + 1)^2 dx from 0 to 2
= π ʃ (e^2x + 2 e^x + 1)dx from 0 to 2
= π[ (1/2)e^(2x) + 2e^x + x] from 0 to 2
= π( e^4/2 + 2e^2 + 2 - (1/2 + 2 + 0) ]
= .....
from that subtract the cylinder with radius 2, height of 2
= π(4)(2) = 8π
what is the integral to find the volume of the solid that is formed when the region bounded by the graphs of y = e^x, x = 2, and y = 1 is revolved around the line y = -1.
I got pi[integral from 0 to 2] e^2x-1^2 dx.
I'm not sure about the 1^2 though because that function is in terms of y I think. Also, the other answer choices all had 2^2 in place of 1^2 so I guess I've started to doubt myself. Is correct answer my answer? Or does it have 2^2 instead?
2 answers
The corners of the region are (0,1), (1,1), and (1,e)
You found the volume rotated around the x-axis (y=0)
Using washers of thickness dx, we have
v = ∫[0,2] π(R^2-r^2) dx
where R=e^x+1 and r=2
v = ∫[0,2] π((e^x+1)^2 - 2^2) dx = π/2 (e^4+4e^2-17)
Using shells of thickness dy,
v = ∫[1,e^2] 2πrh dy
where r=y+1 and h=2-lny
v = ∫[1,e^2] 2π(y+1)(2-lny) dy = π/2 (e^4+4e^2-17)
You found the volume rotated around the x-axis (y=0)
Using washers of thickness dx, we have
v = ∫[0,2] π(R^2-r^2) dx
where R=e^x+1 and r=2
v = ∫[0,2] π((e^x+1)^2 - 2^2) dx = π/2 (e^4+4e^2-17)
Using shells of thickness dy,
v = ∫[1,e^2] 2πrh dy
where r=y+1 and h=2-lny
v = ∫[1,e^2] 2π(y+1)(2-lny) dy = π/2 (e^4+4e^2-17)
2 answers