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?
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
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)
To find the volume of the solid formed by revolving the region bounded by the graphs of y = e^x, x = 2, and y = 1 around the line y = -1, you can use the method of cylindrical shells. Here's how you can set up the integral correctly:
1. First, sketch the region bounded by the graphs. The region is a rectangle with the base on the x-axis and the top curve being the graph of y = e^x.
2. To find the height of the cylindrical shell, you need to consider the distance from the line y = -1 to the curve y = e^x. You can calculate this difference as h = (e^x) - (-1) = e^x + 1.
3. The circumference of the cylindrical shell is given by 2*pi*r, where r is the distance from the y-axis to the curve y = e^x. This distance can be calculated as r = x.
4. Finally, integrate the product of the height and circumference of the cylindrical shell over the interval [0, 2]. The integral setup would be:
V = ∫[from 0 to 2] 2π(x)(e^x + 1) dx
So, the correct integral expression for finding the volume of the solid is:
V = ∫[from 0 to 2] 2πx(e^x + 1) dx
Note that there is no need for (1^2) or (2^2) terms in the integral expression.