using the method of shells, set up, but don't evaluate the integral, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. Y=e^x, x=0, y=2, about y=1
for shells, we need to integrate on y, since the area is rotated about a horizontal line, and the shells have thickness dy
v = ∫[1,2] 2πrh dy
where r = y-1 and h = x = ln y
v = 2π∫[1,2] (y-1)lny dy
Just use integration by parts to evaluate it.
To set up the integral using the method of shells, we need to consider small cylindrical shells that are formed by rotating a vertical line segment around the specified axis.
Given curves:
y = e^x,
x = 0,
y = 2
Axis of rotation:
y = 1
First, we need to find the limits of integration for the variable used in the integral. In this case, we will integrate with respect to y.
To find the limits of integration for y, we need to determine the y-values at which the curves intersect the axis of rotation (y = 1).
For the curve y = e^x:
e^x = 1
x = ln(1) = 0
For the curve x = 0:
y = e^0 = 1
Therefore, the limits of integration for y are from y = 1 to y = 2.
Next, we consider a small vertical strip with width Δy (delta y) and height h. The height h can be calculated as the difference between the y-values of the curves, i.e., h = (2 - e^x).
The radius of the shell is the distance from the axis of rotation (y = 1) to the curve y = e^x, which is given by r = (e^x - 1).
The volume of the small cylindrical shell is approximately given by:
dV = 2πrh Δy,
where r represents the radius, h represents the height, and Δy represents the width of the shell.
Therefore, the integral for the volume of the solid obtained by rotating the region bounded by the curves y = e^x, x = 0, y = 2, about the axis y = 1 is:
V = ∫(1 to 2) 2π(e^x - 1) * (2 - e^x) dy
To set up the integral using the method of shells, we need to consider thin vertical shells that are formed by rotating a small horizontal strip of the region around the given axis.
The region bounded by the curves y = e^x, x = 0, and y = 2 is shown below:
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| x
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We are rotating this region around the horizontal line y = 1.
The height of each shell will be the difference between the y-coordinates of the upper and lower curves at a particular x-value. In this case, the upper curve is y = 2 and the lower curve is y = e^x.
The range of x-values for the region is 0 ≤ x ≤ ln(2).
The circumference of each shell is given by 2π(radius), where the radius is the distance from the axis of rotation (y = 1) to the x-axis.
So, the integral for the volume of the solid will be given by:
∫[0, ln(2)] 2π(1 - e^x) * circumference * dx
To find the circumference, we need to calculate the distance from the axis of rotation to the x-axis, which is 1.
Therefore, the integral for the volume of the solid obtained by rotating the region bounded by y = e^x, x = 0, y = 2, about the axis y = 1 is:
∫[0, ln(2)] 2π(1 - e^x) * 2π * dx.