Answer:
Step-by-step explanation:
Given that the region bounded by x axis, x=e^2 and y = lnx be rotated about x axis
The limits of x are [tex]1<x<e^2[/tex]
Volume generated = [tex]\int\limits^{e^2} _{1} \pi {y^2} \, dx \\\\=\int\limits^{e^2} _{1} \pi {(lnx)^2} \, dx \\\pi[x (lnx)^2 -\int\limits^{e^2} _{1} x(2lnx)dx]\\=2(e^1-1)\pi[/tex]
We will see that the volume is 46.403 cubic units.
The volume will be equal to 2*pi times the area of the region.
The area of the region is given by the integral of the function between the given interval, which is 0 < x < e^2
Then the area is given by:
[tex]A = \int\limits^{e^2}_0 {ln(x)} \, dx =( e^2*ln(e^2) - e^2 - 0*ln(0) + 0)\\\\\\A = e^2*ln(e^2) - e^2 = e^2 = 7.389[/tex]
To get the volume, we multiply that by 2*pi = 2*3.14, so we get:
V = (7.389)*2*3.14 = 46.403
So the volume is 46.403 cubic units.
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