Question

The curve y = e−x for 0  x  2 is rotated around the x-axis to produce a 3-dimensional solid whose volume is V =  2 (1−e−4). The centre of mass lies somewhere on the x-axis - where?

Answer 1

make sure all the parts posted correctly in the question ....

Answer 2

yea.. they are correct

Answer 3

y = e−x for 0 x 2

notationally this is unreadable ...

Answer 4

wait a sec ... will check it again!!

Answer 5

The curve y = e^−x for 0 <x< 2 is rotated around the x-axis to produce a 3-dimensional solid whose volume is V =pie/ 2 (1−e^−4). The centre of mass lies somewhere on the x-axis - where?

Answer 6

center of mass would most likely be situated when we have half the volume

Answer 7

id prolly integrate it from 0 to b; and equate that to half the volume that is stated to solve for b

Answer 8

\[\int_{a}^{n}f(x)~dx=\frac{V}{2}=F(n)-F(a)\]
\[\frac{V}{2}+F(a)=F(n)\]
and invert for n

Answer 9

there is also some sort of moment calculations that are a bit foggy ... since it might be hard to integrate e^(-2x) in general

Answer 10

i got this e^-x ?? :/

Answer 11

please solve it completely!!!

Answer 12

i was thinking e^(x^2) ... this is simpler

\[-\frac{1}{2}\pi e^{-2(n)}+\frac{1}{2}\pi e^{-2(0)}=\frac12*\frac{\pi}{2}(1-e^{-4})\]
\[-\pi e^{-2(n)}+\pi e^{-2(0)}=\frac{\pi}{2}(1-e^{-4})\]
\[-e^{-2(n)}+ 1=\frac{1}{2}(1-e^{-4})\]
\[e^{-2(n)}=1-\frac{1}{2}(1-e^{-4})\]
\[-2n=ln(1-\frac{1}{2}(1-e^{-4}))\]
\[n=-\frac12ln(1-\frac{1}{2}(1-e^{-4}))\]

Answer 13

chk the algebra ... but that is what i get too