Volume using Cylindrical Shells: y = e^(-2x), y = 0, x = 0,

[skatru]

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis.

y = e[supgttlfra]-2x[/supgttlfra]
y=0
x=0
x=1

What I know:
V = circumference * height* thickness
$\displaystyle V = 2\pi \int_0^1 x y dx$
How do I integrate e to the -x squared?

I'm thinking substitution but I'm not sure if that is the right thing to do.

 

[galactus]

Re: Volume using Cylindrical Shells

You're on the right track. Just put $\displaystyle e^{-2x}$ in place of y and integrate

 

[skatru]

Re: Volume using Cylindrical Shells

I'm not sure how to find the integral of e[sup:1n780gt1]-2x[/sup:1n780gt1]

\(\displaystyle \int_\)e[sup:1n780gt1]x[/sup:1n780gt1] dx = e[sup:1n780gt1]x[/sup:1n780gt1]

what happens when e[sup:1n780gt1]-2x[/sup:1n780gt1]?

 

[tkhunny]

Re: Volume using Cylindrical Shells

Some general rules.

$\displaystyle \frac{d}{dx}e^{ax} = ae^{ax}$


$\displaystyle \int e^{ax} dx = \frac{e^{ax}}{a} + C$

You should see the relationship between the two.

 

[skeeter]

Re: Volume using Cylindrical Shells

$\displaystyle \int_0^1 xe^{-2x} \, dx$ has to be done using integration by parts.

 

[skatru]

Re: Volume using Cylindrical Shells

I've found this formula and when I use it... I don't get the right answer

$\displaystyle \int_b^c xe^{ax} \, dx = \frac{1}{a^{2}}(ax-1)e^{ax}$


$\displaystyle 2\pi\int_0^1 xe^{-2x} \, dx =2\pi( \frac{1}{-2^{2}}(-2x-1)e^{-2x})$


$\displaystyle 2\pi((\frac{1}{4}(-2-1)e^{-2}) - (\frac{1}{4}(0-1)e^{0}))$



$\displaystyle 2\pi((\frac{-3}{4}e^{-2}) - (\frac{-1}{4})$

but the answer in my book is
$\displaystyle \pi(1-\frac{1}{e})$

 

[skeeter]

Re: Volume using Cylindrical Shells

the book answer is incorrect.

your solution simplifies to ...

$\displaystyle V = \frac{\pi(e^2 - 3)}{2e^2}$

which is correct.

 

[galactus]

Re: Volume using Cylindrical Shells

It's something how many erroneous answers are in textbooks these days.