Cylindrical Shell Method

**ch2kb0x** · March 18th 2008, 08:40 AM

I need help asap. Finals in couple hours.

f(x) = SquareRoot(x^2 + 9) over [0,3]

I got 2pi integral from 0,3 and inside of integral i put x(SquareRoot(x^2 + 9)dx.

I know that I'm suppose to multiply it out but I dont know how to im suppose to multiply that because of the square root. help would be appreciated.

Sorry for not using the symbolic characters.

**galactus** · March 18th 2008, 09:38 AM

You didn't say which axis you have to revolve it around. y axis?.

**ch2kb0x** · March 18th 2008, 09:52 AM

yes, sorry. the y-axis.

**galactus** · March 18th 2008, 10:08 AM

In that event, you set up is correct.

$2{\pi}\int_{0}^{3}x\sqrt{x^{2}+9}dx$

Now, let $u=x^{2}+9, \;\ du=2xdx, \;\ \frac{1}{2}du=xdx$

Don't forget to change your limits of integration:

$(3)^{2}+9=18, \;\ (0)^{2}+9=9$

So, your new limits are 9 to 18 and you have:

${\pi}\int\sqrt{u}du$

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