How to integrate a function to find the volume:

• Feb 1st 2010, 05:44 AM
appleseed
How to integrate a function to find the volume:
Hello!

I've been pulling my hair out over this question, could you help me please?

the region between the curve y= (2x-1)^(-1) and the lines x=1 and x=b (where b>1) is rotated through 360 about the x axis

a) find, in terms of b, the volume of the solid generated.
b)show that, for large values of b, this volume is approximately pi/2

Ok so this what I've been doing for a)
http://img519.imageshack.us/img519/5567/calculus.png

while the actual answer for a is

so as you can see I am quite far off, I'd really appreciate it if you could explain how to get the right answer. You see I have either moved in the complete wrong direction or i just dont know how to get from the (ln) to the next level

Thank you!
• Feb 1st 2010, 06:06 AM
tonio
Quote:

Originally Posted by appleseed
Hello!

I've been pulling my hair out over this question, could you help me please?

the region between the curve y= (2x-1)^(-1) and the lines x=1 and x=b (where b>1) is rotated through 360 about the x axis

a) find, in terms of b, the volume of the solid generated.
b)show that, for large values of b, this volume is approximately pi/2

Ok so this what I've been doing for a)
http://img519.imageshack.us/img519/5567/calculus.png

while the actual answer for a is

so as you can see I am quite far off, I'd really appreciate it if you could explain how to get the right answer. You see I have either moved in the complete wrong direction or i just dont know how to get from the (ln) to the next level

Thank you!

$\int \frac{1}{2x-1}\,dx=\frac{1}{2}\ln|2x-1|+C\,,\,\,C=$ a constant. Where did you get your value for the integral from??

And the answer you gave is wrong, it should be $\frac{\pi}{2}\ln(2b-1)$ and this does approach $\infty$ for large values of b so perhaps you wrote down incorrectly your function

Tonio
• Feb 1st 2010, 06:22 AM
HallsofIvy
I see what you did wrong. You are trying to use the "chain rule" in reverse:
If y= f(u(x)) then dy/dx= (df/du)(du/dx) and you are trying to say that
$\int f(u(x))dx= \frac{\int f(u)du}{u'(x)}$
That doesn't work as you can see by now differentiating on both sides.

The "chain rule in reverse" is "integration by substitution" but you have to have u' already in the integral if it is not a constant.