# Volume of rotation (y-axis)

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• Feb 1st 2010, 04:20 PM
iheartmath
Volume of rotation (y-axis)
The function is y(x) = 7/(x^2 +5x +6) evaluated from x = 0 to x = 1. Obviously it cannot be written as x(y). I tried changing the bounds of integration (from x values to their corresponding y values on the function), but this didn't work.

How do you do this? Please explain this in steps. Thanks!
• Feb 1st 2010, 04:48 PM
Archie Meade
Quote:

Originally Posted by iheartmath
The function is y(x) = 7/(x^2 +5x +6) evaluated from x = 0 to x = 1. Obviously it cannot be written as x(y). I tried changing the bounds of integration (from x values to their corresponding y values on the function), but this didn't work.

How do you do this? Please explain this in steps. Thanks!

You can split the denominator into partial fractions using

$\displaystyle x^2+5x+6=(x+3)(x+2)$

$\displaystyle \frac{1}{(x+3)(x+2)}=\frac{A}{x+3}+\frac{B}{x+2}$

which gives

A+B=0
2A+3B=1

Simply solve for A and B.

Then you can develop the volume of revolution using shells.
• Feb 1st 2010, 05:51 PM
skeeter
Quote:

Originally Posted by iheartmath
The function is y(x) = 7/(x^2 +5x +6) evaluated from x = 0 to x = 1. Obviously it cannot be written as x(y). I tried changing the bounds of integration (from x values to their corresponding y values on the function), but this didn't work.

How do you do this? Please explain this in steps. Thanks!

rotating about the y-axis, using the method of cylindrical shells ...

$\displaystyle V = 2\pi \int_0^2 \frac{7x}{x^2+5x+6} \, dx$

partial fractions ...

$\displaystyle \frac{7x}{(x+3)(x+2)} = \frac{21}{x+3} - \frac{14}{x+2}$