# Cylindrical shell method to find volume

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• May 16th 2010, 09:57 PM
olski1
Cylindrical shell method to find volume
Hi I have been asked to find the volume generated by rotating about the y axis the area in the first quadrant lying over the x-iterval [0,a] and under the graph of $y=\sqrt{x}$.

the formula i used for this was the cylindircal shell method:
$
$
$v=\int^b_{a} (2\pi{x}*f{(x)})dx

$

So i evaluated the volume to be:
$
$
$v=\int^a_{0} (2\pi{x}*\sqrt{x})dx

$

$
$
$v=\int^a_{0} (2\pi*{x}^{3/2})dx

$

$
$
$v=\frac{4\pi*{x}^{5/2}}5

$

$
$
$v=\frac{4\pi*{a}^{5/2}}5

$

But my friend got the result :

$
\frac{\pi{a}^{5/2}}{5}
$

so where did i go wrong? Can i not use this method?

• May 16th 2010, 10:08 PM
adkinsjr
Quote:

Originally Posted by olski1
Hi I have been asked to find the volume generated by rotating about the y axis the area in the first quadrant lying over the x-iterval [0,a] and under the graph of $y=\sqrt{x}$.

the formula i used for this was the cylindircal shell method:
$
$
$v=\int^b_{a} (2\pi{x}*f{(x)})dx$ $

$

So i evaluated the volume to be:
$
$
$v=\int^a_{0} (2\pi{x}*\sqrt{x})dx$ $

$

$
$
$v=\int^a_{0} (2\pi*{x}^{3/2})dx$ $

$

$
$
$v=\frac{4\pi*{x}^{5/2}}5$ $

$

$
$
$v=\frac{4\pi*{a}^{5/2}}5$ $

$

But my friend got the result :

$
\frac{\pi{a}^{5/2}}{5}
" alt="
\frac{\pi{a}^{5/2}}{5}
" />

so where did i go wrong? Can i not use this method?

You're correct. You can test it by simply choosing value of "a". If you plug $\int_0^22\pi x\sqrt{x}dx$ into a calculator, you obtain 14.217. Now plug 2 into the equation you obtained, and you'll get the same result.