Thread: Cylindrical shell method to find volume

1. 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 $\displaystyle y=\sqrt{x}$.

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

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

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

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

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

But my friend got the result :

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

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

2. 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 $\displaystyle y=\sqrt{x}$.

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

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

$\displaystyle $$\displaystyle v=\int^a_{0} (2\pi*{x}^{3/2})dx$$\displaystyle$
$\displaystyle $$\displaystyle v=\frac{4\pi*{x}^{5/2}}5$$\displaystyle$

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

But my friend got the result :

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

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 $\displaystyle \int_0^22\pi x\sqrt{x}dx$ into a calculator, you obtain 14.217. Now plug 2 into the equation you obtained, and you'll get the same result.