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?