# Cylindrical shell method to find volume

#### 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:
$$\displaystyle$$
$$\displaystyle v=\int^b_{a} (2\pi{x}*f{(x)})dx$$

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?

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.

• olski1