Cylindrical shell method to find volume

Apr 2010
65
2
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?


 
Jun 2009
696
170
United States
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.
 
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