# solid of revolution

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis.

$y= 27e^x$, $y= 27e^{-x}$, $x = 1$.

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Where are you having difficulty with this? Is this a homework problem? What have you tried? Have you sketched the region? – Matthew Conroy Feb 28 '11 at 7:52
I am randomly trying calculus problems from a book. I would appreciate if you can explain it step by step. it has been a while since i have done this kind of problem – user7591 Feb 28 '11 at 9:34

Consider the approximately rectangular vertical strip running from the curve $y=27e^{-x}$ to the curve $y=27e^x$ at a distance $x$ from the $y$-axis and of width $\delta x.$ The height of the strip is about $27(e^x-e^{-x})$ and so its area is approximately

$$27(e^x-e^{-x}) \delta x.$$

Now consider rotating this about the $y$-axis one complete revolution. The rectangle will sweep out a volume of approximately

$$2 \pi x \times 27(e^x-e^{-x}) \delta x.$$

You then need to sum up all these elements of volume from $x=0$ to $x=1$ in the limit as $\delta x \rightarrow 0.$ This gives you the required integral.

Be sure to draw a diagram to see what is going on.

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