# Volume of revolution - shell method

A mould for a circular fish pond is made by rotating the region bounded by the curve$y = 2-\cos^2 x$ and the $x$-axis between $x = \displaystyle -\frac{\pi}{4}$ and $x = \displaystyle \frac{\pi}{4}$ through one complete revolution about the line $x=1$.

Use the method of cylindrical shells to show that the volume of the fish pond is given by:

$$V=\pi \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{(1-x)\cos(2x)\, dx}$$

Basically I know that a cylinder is formed with inner radius of $1-x$ and outer radius of $1-x-\delta(x)$ and height $y$. I also know the formula is integration $2\pi xy \mathrm\ {d}x$ using appropriate borders.

Thanks for the help :)

• Could you define y properly with format. Is it $2-(cosx)^2$ or $2-cos2x$ – Satish Ramanathan Feb 21 '15 at 7:21
• yep just did, are you able to help me out? – betty_carr Feb 21 '15 at 7:24
• Your approach is correct, somehow (2-cos^2(x)) must be equal to (cos2x/2). But it does not equal. I am suspecting that you copied down y wrongly. Are you sure it is what it is? – Satish Ramanathan Feb 21 '15 at 7:32
• yep i am absolutely sure, that's where I am stuck – betty_carr Feb 21 '15 at 7:39
• It also looks weird to me that the region is revolved around $x=1$, a vertical line, but the integration is with respect to $x$. – KittyL Feb 21 '15 at 10:39

This picture helps. So the pond is between $y=2-(\cos{x})^2$ and $y=2-(\cos{\pi/4})^2=\frac{3}{2}$.
$$2\pi \int ^{\pi/4}_{-\pi/4} (1-x)(\frac{3}{2}-y)dx=2\pi \int ^{\pi/4}_{-\pi/4} (1-x)(-\frac{1}{2}+(\cos{x})^2)dx\\=2\pi \int ^{\pi/4}_{-\pi/4} (1-x)(-\frac{1}{2}+\frac{\cos{2x}+1}{2})dx=2\pi \int ^{\pi/4}_{-\pi/4} (1-x)\cos{2x}dx$$