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 :)
well that is the question, that may help you out :)
This picture helps. So the pond is between $y=2-(\cos{x})^2$ and $y=2-(\cos{\pi/4})^2=\frac{3}{2}$.
Then the volume should be
$$2\pi \int ^{\pi/4}_{-\pi/4} (1-x)(\frac{3}{2}-y)dx=2\pi \int ^{\pi/4}_{-\pi/4} (1-x)(-\frac{1}{2}+(\cos{x})^2)dx\\=2\pi \int ^{\pi/4}_{-\pi/4} (1-x)(-\frac{1}{2}+\frac{\cos{2x}+1}{2})dx=2\pi \int ^{\pi/4}_{-\pi/4} (1-x)\cos{2x}dx$$