suppose the area of the region is bound by the curve y= (e^x + e^-x)/2 , the lines x=-1 and x=1 and the x-axis. Find the volume of the solid of revolution obtained when the region is rotated about the x-axis.
The answer given is π/4(e^2 +4-e^-2)units³ how can i obtain the answer? I'm stuck.
Let $y=f(x)\ge 0$ be a curve. Let $K$ be the region below $y=f(x)$, above the $x$-axis, from $x=a$ to $x=b$.
Then the volume obtained by rotating the region $K$ about the $x$-axis is $$\int_a^b \pi(f(x))^2\,dx.$$ Use $f(x)=\frac{e^x+e^{-x}}{2}$, $a=-1$ and $b=1$. Then $(f(x))^2=\frac{1}{4}(e^{2x}+2+e^{-2x})$. Thus the volume is $$\frac{\pi}{4}\int_{-1}^1(e^{2x}+2+e^{-2x})\,dx.$$ It is convenient (but not necessary) to exploit the symmetry, and conclude that the volume is $$\frac{2\pi}{4}\int_{0}^1(e^{2x}+2+e^{-2x})\,dx.$$ I expect you will find the integration straightforward.