Question

Find the volume of the solid generated when the region bounded by y= x and y= 4√x is revolved about the x-axis

Answer 1

Check the picture below, that enclosed area is pretty much our "washer".

now, to get the outer radius, or "farther" radius from the axis of rotation, what I usually do is, use the f(x) - g(x) to get the area under the curve, using the axis of rotation for g(x).

$\bf \stackrel{\textit{top, f(x)}}{4\sqrt{x}}~~-~~\stackrel{\textit{bottom, g(x)}}{(0)}\implies \stackrel{\textit{farthest radius}}{4\sqrt{x}-0\implies 4\sqrt{x}} \\\\\\ \stackrel{\textit{top, f(x)}}{x}~~-~~\stackrel{\textit{bottom, g(x)}}{(0)}\implies \stackrel{\textit{closest radius}}{x-0\implies x} \\\\[-0.35em] ~\dotfill\\\\ 4\sqrt{x}=x\implies \stackrel{\textit{squaring both sides}}{16x=x^2\implies }16x-x^2=0$

$\bf x(16-x)=0\implies x= \begin{cases} 0\\ 16 \end{cases}\qquad \qquad \impliedby \textit{these are the bounds} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int\limits_{0}^{16}~\pi [(4\sqrt{x})^2-(x)^2]dx\implies \pi \int\limits_{0}^{16}[16x-x^2]dx\implies \pi \int\limits_{0}^{16}16x~~-~~\pi \int\limits_{0}^{16}x^2$

$\bf \pi \left. 16\cdot \cfrac{x^2}{2} \right]_{0}^{16}-\left. \cfrac{\pi x^3}{3} \right]_{0}^{16}\implies \left. \pi 8x^2 \cfrac{}{}\right]_{0}^{16}-\left. \cfrac{\pi x^3}{3} \right]_{0}^{16} \\\\\\ \left( \pi [2048]-\pi [0] \right)-\left(\left[ \cfrac{4096\pi }{3} \right]-[0] \right)\implies 2048\pi -\cfrac{4096\pi }{3} \implies \cfrac{2048\pi }{3}$