I came across this question from Schaum's Outlines of Advanced Calculus:
Show that the volume generated by revolving the region in the first quadrant bounded by the parabolas $y^2=x$, $y^2=8x$, $x^2=y$, $x^2=8y$ about the $x$-axis is $\dfrac{279\pi}{2}$.
My solution was to divide the region into three parts:
$$\pi\int_{1}^{2}(x^2)^2 - (\sqrt{x})^2\;dx = 4.7\pi$$
$$\pi\int_{4}^{8}(\sqrt{8x})^2 - (\sqrt{x})^2\;dx = 42\pi$$
$$\pi\int_{2}^{4} (x^2)^2 - \left(\frac{1}{8}x^2 \right)^2\;dx = 44.9\pi.$$
Hence the total volume will be
$$\pi(4.7 + 42 + 44.9) = 91.6\pi$$
which is wrong.
Can someone point out my mistake?
Your first term is correct, but the second and third are incorrect. The correct volume expression should read
$$V = \pi \left(\int_{x=1}^2 (\color{magenta}{x^2})^2 - (\color{orange}{\sqrt{x}})^2 \, dx + \int_{x=2}^4 (\color{green}{\sqrt{8x}})^2 - (\color{orange}{\sqrt{x}})^2 \, dx + \int_{x=4}^8 (\color{green}{\sqrt{8x}})^2 - (x^2/8)^2 \, dx \right).$$
It is worth noting that this is the only region in the first quadrant that is bounded by all four curves. Admittedly, it is not the most precisely phrased question, but it is reasonable to infer that this is the intended volume.