Find the volume of the solid that results when the region enclosed by the curves is revolved about the x-axis.

Answer:
A
Step-by-step explanation:
[tex]volume=\pi \int\limits^0_{-1} {[(-x^2+7)^2-({x+5)^2}] } \, dx \\\\=\pi\int\limits^0_{-1}[( {x^4-14x^2+49})-(x^2+10x+25)] \, dx \\=\pi \int\limits^0_{-1} {[x^4-15x^2+10x+24] } \, dx =\pi [(\frac{x^5}{5} -15\frac{x^3}{3}+10 \frac{x^2}{2} +24x)]| x: ~-1 \rightarrow 0\\=\pi [\frac{1}{5}(0-(-1))-5(0-(-1))+5(0-(-1))+24(0-(-1)]\\=\pi [\frac{1}{5}-5+5+24]\\=24\frac{1}{5} \pi\\=\frac{121 \pi}{5}\\\approx ~76.02 ~units~cubed[/tex]