An engineer is designing a precision acoustic resonator. The internal radius of the resonator, RRR cm, at a distance xxx cm from the origin is modeled by the function
R(x)=6xe−13x0≤x≤6 R(x) = 6x e^{-\frac{1}{3}x} \quad 0 \le x \le 6 R(x)=6xe−31x0≤x≤6A cross-section of the resonator's interior, denoted by the region SSS, is bounded by the curve, the xxx-axis, and the line with equation x=6x = 6x=6.
The solid of revolution for the resonator's internal chamber is formed by rotating the region SSS through 2π2\pi2π radians about the xxx-axis.
Show that the internal volume, VVV, of this chamber is given by
V=k∫06x2e−23x dx V = k \int_{0}^{6} x^2 e^{-\frac{2}{3}x} \, dx V=k∫06x2e−32xdxwhere kkk is a constant to be determined.
Find ∫x2e−23x dx\int x^2 e^{-\frac{2}{3}x} \, dx∫x2e−32xdx.
A complete resonator is constructed by joining two of these chambers end-to-end at their widest faces. The resulting device has a mass of 0.8 kg and a total length of 12 cm.
Given that density=massvolume\text{density} = \frac{\text{mass}}{\text{volume}}density=volumemass,
find the density of this resonator. Give your answer in g/cm3\text{g/cm}^3g/cm3 to 3 significant figures.