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.
864 exam-style questions on Edexcel A Level Maths Integration, covering 11.1 Integrating Standard Functions, 11.2 Integrating f(ax + b), 11.3 Using Trigonometric Identities, 11.4 Reverse Chain Rule, 11.5 Integration by Substitution, 11.6 Integration by Parts, 11.7 Partial Fractions, 11.8 Finding Areas, 11.9 The Trapezium Rule, 11.10 Solving Differential Equations, and 11.11 Modelling with Differential Equations. Each one has a worked solution and a mark scheme showing where the marks go.