Question

(a) Find the volume of the solid generated by revolving the region bounded by the curves
y = x² and y = 4x - x² about the line y = 6
(b) Sketch the graph of a continuous function f(x) satisfying the following properties:
(i) the graph of f goes through the origin
(ii) f¹(-2) = 0 and f¹(3) = 0.
(iii) f¹(x) > 0 on the intervals (-∞, -2) and (-2; 3).
(iv) f¹(x) < 0 on the interval (3;∞).
Label all important points.

Answer 1

a)

$$y=x^{2} \quad ,y=4 x-x^{2} \quad ,y=6 \\[1 em] \therefore x^{2}=4 x-x^{2} \rightarrow x^{2}-2 x=0 \rightarrow x=0,2 \\[1 em] \therefore V=\pi \int_{0}^{2} r_{1}^{2}-r_{2}^{2} d x \\[1 em] r_{1}=6-x^{2} \quad, r_{2}=6-4 x+x^{2} \\[1 em] \therefore V=\pi \int_{0}^{2}\left(6-x^{2}\right)^{2}-\left(6-4 x+x^{2}\right)^{2} dx\\[1 em] =\pi \int_{0}^{2} 36-12 x^{2}+x^{4}-\left(x^{4}-8 x^{3}+28 x^{2}-48 x+36\right) d x \\[1 em] =\pi \int_{0}^{2} 36-12 x^{2}+x^{4}-x^{4}+8 x^{3}-28 x^{2}+48 x-36 d x \\[1 em] =\pi \int_{0}^{2} 8 x^{3}-40 x^{2}+48 x d x \\[1 em] =\pi\left(2 x^{4}-\frac{40}{3} x^{3}+24 x^{2}\right) \bigg|_{0}^{2} \\[1 em] \therefore V\approx 67\\[2 em] b) f^{\prime}(x) >0 ~\text{in} ~(-\infty ,-2),(-2,3) ~\text{therefore the function increasing}\\[1 em] f^{\prime}(x) <0 ~\text{in} ~(3,\infty ) ~\text{therefore the function decreasing}\\[1 em] f^{\prime}(x) =0 ~\text{at} ~x=-2 ,x=3 \\[1 em]\text{Therefore the tangent at these points is horizontal}\\[1 em]$$