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) f(−2) = 0 and f(3) = 0.
(iii) f(x) > 0 on the intervals (−∞,−2) and (−2,3).
(iv) f(x) < 0 on the interval (3,∞). Label all important points.

Answer 1

a)

f(x) = x²

g(x) = 4x - x²

Rotation about y = 6

Inner radius = 6-(4x-x²)= 6-4x+x²

Outer radius = 6-x²

Area of a slit of ring

A(x) = π[(6-x²)² - (6-4x+x²)²]

= π[36-12x²+x⁴-{36+16x²+x⁴-48x-8x³+12x²}]

= π[36-12x²+x⁴-36-16x²-x⁴+48x+8x³-12x²]

= π[8x³-40x²+48x]

Volume of revolution

$\int_0^2π[8x³-40x²+48x]dx$

=π $[\frac {8x4}{4} - \frac {40x³}{3} +\frac {48x²}{2} ]_0^2$

= π$[\frac {8.2⁴}{4} - \frac {40.2³}{3} +\frac {48.2²}{2} - 0]$

= π[32- $\frac{320}{3}$ + 96]

= $\frac {64π}{3}$

So required volume = $\frac {64π}{3}$ cubic unit = 67.02 cubic unit

b)

The function satisfies the following conditions

1. The graph goes through origin

So x is a factor

2. f(-2)=f(3)=0

So (x+2), (x-3) are factors

3. f(x) > 0 on (-∞, -2) and (-2,3)

So f(x) ≥0 on (-∞,3)

4. f(x) < 0 on (3,∞)

According to properties mentioned above f(x) will cut x-axis at x = -2, 0, 3

Part of the graph is above x-axis on

(-∞,-2), (-2,0),(0,3)

Part of the graph is below the x-axis on (3,∞)

A sketch of the graph is given below satisfying all the above