Question

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) a)

Area enclosed by the graphs, f₁(x) = x² and f₂(x)=4x-x² is rotating about the line y = 6

Solving f₁(x) = f₂(x) we get

x²= 4x-x²=> 2x²=4x=> 2x(x-2)=0=> x=0,2

Inner radius = 6-(4x-x²) = x²-4x+6

Outer radius = 6-x²

Area of a slit of ring

=π[(Outer radius)² -(inner radius)²]

=  π[(6-x²)²-(x²-4x+6)²]

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

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

= π[8x³-40x²+48x]

Volume generated by rotation

= $\int_0^2$ π[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 - $\frac{320}{3} + 96]$

= $\frac{64π}{3}$

$\mathbf{Answer}$

Exact value of volume of revolution is $\frac{64π}{3}$ cubic unit and approximately it is 67.02 cubic unit

b)

The function to sketch has the following properties

1. Passes through origin and hence x is a factor

2. f(-2)=f(3)=0 and hence (x+2)(x-3) is factor

3.f(x) > 0 on (-∞, -2) and (-2,3). So f(x)≥0 on (-∞,3)

4. f(x) < 0 on (3,∞)

As f(0)=f(-2)=f(3)=0, the graph of f(x) will cut x-axis at x=-2,0,3

As f(x)≥0 on (-∞,3) , parts of the graph on (-∞,-2), (-2,0),(0,3) are above x-axis

As f(x) < 0 on (3,∞), part of the graph on (3,∞) is below x-axis

Satisfying all of the above a sketch of the graph is attached.