Question

3. (a) Find the volume of the solid generated by revolving the region bounded by the curves y=x^2 and y=4x−x^2 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

ANSWER (a) : $\frac { 64\pi }{ 3 }$

Explanation. We pass to the new coordinate system ${ y }_{ 1 }=y-6\ ,\ { x }_{ 1 }=x\quad$.The volume of the solid generated by revolving the region bounded by the curves $y=f(x),y=6, x=a,x=b(a\le x\le b)$ around the line $y=6$ is calculated by the formula $V= \pi \int _{ a }^{ b }{ { \left( f(x)-6 \right) }^{ 2 } } dx\quad$. Therefore , $V=\pi \int _{ 0 }^{ 2 }{ \left[ { { \left( { x }^{ 2 }-6 \right) }^{ 2 } }^{ \ }-{ \left( 4x-{ x }^{ 2 }-6 \right) }^{ 2 } \right] dx }$$=\pi \int _{ 0 }^{ 2 }{ ({ x }^{ 2 }-6-4x+{ x }^{ 2 } } +6)({ x }^{ 2 }-6+4x-{ x }^{ 2 }-6)dx=$ $=\pi \int _{ 0 }^{ 2 }{ ({ 2x }^{ 2 }\ -4x)(4x-12)dx=8\pi \int _{ 0 }^{ 2 }{ ( } { x }^{ 3 } } -5{ x }^{ 2 }+6x)dx=$ $=8\pi \left[ \frac { { x }^{ 4 } }{ 4 } -\frac { 5{ x }^{ 3 } }{ 3 } +3{ x }^{ 2 }\quad \right] \begin{matrix} 2 \\ 0 \end{matrix}=$ $=8\pi \left( 4-\frac { 40 }{ 3 } +12 \right) \ =\frac { 64\pi }{ 3 }$

ANSWER (b)

(i) for example :$f(x)=-{ (x+2) }^{ 3 }\left( x-\frac { 14 }{ 3 } \right) -\frac { 112 }{ 3 } \quad$

(ii) the line $y=f(-2)$ is the tangent line at the point $(-2,f(-2))$ , the line $y=f(3)$ is the tangent line at the point $(3,f(3))$

(iii) function increases in intervals $\left( -\infty ,-2 \right), (-2,3)$ . So $x=-2$ is not an extremum point.(iv)function decreases in the interval $(3,+\infty )$ and increases in the interval (-2,3) . So, x=3 is the maximum point