Let and be differentiable functions that satisfy for every in

MATHEMATICS · Single correct · CBSE · Mathematics · Applications of Integrals · Applications of Integrals

Problem

Let

g

and

h

be differentiable functions that satisfy

0\le g(x)\le h(x)

for every

x

[0,3]

. The regions under

y=g(x)

and under

y=h(x)

[0,3]

are each revolved about the

x

-axis, producing solids with volumes

V_g

and

V_h

, respectively. Which of the following best explains why

V_h>V_g

? (a) Because

h

increases faster than

g

near

x=3

, the total cross-sectional area of the solid generated by

h

grows linearly with height, making its volume larger. (b) The difference in volumes equals

2\pi\displaystyle\int_{0}^{3}\!\bigl(h(x)-g(x)\bigr)\,dx

, which is positive since

h(x)>g(x)

[0,3]

, so

V_h

must exceed

V_g

.},{ (c) For every

x

[0,3]

, the radius of the cross-section generated by

h

is greater than that generated by

g

, and the cross-sectional area depends on the square of this radius, so

V_h>V_g

. (d) The interval of revolution for

h

is longer than that for

g

, so the solid generated by

h

must have the greater volume.