Question

Determine whether the following statements are true and give an explanation or counterexample. If f is not one-to-one on the interval [a, b], then the are of the surface generated when the graph of f on [a, b] is revolved about the x-axis is not defined.

Answer 1

Answer:

Step-by-step explanation:

The missing equation is:

$S = \int ^b_a 2 \pi f(y) \sqrt{1 + f' (y)^2 } \ dy$

Suppose f = nonnegative function whose first derivative is within (a,b) and it is continuous, Then the area of the surface generated is revolved about the x-axis:

The area of the surface revolution is:

$S = \int ^b_a 2 \pi f(x) \sqrt{1 + f' (x)^2 } \ dx$

So; if  $x = g(y)$ for $y \ \varepsilon \ [c,d]$

Then; substituting a with c and b with d; f(x) also with g(y) and dx with dy;

Then;

$S = \int ^d_c 2 \pi g(y) \sqrt{1 + g' (y)^2 } \ dy$

SInce;

$g(y) \ne f(y)$; Then the statement is false.

Provided that the semicircle $y = \sqrt{1-x^2}$ isn't on $[-1,1]$ interval.

Then, the solid generated by the revolution about the x-axis is a sphere.

However, the surface is well defined and the statement is false.