Answer:
Step-by-step explanation:
The missing equation is:
[tex]S = \int ^b_a 2 \pi f(y) \sqrt{1 + f' (y)^2 } \ dy[/tex]
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:
[tex]S = \int ^b_a 2 \pi f(x) \sqrt{1 + f' (x)^2 } \ dx[/tex]
So; if [tex]x = g(y)[/tex] for [tex]y \ \varepsilon \ [c,d][/tex]
Then; substituting a with c and b with d; f(x) also with g(y) and dx with dy;
Then;
[tex]S = \int ^d_c 2 \pi g(y) \sqrt{1 + g' (y)^2 } \ dy[/tex]
SInce;
[tex]g(y) \ne f(y)[/tex]; Then the statement is false.
Provided that the semicircle [tex]y = \sqrt{1-x^2}[/tex] isn't on [tex][-1,1][/tex] 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.
