If f is not​ one-to-one on the interval​ [a,b], then the area of the surface generated when the graph of f on​ [a,b] is revolved about the​ x-axis is undefined.


(a) False. A counter example would be the graph of ​f(x)=x^2 on ​[-​1,1]. It is not​ one-to-one on the interval but its surface area when revolved around the​ x-axis is approximately 7.62 square units.

(b) False. A counter example would be the graph of ​f(x)=x^2 on ​[​-1,1]. It is not​ one-to-one on the​ interval, but its surface area when revolved around the​ x-axis is approximately -7.62 square units.

(c) True. Using the surface integral to calculate the surface area of a function that is​ one-to-one on an interval​ [a,b] always yields zero. This is​ counter-intuitive as two dimensional objects must have some kind of surface area.​ Therefore, the surface areas of these​ one-to-one functions are considered undefined.

(d) True. Using the surface integral to calculate the surface area of a function that is​ one-to-one on an interval​ [a,b] always yields a negative number. Since surface area cannot be​ negative, the surface areas of these functions are considered undefined.