Answer Happy

b. 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. the surface areas of these functions are considered undefined. B. False. A counter example would be the graph of f(x)=−x2 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. False. A counter example would be the graph of f(x)=x2 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. 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 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.