26. Let S be the closed region in the first quadrant of the xy-plane bounded by y = sin(
p x/2) and y = x for 0 £ x £ 1. What is the volume of the closed region in R3 obtained by revolving S about the x-axis?|
A. 2 – ( p /2) |
B. p /6 |
C. p /3 |
|
D. p /2 |
E. (2 p )/3 |
|
Solution: The answer is B
The graph of both curves on the same plane illustrates the closed region S.
By rotating a vertical differential piece of thickness dx of the region S about the x-axis we obtain a thin washer whose volume is given by the product of its area and thickness. Where the area of the washer is found by subtracting the area of the missing inner circle from the area of the bigger circle as if the bigger circle were complete. That is,
Thus, to find the entire volume, we would have to sum all the washers of differential thickness dx that can be formed by the entire region S. Which implies that we integrate the right hand side of the above equation with respect to x over the interval [0,1]. As follows,
