Problem : Apply the disk method to find the volume of the unit sphere obtained by revolving the region below the graph of f (x) = from x = - 1 to 1 about the x -axis.
Using the disk method, we have
Π ()^{2} dx | = | Π (1 - x ^{2})dx | |
= | Π - | ||
= |
Problem : Apply the shell method to find the volume of the solid obtained by revolving the region below the graph of f (x) = x ^{2} from x = 0 to 1 about the y -axis (this solid looks like a cylinder with a bowl carved out of the top).
The formula from the shell method gives
2Π x(x ^{2})dx | = | 2Π x ^{3} dx | |
= | 2Π |_{0} ^{1} | ||
= |
Problem : Compute the volume of the square pyramid with base in the x = 0 plane with sides of length 10 and vertex at the point (5, 0) , using the cross-sectional area method.
The cross-section of the pyramid in the plane perpendicular to the x -axis with a particular x -coordinate in the interval [0, 5] is a square with sides of length s(x) = 10 - 2x . Such a square has area equal to A(x) = s(x)^{2} = (10 - 2x)^{2} = 4x ^{2} - 40x + 100 , so the volume of the pyramid is given by
A(x)dx | = | (4x ^{2} - 40x + 100)dx | |
= | -20x ^{2} + 100x | ||
= |