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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()2dx=Π(1 - x2)dx  
 =Π -  
 =  

as expected.

Problem : Apply the shell method to find the volume of the solid obtained by revolving the region below the graph of f (x) = x2 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


2Πx(x2)dx=2Πx3dx  
 =2Π|01  
 =  

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 = 4x2 - 40x + 100, so the volume of the pyramid is given by


A(x)dx=(4x2 - 40x + 100)dx  
 = -20x2 + 100x  
 =  

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