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Applications of the Integral

Problems

Volumes of Solids

Average Value of a Function

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 -  
  =  

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) = 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  
  =  

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