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 yaxis (this solid looks
like a cylinder with a bowl carved out of the top).
The formula from the shell method gives
2Πx(x^{2})dx  =  2Πx^{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 crosssectional area method.
The crosssection of the pyramid in the plane perpendicular to the
xaxis with a
particular
xcoordinate 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 

 =  
