Part 3: Volumes by Slicing over Type I Regions

Let g, h be continuous on [ a,b] and suppose that g( x) £ h( x) for x in [ a,b] . If R is a region in the xy-plane which is bounded by the curves  y = g(x)  to  y = h(x) for x in [a,b] ,

then R is said to be a type I region. Let's find the volume of the solid between the graph of f( x,y) and the xy-plane over a type I region R when f( x,y) ³ 0.

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To do so, let's notice that if the solid is sliced with a plane parallel to the xz-plane, then its area is

A( x) = ó õ h( x) g( x)   f(x,y) dy

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It follows that if { x_j ,t_j} , j = 1,¼,n, is a tagged partition of [ a,b] , then the volume of the solid under the graph of f( x,y) and over the region R is

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A limit of such simple function approximations yields the volumes by slicing formula

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After combining this with the definition of A( x) , the result is the iterated integral

EXAMPLE 5    Find the volume of the solid under the graph of f( x,y) = 2-x²-y² over the type I region

x in [0,1] y = 0  to y = x

Solution: According to (1), the volume of the solid is

V = ó õ 1 0  é ë ó õ x 0  ( 2-x2-y2) dy ù û dx

We evaluate the resulting type I iterated integral by first evaluating the innermost integral:

V = ó õ 1 0  é ë 2y-x2y-  y3 3 ê ê x 0  ù û dx = ó õ 1 0  é ë 2x-  4 3 x3 ù û dx =  2 3

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