Part 2: Volume

If f( x,y) ³ 0 on [ a,b] ×[c,d] , then f( r_j,t_k) Dx_jDy_k is the volume of a "box" over a rectangle determined by partitions of [a,b] and [c,d], respectively.

 
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Consequently, the Riemann sum is an approximation of the volume of the solid under z = f( x,y) and over the rectangle [a,b] ×[ c,d] . 

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Thus, if f( x,y) ³ 0 over R, then the volume of the solid below z = f( x,y) and above R is

V = R   f( x,y) dA

It follows from the previous section that if R is a type I region bounded by x = a, x = b, y = h( x) , y = g( x) , then

R f( x,y) dA = ó õ b a  ó õ g(x) h( x)   f( x,y) dydx

and if R is a type II region bounded by y = c, y = d, x = q( y) , x = p( y) , then

R   f( x,y) dA = ó õ d c  ó õ p(y) q( y)   f( x,y) dxdy

EXAMPLE 2    Find the volume of the region below z = x²y and over the region

R: x = 0 y = x x = 1 y = 1

Solution: Since the region is a type I region, we obtain

V = R  x2y  dA = ó õ 1 0  ó õ 1 x  x2y  dydx = ó õ 1 0   x2y2 2 ê ê 1 x  dx = ó õ 1 0  æ è  x2 2 -  x4 2 ö ø dx =  1 15

In general, if f( x,y) ³ g( x,y) over a region R,

then the volume of the solid between z = f( x,y) and z = g(x,y) over R is

V = R  [ f( x,y) -g( x,y) ] dA

(5)

EXAMPLE 3    Find the volume of the solid between z = x+y and z = x-y over the region

R: y = 0 x = y2 y = 1 x = y

Solution: According to (5), the volume of the solid is

V = R  ( ( x+y) -( x-y) )dA = R  2y dA

which transforms into the type II iterated integral

V = ó õ 1 0  ó õ y y2  2y  dxdy

Evaluating the inside integral results in

V = ó õ 1 0  2y x y y2 dy = ó õ 1 0  ( 2y·y-2y·y2) dy

It then follows that

V = ó õ 1 0  ( 2y2-2y3) dy =  1 6