1. Suppose f(x,y,z) is bounded for all (x,y,z) in a box B = [a,b]  [c,d]  [p,q]. z y x (b,c,q) (b,d,p) (b,d,q) (a,d,q) (a,c,p) (b,c,p) (a,d,p) (a,c,q) Just as in the development of the double integral, we can partition [a,b] into n equal sub-intervals each with length  x, partition [c,d] into n equal sub-intervals each with length  y, partition [p,q] into n equal sub-intervals each with length  z. For each of the n 3 sub-boxes, we may multiply a value of the function f(x,y,z) anywhere on the sub-box by the volume  V =  x  y  z. If the limit as n  exists for the sum of these products (which will be true for continuous functions), then we say that f is integrable over B, and this limit is called the triple integral

2. f(x,y,z) dx dy dz =f(x,y,z) dy dx dz = a b c d p q db q cap etc. ExampleIntegrate f(x,y,z) = (3x + 2y + z) 2 over the box defined by [0,1/3]  [–1/2,0]  [0,1] using any two different orders of integration. 1/301 0–1/20 (3x + 2y + z) 2 dx dy dz =1/12 f(x,y,z) dx dy dz =f(x,y,z) dV = BB

3. An elementary region in R 3 is one which can be defined by restricting one of the variables to be between two functions of the remaining two variables, where the domains of these functions is an elementary region in R 2. Similar to the way the integral of a function f(x,y) over an elementary region in R 2 is defined, we can define the integral of a function f(x,y,z) over an elementary region W in R 3 to be f(x,y,z) dx dy dz = f(x,y,z) dx dy dz WB where B is a box containing the region W, and the function f(x,y,z) is defined to be equal to f(x,y,z) on the region W and equal to 0 (zero) outside the region W. Note that the volume of an elementary region W in R 3 can be found from W dx dy dz which can be written as a double integral of the difference between two functions of two variables.

4. ExampleFind the volume inside the sphere of radius a defined by x 2 + y 2 + z 2 = a 2. One way the region inside the sphere can be described is dz dy dx = –  a 2 – x 2 – y 2  a 2 – x 2 – y 2 –  a 2 – x 2  a 2 – x 2 – a a – a  x  a –  a 2 – x 2  y   a 2 – x 2 –  a 2 – x 2 – y 2  z   a 2 – x 2 – y 2

5. –  a 2 – x 2  a 2 – x 2 – a a z dy dx = z = –  a 2 – x 2 – y 2  a 2 – x 2 – y 2 2  a 2 – x 2 – y 2 dy dx –  a 2 – x 2  a 2 – x 2 – a a Note: One way of evaluating this integral is to use a table of integrals. Another way of evaluating this integral is to realize that – b b  b 2 – t 2 dt is the area of a semicircle of radius b, which is equal to  b 2 /2.

6. We see then (with b =  a 2 – x 2 and t = y) that 2  a 2 – x 2 – y 2 dy dx = –  a 2 – x 2  a 2 – x 2 – a a  (a 2 – x 2 ) dx = – a a  (a 2 x – x 3 /3)= x = – a a 4a 3  —— 3

7. Example Integrate the function xyze over the region between the xy plane, the plane y = x, the plane x = 4, and the cone z 2 = x 2 + y 2 in the region where x, y, and z are all non-negative. z2z2 First, we observe that this region can be described by  x   y  y   z  z  04 0x 0  x 2 + y 2 z y x 4 0 x 0 0 xyze dz dy dx = z2z2 4 0 x 0  x 2 + y 2 z = 0 xye —— dy dx = 2 z2z2 x = 4 y = x z 2 = x 2 + y 2

8. 4 0 x 0 xyexy ——— –— dy dx = 2 2 x 2 + y 2 4 0 xexy 2 ——— –— dx = 4 4 x 2 + y 2 x y = 0 4 0 xe x 3 xe ——— –— – —— dx = 4 4 4 2x22x2 x2x2 e x 4 e —— –— – ——= 1616 8 2x22x2 x2x2 x = 0 4 e 32 – 256 – 2e 16 – 1 + 2 e 32 – 2e 16 – 255 ——————————=——————— 1616