Triple integrals in Cartesian coordinates (Sect. 15.5) Review: Triple integrals in arbitrary domains
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1 Triple integrals in Cartesian coordinates (Sect. 5.5) Review: Triple integrals in arbitrar domains. s: Changing the order of integration. The average value of a function in a region in space. Triple integrals in arbitrar domains. Review: Triple integrals in arbitrar domains Theorem If f : D R R is continuous in the domain D = { [, ], [h (), h ()], [g (, ), g (, )] }, where g, g : R R and h, h : R R are continuous, then the triple integral of the function f in the region D is given b D f dv = h () g (,) h () g (,) f (,, ) d d d. In the case that D is an ellipsoid, the figure represents the graph of functions g, g and h, h. = h ( ) = g (, ) = g (, ) = h ( )
2 Triple integrals in Cartesian coordinates (Sect. 5.5) Review: Triple integrals in arbitrar domains. s: Changing the order of integration. The average value of a function in a region in space. Triple integrals in arbitrar domains. Changing the order of integration Change the order of integration in the triple integral (/) d d d. (/) Solution: First: Sketch the integration region. Start from the outer integration limits to the inner limits. Limits in : [, ]. Limits in :, so, +. The limits in :, so, + +.
3 Changing the order of integration Change the order of integration in the triple integral (/) d d d. (/) Solution: Region: + +. We conclude: (/) d d d. (/) (/) (/) (/) (/) (/) d d d. (/) (/) (/) d d d. (/) (/) Changing the order of integration Interchange the limits in / / d d d. Solution: Recall: Sketch the integration region starting from the outer integration limits to the inner integration limits. [, ]. [, ] so the upper limit is the line =. [, ] so the upper = / = (6 6) / = limit is the plane =. This plane contains the points (,, ), (,, ) and (,, ). = /
4 Changing the order of integration Interchange the limits in / / d d d. Solution: The region:,, and = / = (6 6) / = = / / / / / / / / / / d d d. d d d. d d d. d d d. Triple integrals in Cartesian coordinates (Sect. 5.5) Review: Triple integrals in arbitrar domains. s: Changing the order of integration. The average value of a function in a region in space. Triple integrals in arbitrar domains.
5 Average value of a function in a region in space Definition (Review: -variable) The average of a function f : [a, b] R on the interval [a, b], denoted b f, is given b f = b f () d. (b a) a f f() a b Definition The average of a function f : R R R on the region R with volume V, denoted b f, is given b f = f dv. V R Average value of a function in a region in space Find the average of f (,, ) = in the first octant bounded b the planes =, =, =. Solution: The volume of the rectangular integration region is d d d 6. The average of function f is: f = 6 d d d = 6 [ f = ( )( )( ) = ( ) ( 4 ) ( ) ][ ][ ] d d d f = /4.
6 Triple integrals in Cartesian coordinates (Sect. 5.5) Review: Triple integrals in arbitrar domains. s: Changing the order of integration. The average value of a function in a region in space. Triple integrals in arbitrar domains. Triple integrals in arbitrar domains Compute the triple integral of f (,, ) = in the region bounded b,,, and 9 +. Solution: Recall: Sketch the integration region. The integration region is in the first octant. It is inside the clinder + = 9. + = 9 It is on one side of the plane =. The plane has normal vector n =,, and contains (,, ). n =
7 Triple integrals in arbitrar domains Compute the triple integral of f (,, ) = in the region bounded b,,, and 9 +. Solution: We have found the region: = We obtain I = = 9 = / 9 d d d. The integration limits are: Limits in : 9. Limits in : /. Limits in :. Triple integrals in arbitrar domains Compute the triple integral of f (,, ) = in the region bounded b,,, and 9 +. Solution: Recall: / 9 d d d. Just for practice, let us change the integration order to d d d: The result is: I = 9 d d d. = = = 9
8 Triple integrals in arbitrar domains Compute the triple integral of f (,, ) = in the region bounded b,,, and 9 +. Solution: Recall: I = We now compute the integral: I = 9 d d d. ( I = I = [ ( 9 9 ) d d, (9 )d d, ) ( )] d. Triple integrals in arbitrar domains Compute the triple integral of f (,, ) = in the region bounded b,,, and 9 +. Solution: Recall: I = Therefore, I = I = 9 [ ( 9 ) ( )] d. [ 7( ) 9( ) ] d, [ ( ) ( ) ] d. Substitute u =, then du = d, so, I = 9 (u u )du.
9 Triple integrals in arbitrar domains Compute the triple integral of f (,, ) = in the region bounded b,,, and 9 +. Solution: Recall: I = 9 (u u )du. I = 9 (u u )du, We conclude I = 9 [ ( u ) ( u 4 )] 4 D f dv = = 9 ( 4).
Triple integrals in Cartesian coordinates (Sect. 15.5)
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