Triple integrals in Cartesian coordinates (Sect. 5.5) Triple integrals in rectangular boes. Triple integrals in arbitrar domains. Volume on a region in space. Triple integrals in rectangular boes Definition The triple integral of a function f : in the rectangular bo = [ˆ, ˆ ] [ŷ, ŷ ] [ẑ, ẑ ] is the number f (,, ) d d d = lim n i= j= k= f ( i, j, k ) where i [ i, i+ ], j [ j, j+ ], k [ k, k+ ] are sample points, while { i }, { j }, { k }, with i, j, k =,, n, are partitions of the intervals [ˆ, ˆ ], [ŷ, ŷ ], [ẑ, ẑ ], respectivel, and = (ˆ ˆ ) n, = (ŷ ŷ ) n, = (ẑ ẑ ). n
Triple integrals in rectangular boes emark: A finite sum S n below is called a iemann sum, where S n = f (i, j, k ). Then holds Theorem (Fubini) i= j= k= f (,, ) d d d = lim n S n. If function f : is continuous in the rectangle = [, ] [, ] [, ], then holds f (,, ) d d d = f (,, ) d d d. Furthermore, the integral above can be computed integrating the variables,, in an order. Triple integrals in rectangular boes eview: The iemann sums and their limits. Single variable functions in [ˆ, ˆ ]: lim n i= ˆ f (i ) = f ()d. ˆ Two variable functions in [ˆ, ˆ ] [ŷ, ŷ ]: (Fubini) lim n i= j= ˆ ŷ f (i, j ) = f (, ) d d. ˆ ŷ Three variable functions in [ˆ, ˆ ] [ŷ, ŷ ] [ẑ, ẑ ]: (Fubini) lim n i= j= k= ˆ ŷ f (i, j, k ) = ˆ ŷ ẑ ẑ f (,, )d d d
Triple integrals in rectangular boes Compute the integral of f (,, ) = on the domain = [, ] [, ] [, ]. Solution: It is useful to sketch the integration region first: = {(,, ) : [, ], [, ], [, ]}. The integral we need to compute is f dv = d d d, where we denoted dv = d d d. Triple integrals in rectangular boes Compute the integral of f (,, ) = on the domain = [, ] [, ] [, ]. Solution: f dv = d d d. We have chosen a particular integration order. (ecall: Since the region is a rectangle, integration limits are simple to interchange.) f dv = We conclude: f dv = 9 ( f dv = 9. ( ) d d = 7 ) d = 8 d = 9. d d.
Triple integrals in Cartesian coordinates (Sect. 5.5) Triple integrals in rectangular boes. Triple integrals in arbitrar domains. Volume on a region in space. Triple integrals in arbitrar domains Theorem If f : D is continuous in the domain D = { [, ], [h (), h ()], [g (, ), g (, )] }, where g, g : and h, h : are continuous, then the triple integral of the function f in the region D is given b D f dv = h () g (,) h () g (,) In the case that D is an ellipsoid, the figure represents the graph of functions g, g and h, h. f (,, ) d d d. = h ( ) = g (, ) = g (, ) = h ( )
Triple integrals in Cartesian coordinates (Sect. 5.5) Triple integrals in rectangular boes. Triple integrals in arbitrar domains. Volume on a region in space. Volume on a region in space emark: The volume of a bounded, closed region D is V = dv. Find the integration limits needed to compute the volume of the ellipsoid + + =. D Solution: We first sketch the integration domain.
Volume on a region in space Find the integration limits needed to compute the volume of the ellipsoid + + =. Solution: The functions = g and = g are, respectivel, =, =. The functions = h and = h are defined on =, and are given b, = and =, respectivel. The limits on are defined at =, = : = ±. Hence, V = (/) (/) d d d. Volume on a region in space Use Cartesian coordinates to find the integration limits needed to compute the volume between the sphere + + = and the cone = +. Solution: The top surface is the sphere, =. + = / / + The bottom surface is the cone, = +. The limits on are obtained projecting the -dimensional figure onto the plane =. We obtain the disk + = /. (The polar radius at the intersection cone-sphere was r = /.)
Volume on a region in space Use Cartesian coordinates to find the integration limits needed to compute the volume between the sphere + + = and the cone = +. Solution: ecall: =, = +. The -top of the disk is, = /. + = / We conclude: V = / + / / The -bottom of the disk is, = /. / d d d. + / Volume on a region in space Compute the volume of the region given b,, and + 6 + 6. Solution: The region is given b the first octant and below the plane + 6 + = 6. (6 6) / This plane contains the points (,, ), (,, ) and (,, ). In the limits are = (6 6)/ and =. = /
Volume on a region in space Compute the volume of the region given b,, and + 6 + 6. Solution: In the limits are = (6 6)/ and =. At = the projection of the region is the triangle,, and +. In the limits are = / and =. (6 6) / = / We conclude: V = / / d d d. Volume on a region in space Compute the volume of the region given b,, and + 6 + 6. Solution: ecall: V = V = V = V = / / / d d d. ( ) d d, [( )( ( /) ) ( ( /) [ ( ) ( ) ( ) ] d. We onl need to compute: V = ( ) d. )] d,
Volume on a region in space Compute the volume of the region given b,, and + 6 + 6. Solution: ecall: V = ( ) d. Substitute u = /, then du = d/, so ( u V = u du = u du = ) We conclude: V =. Triple integrals in arbitrar domains Compute the triple integral of f (,, ) = in the first octant and bounded b,, and + 9. Solution: The upper surface is = 9, the bottom surface is =. = = 9 = The coordinate is bounded below b the line = and above b =. (Because of the clinder equation at =.)
Triple integrals in arbitrar domains Compute the triple integral of f (,, ) = in the first octant and bounded b,, and + 9. Solution: ecall: 9, and f =. I = D I = f dv = ( 9 d d d, 9 ) d d, I = (9 )d d, I = [ ( 7( ) )] d. Triple integrals in arbitrar domains Compute the triple integral of f (,, ) = in the first octant and bounded b,, and + 9. Solution: ecall: I = I = I = 9 [ 7( ) ( [ 7( ) 9( ) ] d, [ ( ) ( ) ] d. Substitute u =, then du = d, so, I = 9 (u u )du. )] d. So,
Triple integrals in arbitrar domains Compute the triple integral of f (,, ) = in the first octant and bounded b,, and + 9. Solution: ecall: I = 9 (u u )du. I = 9 [ (u ) (u 4 )], 4 I = 9 ( 4). We conclude: D f dv = 45 8.