Triple integrals in Cartesian coordinates (Sect. 15.4) Review: Triple integrals in arbitrary domains.


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1 Triple integrals in Cartesian coordinates (Sect. 5.4 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. 5.4 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: 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 limit is the plane =. This plane contains the points (,,, (,, and (,,. = / = (6 6 / = / =
4 Changing the order of integration. Interchange the limits in / / d d d. Solution: The region:,, and We conclude: = (6 6 / = / = = / / / / / / / / / / d d d. d d d. d d d. d d d. Triple integrals in Cartesian coordinates (Sect. 5.4 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. Review: The average of a single variable function. Definition 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 [ ][ ][ ] d d d f = ( ( ( = ( ( 4 ( We conclude: f = /4.
6 Triple integrals in Cartesian coordinates (Sect. 5.4 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: Sketch the integration region. The integration region is in the first octant. It is inside the clinder + = 9. = It is on one side of the plane =. The plane has normal vector n =,, and contains (,,. n + = 9
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. For practice purpose onl, let us change the integration order to d d d: = 9 = The result is: I = 9 d d d.
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: D f dv = = = 9 d d d. ( [ 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: f dv = [ ( 9 Therefore, D D f dv = = 9 ( [ 7( 9( ] d, [ ( ( ] d. Substitute u =, then du = d, so, f dv = 9 (u u du. D ] d.
9 Triple integrals in arbitrar domains. Compute the triple integral of f (,, = in the region bounded b,,, and 9 +. Solution: We conclude D D f dv = 9 (u u du, = 9 [ ( u ( u 4 4 = 9 (. 4 f dv = ], Integrals in clindrical, spherical coordinates (Sect Integration in clindrical coordinates. Review: Polar coordinates in a plane. Clindrical coordinates in space. Triple integral in clindrical coordinates. Net class: Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates.
10 Review: Polar coordinates in plane. Definition The polar coordinates of a point P R is the ordered pair (r, θ defined b the picture. r P = ( r, Theorem (Cartesianpolar transformations The Cartesian coordinates of a point P = (r, θ in the first quadrant are given b = r cos(θ, = r sin(θ. The polar coordinates of a point P = (, in the first quadrant are given b r = ( +, θ = arctan. Recall: Polar coordinates in a plane. Epress in polar coordinates the integral I = Solution: Recall: = r cos(θ and = r sin(θ. More often than not helps to sketch the integration region. The outer integration limit: [, ]. Then, for ever [, ] the coordinate satisfies [, ]. = = d d. The upper limit for is the curve =. Now is simple to describe this domain in polar coordinates: The line = is θ = π/4; the line = is θ = π/.
11 Recall: Polar coordinates in a plane. Epress in polar coordinates the integral I = d d. Solution: Recall: = r cos(θ, = r sin(θ, θ = π/4, θ = π/. The lower integration limit in r is r =. The upper integration limit is =, that is, = = r sin(θ. Hence r = / sin(θ. = = r sin ( 9 45 We conclude: d d = π/ / sin(θ π/4 r cos(θ(r dr dθ. Integrals in clindrical, spherical coordinates (Sect Integration in clindrical coordinates. Review: Polar coordinates in a plane. Clindrical coordinates in space. Triple integral in clindrical coordinates.
12 Clindrical coordinates in space. Definition The clindrical coordinates of a point P R is the ordered triple (r, θ, defined b the picture. Remark: Clindrical coordinates are just polar coordinates on the plane = together with the vertical coordinate. Theorem (Cartesianclindrical transformations The Cartesian coordinates of a point P = (r, θ, in the first quadrant are given b = r cos(θ, = r sin(θ, and =. The clindrical coordinates of a point P = (,, in the first quadrant are given b r = +, θ = arctan(/, and =. r P Clindrical coordinates in space. Use clindrical coordinates to describe the region R = {(,, : + (, + }. Solution: We first sketch the region. The base of the region is at =, given b the disk + (. The top of the region is the paraboloid = +. In clindrical coordinates: = + = r, and = + + ( = + + r r sin(θ r sin(θ Hence: R = {(r, θ, : θ [, π], r [, sin(θ], [, r ]}.
13 Integrals in clindrical, spherical coordinates (Sect Integration in clindrical coordinates. Review: Polar coordinates in a plane. Clindrical coordinates in space. Triple integral in clindrical coordinates. Triple integrals using clindrical coordinates. Theorem If the function f : R R R is continuous, then the triple integral of function f in the region R can be epressed in clindrical coordinates as follows, f dv = f (r, θ, r dr dθ d. Remark: R R Clindrical coordinates are useful when the integration region R is described in a simple wa using clindrical coordinates. Notice the etra factor r on the righthand side.
14 Triple integrals using clindrical coordinates. Find the volume of a clinder of radius R and height h. Solution: R = {(r, θ, : θ [, π], r [, R], [, h]}. R h = h π R h π R = h R π d (r dr dθ, r dr dθ, dθ, = h R π, We conclude: πr h. Triple integrals using clindrical coordinates. Find the volume of a cone of base radius R and height h. { Solution: R = θ [, π], r [, R], [, h ]} R r + h. h R = h = h π R h( r/r π R π R ( R = h R R ( r R (r r d (r dr dθ, r dr dθ, dr dθ, R π dθ = πhr 6. We conclude: πr h.
15 Triple integrals using clindrical coordinates. Sketch the region with volume π/ 9 r rd dr dθ. Solution: The integration region is R = {(r, θ, : θ [, π/], r [, ], [, 9 r ]}. We upper boundar is a sphere, since = 9 r = 9 r + + =. 5 The upper limit for r is r =, so = 9 = 5. Triple integrals using clindrical coordinates. Find the centroid vector r =,, of the region in space R = {(,, : +, + 4}. Solution: 4 = + The smmetr of the region implies = and =. (We verif this result later on. We onl need to compute. Since = dv, we start V R computing the total volume V. We use clindrical coordinates. π 4 r d rdr dθ = π ( 4 rdr = π r (4r r dr.
16 Triple integrals using clindrical coordinates. Find the centroid vector r =,, of the region in space R = {(,, : +, + 4}. Solution: π [ ( r (4r r dr = π 4 ( r 4 4 Hence π(8 4, so 8π. Then, is given b, = 8π = 8 π 4 = 8 r d rdr dθ = π 8π ]. ( 4 rdr; r (6r r 5 dr = [ ( r 6 ( r ( 64 6 = 4 4 = 8. ] ; Triple integrals using clindrical coordinates. Find the centroid vector r =,, of the region in space R = {(,, : +, + 4}. Solution: We obtained = 8. It is simple to see that = and =. For eample, π 4 = [ ] r cos(θ d rdr dθ 8π r = [ π ][ 4 cos(θdθ d r dr 8π r ]. But π cos(θdθ = sin(π sin( =, so =. A similar calculation shows =. Hence r =,, 8/.
17 Triple integrals using clindrical coordinates. Change the integration order and compute the integral I = π / Solution: r = / r dr d dθ. I = π = π = π ( r 4 = 6π ( r r d r dr dθ r dr (r r 4 dr 4 r 5 5. So, I = 6π π, that is, I =. Integrals in clindrical, spherical coordinates (Sect Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates.
18 Clindrical coordinates in space. Definition The clindrical coordinates of a point P R is the ordered triple (r, θ, defined b the picture. Remark: Clindrical coordinates are just polar coordinates on the plane = together with the vertical coordinate. Theorem (Cartesianclindrical transformations The Cartesian coordinates of a point P = (r, θ, in the first quadrant are given b = r cos(θ, = r sin(θ, and =. The clindrical coordinates of a point P = (,, in the first quadrant are given b r = +, θ = arctan(/, and =. r P Integrals in clindrical, spherical coordinates (Sect Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates.
19 Spherical coordinates in R Definition The spherical coordinates of a point P R is the ordered triple (ρ, φ, θ defined b the picture. rho Theorem (Cartesianspherical transformations The Cartesian coordinates of P = (ρ, φ, θ in the first quadrant are given b = ρ sin(φ cos(θ, = ρ sin(φ sin(θ, and = ρ cos(φ. The spherical coordinates of P = (,, in the first quadrant are ρ = ( ( + +, θ = arctan, and φ = arctan +. Spherical coordinates in R Use spherical coordinates to epress region between the sphere + + = and the cone = +. Solution: ( = ρ sin(φ cos(θ, = ρ sin(φ sin(θ, = ρ cos(φ. = = + The top surface is the sphere ρ =. The bottom surface is the cone: ρ cos(φ = ρ sin (φ cos(φ = sin(φ, + = / Hence: R = / so the cone is φ = π 4. { [ (ρ, φ, θ : θ [, π], φ, π ] }, ρ [, ]. 4
20 Integrals in clindrical, spherical coordinates (Sect Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates. Triple integral in spherical coordinates. Theorem If the function f : R R R is continuous, then the triple integral of function f in the region R can be epressed in spherical coordinates as follows, f dv = f (ρ, φ, θ ρ sin(φ dρ dφ dθ. R R Remark: Spherical coordinates are useful when the integration region R is described in a simple wa using spherical coordinates. Notice the etra factor ρ sin(φ on the righthand side.
21 Triple integral in spherical coordinates. Find the volume of a sphere of radius R. Solution: Sphere: S = {θ [, π], φ [, π], ρ [, R]}. π π R [ π ][ π = dθ [ = π cos(φ ρ sin(φ dρ dφ dθ, π ][ R sin(φ dφ ] R, = π [ cos(π + cos( ] R ; ] ρ dρ, hence: 4 πr. Triple integral in spherical coordinates. Use spherical coordinates to find the volume below the sphere + + = and above the cone = +. { [ Solution: R = (ρ, φ, θ : θ [, π], φ, π ] }, ρ [, ]. 4 The calculation is simple, the region is a simple section of a sphere. π π/4 [ π ][ π/4 = dθ [ = π cos(φ = π [ ρ sin(φ dρ dφ dθ, π/4 + ] ][ sin(φ dφ ]( ρ, ] ρ dρ, π (.
22 Triple integral in spherical coordinates. Find the integral of f (,, = e ( + + / in the region R = {,,, + + } using spherical coordinates. { [ Solution: R = θ, π ] [, φ, π ] }, ρ [, ]. Hence, R f dv = = π/ π/ [ π/ ] [ π/ dθ e ρ ρ sin(φ dρ dφ dθ, ] [ ] sin(φ dφ e ρ ρ dρ. Use substitution: u = ρ, hence du = ρ dρ, so R f dv = π [ cos(φ π ] e u du f dv = π (e. R 6 Triple integral in spherical coordinates. Change to spherical coordinates and compute the integral I = d d d. Solution: ( = ρ sin(φ cos(θ, = ρ sin(φ sin(θ, = ρ cos(φ. Limits in : ; Limits in : 4, so the positive side of the disk + 4. Limits in : 4, so a positive quarter of the ball
23 Triple integral in spherical coordinates. Change to spherical coordinates and compute the integral I = d d d. Solution: ( = ρ sin(φ cos(θ, = ρ sin(φ sin(θ, = ρ cos(φ. Limits in θ: θ [.π]; Limits in φ: φ [, π/]; Limits in ρ: ρ [, ]. The function to integrate is: f = ρ sin(φ sin(θ. I = π π/ ρ sin(φ sin(θ ( ρ sin(φ dρ dφ dθ. Triple integral in spherical coordinates. Change to spherical coordinates and compute the integral I = Solution: I = [ π I = 4 π π/ ( = cos(θ d d d. ][ π/ sin(θ dθ π [ π/ ρ sin(φ sin(θ ( ρ sin(φ dρ dφ dθ. = [( π ( sin(φ ][ sin ] (φ dφ ρ 4 dρ, ( ]( ρ 5 cos(φ dφ π/ ] 5 5 5, I = 4 π 5.
24 Triple integral in spherical coordinates. Compute the integral I = π π/ sec(φ ρ sin(φ dρ dφ dθ. Solution: Recall: sec(φ = / cos(φ. I = π = π π/ π/ (ρ ( sec(φ sin(φ dφ, cos (φ sin(φ dφ In the second term substitute: u = cos(φ, du = sin(φ dφ. [ I = π ( cos(φ π/ + / du ] u. Triple integral in spherical coordinates. Compute the integral I = π π/ sec(φ ρ sin(φ dρ dφ dθ. [ Solution: I = π ( cos(φ π/ + / du ] u. I = π [ ( ] [ + u du = π 4 / ( u / ], [ I = π 4 + ( u / ] [ = π 4 + ( ] [ 8 = π ] We conclude: I = 5π.
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