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


 Clemence Gaines
 1 years ago
 Views:
Transcription
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.
Integrals in cylindrical, spherical coordinates (Sect. 15.7)
Integrals in clindrical, spherical coordinates (Sect. 15.7) Integration in clindrical coordinates. eview: Polar coordinates in a plane. Clindrical coordinates in space. Triple integral in clindrical coordinates.
More informationEngineering Mathematics 233 Solutions: Double and triple integrals
Engineering Mathematics s: Double and triple integrals Double Integrals. Sketch the region in the plane bounded b the curves and, and find its area. The region is bounded b the parabola and the straight
More informationMath 32B Practice Midterm 1
Math B Practice Midterm. Let be the region bounded b the curves { ( e)x + and e x. Note that these curves intersect at the points (, ) and (, e) and that e
More informationMath 21a Old Exam One Fall 2003 Solutions Spring, 2009
1 (a) Find the curvature κ(t) of the curve r(t) = cos t, sin t, t at the point corresponding to t = Hint: You ma use the two formulas for the curvature κ(t) = T (t) r (t) = r (t) r (t) r (t) 3 Solution:
More informationLines and planes in space (Sect. 12.5) Review: Lines on a plane. Lines in space (Today). Planes in space (Next class). Equation of a line
Lines and planes in space (Sect. 2.5) Lines in space (Toda). Review: Lines on a plane. The equations of lines in space: Vector equation. arametric equation. Distance from a point to a line. lanes in space
More informationChange of Coordinates in Two Dimensions
CHAPTER 5 Change of Coordinates in Two Dimensions Suppose that E is an ellipse centered at the origin. If the major and minor aes are horiontal and vertical, as in figure 5., then the equation of the ellipse
More informationMATH 52 FINAL EXAM SOLUTIONS (AUTUMN 2003)
MATH 5 FINAL EXAM OLUTION (AUTUMN 3). Evaluate the integral by reversing the order of integration olution. 3 3y e x dxdy = = 3 3y 3 x/3 3 = ex 3 [ e x y e x dxdy. e x dxdy ] y=x/3 y= = e9 dx =. Rewrite
More information2 Topics in 3D Geometry
2 Topics in 3D Geometry In two dimensional space, we can graph curves and lines. In three dimensional space, there is so much extra space that we can graph planes and surfaces in addition to lines and
More informationScalar functions of several variables (Sect. 14.1).
Scalar functions of several variables (Sect. 14.1). Functions of several variables. On open, closed sets. Functions of two variables: Graph of the function. Level curves, contour curves. Functions of three
More informationTriple Integrals in Cylindrical or Spherical Coordinates
Triple Integrals in Clindrical or Spherical Coordinates. Find the volume of the solid ball 2 + 2 + 2. Solution. Let be the ball. We know b #a of the worksheet Triple Integrals that the volume of is given
More informationWe can use more sectors (i.e., decrease the sector s angle θ) to get a better approximation:
Section 1.4 Areas of Polar Curves In this section we will find a formula for determining the area of regions bounded by polar curves. To do this, wee again make use of the idea of approximating a region
More informationSolutions to Vector Calculus Practice Problems
olutions to Vector alculus Practice Problems 1. Let be the region in determined by the inequalities x + y 4 and y x. Evaluate the following integral. sinx + y ) da Answer: The region looks like y y x x
More information27.4. Changing Coordinates. Introduction. Prerequisites. Learning Outcomes
Changing Coordinates 27. Introduction We have seen how changing the variable of integration of a single integral or changing the coordinate sstem for multiple integrals can make integrals easier to evaluate.
More informationFigure 1: Volume between z = f(x, y) and the region R.
3. Double Integrals 3.. Volume of an enclosed region Consider the diagram in Figure. It shows a curve in two variables z f(x, y) that lies above some region on the xyplane. How can we calculate the volume
More information1.7 Cylindrical and Spherical Coordinates
56 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE 1.7 Cylindrical and Spherical Coordinates 1.7.1 Review: Polar Coordinates The polar coordinate system is a twodimensional coordinate system in which the
More informationAssignment 7  Solutions Math 209 Fall 2008
Assignment 7  Solutions Math 9 Fall 8. Sec. 5., eercise 8. Use polar coordinates to evaluate the double integral + y da, R where R is the region that lies to the left of the yais between the circles
More informationSolutions to Final Practice Problems
s to Final Practice Problems Math March 5, Change the Cartesian integral into an equivalent polar integral and evaluate: I 5 x 5 5 x ( x + y ) dydx The domain of integration for this integral is D {(x,
More informationMath 263 Assignment 7 SOLUTIONS
Problems to turn in: Math 6 Assignment 7 SOLUTIONS In each case sketch the region and then compute the volume of the solid region. a The icecream cone region which is bounded above by the hemisphere z
More informationLecture 8. Metrics. 8.1 Riemannian Geometry. Objectives: More on the metric and how it transforms. Reading: Hobson, 2.
Lecture 8 Metrics Objectives: More on the metric and how it transforms. Reading: Hobson, 2. 8.1 Riemannian Geometry The interval ds 2 = g αβ dx α dx β, is a quadratic function of the coordinate differentials.
More informationMATH 118, LECTURES 14 & 15: POLAR AREAS
MATH 118, LECTURES 1 & 15: POLAR AREAS 1 Polar Areas We recall from Cartesian coordinates that we could calculate the area under the curve b taking Riemann sums. We divided the region into subregions,
More information27.2. Multiple Integrals over Nonrectangular Regions. Introduction. Prerequisites. Learning Outcomes
Multiple Integrals over Nonrectangular Regions 7. Introduction In the previous Section we saw how to evaluate double integrals over simple rectangular regions. We now see how to etend this to nonrectangular
More informationPartial differentiation
Chapter 1 Partial differentiation Example 1.1 What is the maximal domain of the real function g defined b g(x) = x 2 + 3x + 2? : The ke point is that the square root onl gives a real result if the argument
More informationAreas and double integrals. (Sect. 15.3)
Areas and double integrals. (Sect. 5.) Areas of a region on a plane. Average value of a function. More eamples of double integrals. Areas of a region on a plane The area of a closed, bounded region on
More informationLengths in Polar Coordinates
Lengths in Polar Coordinates Given a polar curve r = f (θ), we can use the relationship between Cartesian coordinates and Polar coordinates to write parametric equations which describe the curve using
More informationSolutions to Homework 10
Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x
More information2.1 Three Dimensional Curves and Surfaces
. Three Dimensional Curves and Surfaces.. Parametric Equation of a Line An line in two or threedimensional space can be uniquel specified b a point on the line and a vector parallel to the line. The
More informationCHAPTER 54 SOME APPLICATIONS OF DIFFERENTIATION
CHAPTER 5 SOME APPLICATIONS OF DIFFERENTIATION EXERCISE 0 Page 65. An alternating current, i amperes, is given b i = 0 sin πft, where f is the frequenc in hertz and t the time in seconds. Determine the
More informationMULTIPLE INTEGRALS. h 2 (y) are continuous functions on [c, d] and let f(x, y) be a function defined on R. Then
MULTIPLE INTEGALS 1. ouble Integrals Let be a simple region defined by a x b and g 1 (x) y g 2 (x), where g 1 (x) and g 2 (x) are continuous functions on [a, b] and let f(x, y) be a function defined on.
More informationSection 2.6 Cylindrical and Spherical Coordinates A) Review on the Polar Coordinates
Section.6 Cylindrical and Spherical Coordinates A) Review on the Polar Coordinates The polar coordinate system consists of the origin O,the rotating ray or half line from O with unit tick. A point P in
More informationCalculus Answers. , where π u π
Calculus Answers 13. r(u, v) u cosvi + u sinvj + vk. The parametric equations for the surface are x u cosv y u sin v z v We look at the grid curve first; if we fix v, then x and y parametrize a straight
More informationDouble Integrals in Polar Coordinates
Double Integrals in Polar Coordinates. A flat plate is in the shape of the region in the first quadrant ling between the circles + and +. The densit of the plate at point, is + kilograms per square meter
More information0.1 Linear Transformations
.1 Linear Transformations A function is a rule that assigns a value from a set B for each element in a set A. Notation: f : A B If the value b B is assigned to value a A, then write f(a) = b, b is called
More informationx(θ, r) = r cos(θ) 0 θ 2π y(θ, r) = r sin(θ) 0 r 2 z(θ, r) = r
1. A parametric equation for a cone is: x(θ, r) = r cos(θ) θ 2π y(θ, r) = r sin(θ) r 2 z(θ, r) = r Find a downward facing normal vector for an arbitary point on the cone. Is the cone a smooth surface?
More informationTrigonometric Substitution Created by Tynan Lazarus November 3, 2015
. Trig Identities Trigonometric Substitution November 3, 0 tan(θ) = sin(θ) cos(θ) sec(θ) = cos(θ) cot(θ) = cos(θ) sin(θ) csc(θ) = sin(θ). Trig Integrals sin(θ) dθ = cos(θ) + C sec (θ) dθ = tan(θ) + C sec(θ)
More informationMATHEMATICS SPECIALIST ATAR COURSE FORMULA SHEET
MATHEMATICS SPECIALIST ATAR COURSE FORMULA SHEET 06 Copyright School Curriculum and Standards Authority, 06 This document apart from any third party copyright material contained in it may be freely copied,
More informationChapt. 10 Rotation of a Rigid Object About a Fixed Axis
Chapt. otation of a igid Object About a Fixed Axis. Angular Position, Velocity, and Acceleration Translation: motion along a straight line (circular motion?) otation: surround itself, spins rigid body:
More informationThe line segment can be parametrized as x = 1 + t, y = 1 + 2t, z = 1 + 3t 0 t 1. Then dx = dt, dy = 2dt, dz = 3dt so the integral becomes
MATH rd Midterm Review Solutions. Evaluate xz dx z dy + y dz where is the line segment from (,, ) to (,, ). The line segment can be parametrized as x = + t, y = + t, z = + t where t. Then dx = dt, dy =
More informationFigure 1: u2 (x,y) D u 1 (x,y) yρ(x, y, z)dv. xρ(x, y, z)dv. M yz = E
Contiune on.7 Triple Integrals Figure 1: [ ] u2 (x,y) f(x, y, z)dv = f(x, y, z)dz da u 1 (x,y) Applications of Triple Integrals Let be a solid region with a density function ρ(x, y, z). Volume: V () =
More informationThe graph of. horizontal line between1 and 1 that the sine function is not 11 and therefore does not have an inverse.
Inverse Trigonometric Functions The graph of If we look at the graph of we can see that if you draw a horizontal line between1 and 1 that the sine function is not 11 and therefore does not have an inverse.
More informationAssignment 5 Math 101 Spring 2009
Assignment 5 Math 11 Spring 9 1. Find an equation of the tangent line(s) to the given curve at the given point. (a) x 6 sin t, y t + t, (, ). (b) x cos t + cos t, y sin t + sin t, ( 1, 1). Solution. (a)
More informationCalculating Areas Section 6.1
A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics Calculating Areas Section 6.1 Dr. John Ehrke Department of Mathematics Fall 2012 Measuring Area By Slicing We first defined
More information14.1. Multiple Integration. Iterated Integrals and Area in the Plane. Objectives. Iterated Integrals. Iterated Integrals
14 Multiple Integration 14.1 Iterated Integrals and Area in the Plane Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. Objectives! Evaluate an iterated
More informationWe d like to explore the question of inverses of the sine, tangent, and secant functions. We ll start with f ( x)
Inverse Trigonometric Functions: We d like to eplore the question of inverses of the sine, tangent, and secant functions. We ll start with f ( ) sin. Recall the graph: MATH 30 Lecture 0 of 0 Well, we can
More informationMATHEMATICS 121, Worked Problems
MATHEMATICS, Worked Problems ) Evaluate a) + + + 4 + + 99 b) n Solution. a) + + + 4 + + 99 = i i = d i = d i= j= j j i= = ( )+( ) ( ) = + ( ) The right hand side is not defined for =, but this case ma
More informationM PROOF OF THE DIVERGENCE THEOREM AND STOKES THEOREM
68 Theor Supplement Section M M POOF OF THE DIEGENE THEOEM ND STOKES THEOEM In this section we give proofs of the Divergence Theorem Stokes Theorem using the definitions in artesian coordinates. Proof
More informationChapter 24 Physical Pendulum
Chapter 4 Physical Pendulum 4.1 Introduction... 1 4.1.1 Simple Pendulum: Torque Approach... 1 4. Physical Pendulum... 4.3 Worked Examples... 4 Example 4.1 Oscillating Rod... 4 Example 4.3 Torsional Oscillator...
More informationMATH SOLUTIONS TO PRACTICE FINAL EXAM. (x 2)(x + 2) (x 2)(x 3) = x + 2. x 2 x 2 5x + 6 = = 4.
MATH 55 SOLUTIONS TO PRACTICE FINAL EXAM x 2 4.Compute x 2 x 2 5x + 6. When x 2, So x 2 4 x 2 5x + 6 = (x 2)(x + 2) (x 2)(x 3) = x + 2 x 3. x 2 4 x 2 x 2 5x + 6 = 2 + 2 2 3 = 4. x 2 9 2. Compute x + sin
More informationWe get a r b first. We can now observe that y 0 implies 0 θ π. Also y = x corresponds to θ = π 4 and y = x corresponds to θ = 3π 4.
iscussion 24 solution Thursday, April 4th topic: polar coordinates The trick to working with the polar coordinate change of variables is that we don t worry about the substitution r(x, y) and θ(x, y) to
More informationF = 0. x ψ = y + z (1) y ψ = x + z (2) z ψ = x + y (3)
MATH 255 FINAL NAME: Instructions: You must include all the steps in your derivations/answers. Reduce answers as much as possible, but use exact arithmetic. Write neatly, please, and show all steps. Scientists
More informationx + 5 x 2 + x 2 dx Answer: ln x ln x 1 + c
. Evaluate the given integral (a) 3xe x2 dx 3 2 e x2 + c (b) 3 x ln xdx 2x 3/2 ln x 4 3 x3/2 + c (c) x + 5 x 2 + x 2 dx ln x + 2 + 2 ln x + c (d) x sin (πx) dx x π cos (πx) + sin (πx) + c π2 (e) 3x ( +
More informationMath 241: Laplace equation in polar coordinates; consequences and properties
Math 241: Laplace equation in polar coordinates; consequences and properties D. DeTurck University of Pennsylvania October 6, 2012 D. DeTurck Math 241 002 2012C: Laplace in polar coords 1 / 16 Laplace
More informationComplex Numbers Basic Concepts of Complex Numbers Complex Solutions of Equations Operations on Complex Numbers
Complex Numbers Basic Concepts of Complex Numbers Complex Solutions of Equations Operations on Complex Numbers Identify the number as real, complex, or pure imaginary. 2i The complex numbers are an extension
More informationSTRESS TRANSFORMATION AND MOHR S CIRCLE
Chapter 5 STRESS TRANSFORMATION AND MOHR S CIRCLE 5.1 Stress Transformations and Mohr s Circle We have now shown that, in the absence of bod moments, there are si components of the stress tensor at a material
More informationIf f is continuous on [a, b], then the function g defined by. f (t) dt. is continuous on [a, b] and differentiable on (a, b), and g (x) = f (x).
The Fundamental Theorem of Calculus, Part 1 If f is continuous on [a, b], then the function g defined by g(x) = x a f (t) dt a x b is continuous on [a, b] and differentiable on (a, b), and g (x) = f (x).
More informationSAMPLE MATH 209 FINAL EXAM SOLUTIONS. James D. Lewis. [ ] y sin x
1. Evaluate the integral AMPLE MATH 9 FINAL EXAM OLUTION James D. Lewis 1 1 y sin x x dxdy. olution: The integral in its present form is hard to compute. By graphing the region and reversing the order
More informationEngineering Math II Spring 2015 Solutions for Class Activity #2
Engineering Math II Spring 15 Solutions for Class Activity # Problem 1. Find the area of the region bounded by the parabola y = x, the tangent line to this parabola at 1, 1), and the xaxis. Then find
More informationMATH FINAL EXAMINATION  3/22/2012
MATH 22  FINAL EXAMINATION  /22/22 Name: Section number: About this exam: Partial credit will be given on the free response questions. To get full credit you must show all of your work. This is a closed
More informationBending of Beams with Unsymmetrical Sections
Bending of Beams with Unsmmetrical Sections Assume that CZ is a neutral ais. C = centroid of section Hence, if > 0, da has negative stress. From the diagram below, we have: δ = α and s = αρ and δ ε = =
More informationChapter 5 Applications of Integration
MA111 Application of Integration Asst.Prof.Dr.Supranee Lisawadi 1 Chapter 5 Applications of Integration Section 5.1 Area Between Two Curves In this section we use integrals to find areas of regions that
More informationParametric Surfaces. Solution. There are several ways to parameterize this. Here are a few.
Parametric Surfaces 1. (a) Parameterie the elliptic paraboloid = 2 + 2 + 1. Sketch the grid curves defined b our parameteriation. Solution. There are several was to parameterie this. Here are a few. i.
More information6. ISOMETRIES isometry central isometry translation Theorem 1: Proof:
6. ISOMETRIES 6.1. Isometries Fundamental to the theory of symmetry are the concepts of distance and angle. So we work within R n, considered as an innerproduct space. This is the usual n dimensional
More informationMATH Area Between Curves
MATH  Area Between Curves Philippe Laval September, 8 Abstract This handout discusses techniques used to nd the area of regions which lie between two curves. Area Between Curves. Theor Given two functions
More informationPractice Final Math 122 Spring 12 Instructor: Jeff Lang
Practice Final Math Spring Instructor: Jeff Lang. Find the limit of the sequence a n = ln (n 5) ln (3n + 8). A) ln ( ) 3 B) ln C) ln ( ) 3 D) does not exist. Find the limit of the sequence a n = (ln n)6
More informationAN ANALYTICAL EXPRESSION FOR CROSSING PATH PROBABILITIES IN 2D RANDOM WALKS. Keywords: random walk, central limit theorem, intersections
AN ANALYTICAL EXPRESSION FOR CROSSING PATH PROBABILITIES IN 2D RANDOM WALKS MARC ARTZROUNI DEPARTMENT OF MATHEMATICS UNIVERSITY OF PAU; 64000 PAU; FRANCE Abstract. We investigate crossing path probabilities
More informationThis function is symmetric with respect to the yaxis, so I will let  /2 /2 and multiply the area by 2.
INTEGRATION IN POLAR COORDINATES One of the main reasons why we study polar coordinates is to help us to find the area of a region that cannot easily be integrated in terms of x. In this set of notes,
More informationIntersections of Polar Curves
Intersections of Polar Curves The purpose of this supplement is to find a method for determining where graphs of polar equations intersect each other. Let s start with a fairly straightforward example.
More informationDouble integrals. Notice: this material must not be used as a substitute for attending the lectures
ouble integrals Notice: this material must not be used as a substitute for attending the lectures . What is a double integral? Recall that a single integral is something of the form b a f(x) A double integral
More informationMECHANICS OF MATERIALS Plastic Deformations of Members With a Single Plane of Symmetry
Plastic Deformations of Members With a Single Plane of Smmetr Full plastic deformation of a beam with onl a vertical plane of smmetr. The neutral axis cannot be assumed to pass through the section centroid.
More informationMath 2443, Section 16.3
Math 44, Section 6. Review These notes will supplement not replace) the lectures based on Section 6. Section 6. i) ouble integrals over general regions: We defined double integrals over rectangles in the
More informationMath 220 October 11 I. Exponential growth and decay A. The halflife of fakeium20 is 40 years. Suppose we have a 100mg sample.
Math 220 October 11 I. Exponential growth and decay A. The halflife of fakeium20 is 40 years. Suppose we have a 100mg sample. 1. Find the mass that remains after t years. 2. How much of the sample remains
More informationGeometric Transformations
CS3 INTRODUCTION TO COMPUTER GRAPHICS Geometric Transformations D and 3D CS3 INTRODUCTION TO COMPUTER GRAPHICS Grading Plan to be out Wednesdas one week after the due date CS3 INTRODUCTION TO COMPUTER
More informationModule 1 : A Crash Course in Vectors Lecture 2 : Coordinate Systems
Module 1 : A Crash Course in Vectors Lecture 2 : Coordinate Systems Objectives In this lecture you will learn the following Define different coordinate systems like spherical polar and cylindrical coordinates
More information6. In cylindrical coordinate system, the differential normal area along a z is calculated as: a) ds = ρ d dz b) ds = dρ dz c) ds = ρ d dρ d) ds = dρ d
Electrical Engineering Department Electromagnetics I (802323) G1 Dr. Mouaaz Nahas First Term (14361437), Second Exam Tuesday 07/02/1437 H االسم: الرقم الجامعي: Start from here Part A CLO 1: Students will
More informationUnits and Vectors: Tools for Physics
Chapter 1 Units and Vectors: Tools for Physics 1.1 The Important Stuff 1.1.1 The SI System Physics is based on measurement. Measurements are made by comparisons to well defined standards which define the
More informationSection 16.7 Triple Integrals in Cylindrical Coordinates
Section 6.7 Triple Integrals in Cylindrical Coordinates Integrating Functions in Different Coordinate Systems In Section 6.4, we used the polar coordinate system to help integrate functions over circular
More informationCHANGE OF VARIABLES AND THE JACOBIAN. Contents 1. Change of variables: the Jacobian 1
CHANGE OF VARIABLES AND THE JACOBIAN Contents 1. Change of variables: the Jacobian 1 1. Change of variables: the Jacobian So far, we have seen three examples of situations where we change variables to
More informationIntegration Involving Trigonometric Functions and Trigonometric Substitution
Integration Involving Trigonometric Functions and Trigonometric Substitution Dr. Philippe B. Laval Kennesaw State University September 7, 005 Abstract This handout describes techniques of integration involving
More informationSolutions  Homework sections 17.717.9
olutions  Homework sections 7.77.9 7.7 6. valuate xy d, where is the triangle with vertices (,, ), (,, ), and (,, ). The three points  and therefore the triangle between them  are on the plane x +
More informationn th roots of complex numbers
n th roots of complex numbers Nathan Pflueger 1 October 014 This note describes how to solve equations of the form z n = c, where c is a complex number. These problems serve to illustrate the use of polar
More informationGreen s theorem is true, in large part, because. lim
Recall the statement of Green s Theorem: If is the smooth positively oriented boundary of a simply connected region D in the xyplane, and if F is a vector field in the xyplane with continuous first partial
More informationMVE041 Flervariabelanalys
MVE041 Flervariabelanalys 201516 This document contains the learning goals for this course. The goals are organized by subject, with reference to the course textbook Calculus: A Complete Course 8th ed.
More informationAPPLICATIONS OF THE INTEGRAL
6 APPLICATIONS OF THE INTEGRAL 6. Area Between Two Curves Preliminar Questions b. What is the area interpretation of f g d if f g? a Because f g, b a f g d represents the area of the region bounded between
More informationCHAPTER 13. Definite Integrals. Since integration can be used in a practical sense in many applications it is often
7 CHAPTER Definite Integrals Since integration can be used in a practical sense in many applications it is often useful to have integrals evaluated for different values of the variable of integration.
More informationMATH 2030: MATRICES. Av = 3
MATH 030: MATRICES Introduction to Linear Transformations We have seen that we ma describe matrices as smbol with simple algebraic properties like matri multiplication, addition and scalar addition In
More informationThe Rectangular Coordinate System
The Mathematics Competenc Test The Rectangular Coordinate Sstem When we write down a formula for some quantit,, in terms of another quantit,, we are epressing a relationship between the two quantities.
More informationArea Between Curves. The idea: the area between curves y = f(x) and y = g(x) (if the graph of f(x) is above that of g(x)) for a x b is given by
MATH 42, Fall 29 Examples from Section, Tue, 27 Oct 29 1 The First Hour Area Between Curves. The idea: the area between curves y = f(x) and y = g(x) (if the graph of f(x) is above that of g(x)) for a x
More informationMA 105 D1 Lecture 20
MA 105 D1 Lecture 20 Ravi Raghunathan Department of Mathematics Autumn 2014, IIT Bombay, Mumbai Orientation and the Jordan curve theorem Green s Theorem Various examples Green s Theorem Other forms of
More informationLine and surface integrals: Solutions
hapter 5 Line and surface integrals: olutions Example 5.1 Find the work done by the force F(x, y) x 2 i xyj in moving a particle along the curve which runs from (1, ) to (, 1) along the unit circle and
More information1. AREA BETWEEN the CURVES
1 The area between two curves The Volume of the Solid of revolution (by slicing) 1. AREA BETWEEN the CURVES da = {( outer function ) ( inner )} dx function b b A = da = [y 1 (x) y (x)]dx a a d d A = da
More informationMULTIPLE INTEGRATION NOTES AND EXERCISES. Antony Foster
MULTIPL INTGRATION NOTS AN XRCISS by Antony Foster Overview In this discussion we consider the integral of a function of two variables z f(x, y) over a region in the xyplane (space) and the integral
More informationLesson 03: Kinematics
www.scimsacademy.com PHYSICS Lesson 3: Kinematics Translational motion (Part ) If you are not familiar with the basics of calculus and vectors, please read our freely available lessons on these topics,
More informationCenter of Mass and Centroids :: Guidelines
Center of Mass and Centroids :: Guidelines Centroids of Lines, reas, and Volumes 1.Order of Element Selected for ntegration.continuit.discarding Higer Order Terms.Coice of Coordinates 5.Centroidal Coordinate
More information1. a. standard form of a parabola with. 2 b 1 2 horizontal axis of symmetry 2. x 2 y 2 r 2 o. standard form of an ellipse centered
Conic Sections. Distance Formula and Circles. More on the Parabola. The Ellipse and Hperbola. Nonlinear Sstems of Equations in Two Variables. Nonlinear Inequalities and Sstems of Inequalities In Chapter,
More informationSolutions for Review Problems
olutions for Review Problems 1. Let be the triangle with vertices A (,, ), B (4,, 1) and C (,, 1). (a) Find the cosine of the angle BAC at vertex A. (b) Find the area of the triangle ABC. (c) Find a vector
More informationMath 30G: Calculus Homework 12
Math G: Calculus Homework ue b Frida, Oct. 9, (6 pm O NOT EISTIBUTE THIS SOLUTION FILE. Let be the region inside the unit circle centered at the origin, let be the right half of, and let B be the bottom
More informationProblem 1 (10 pts) Find the radius of convergence and interval of convergence of the series
1 Problem 1 (10 pts) Find the radius of convergence and interval of convergence of the series a n n=1 n(x + 2) n 5 n 1. n(x + 2)n Solution: Do the ratio test for the absolute convergence. Let a n =. Then,
More informationTrigonometric Substitutions.
E Trigonometric Substitutions. E. The Area of a Circle. The area of a circle of radius (the unit circle) is wellknown to be π. We will investigate this with several different approaches, each illuminating
More informationName Class. Date Section. Test Form A Chapter 11. Chapter 11 Test Bank 155
Chapter Test Bank 55 Test Form A Chapter Name Class Date Section. Find a unit vector in the direction of v if v is the vector from P,, 3 to Q,, 0. (a) 3i 3j 3k (b) i j k 3 i 3 j 3 k 3 i 3 j 3 k. Calculate
More informationEven, odd functions. Main properties of even, odd functions. Sine and cosine series. Evenperiodic, oddperiodic extensions of functions.
Sine and Cosine Series (Sect..4. Even, odd functions. Main properties of even, odd functions. Sine and cosine series. Evenperiodic, oddperiodic etensions of functions. Even, odd functions. Definition
More information