MATH-27-Lecture-Guide-Unit-3-for-AY-2018-19 (1).pdf - MATH...

This preview shows page 1 out of 22 pages.

We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
Numerical Analysis
The document you are viewing contains questions related to this textbook.
Chapter 4 / Exercise 2
Numerical Analysis
Burden/Faires
Expert Verified

Unformatted text preview: MATH 27 LECTURE GUIDE UNIT 3. APPLICATIONS OF THE DEFINITE INTEGRAL The definite integral has numerous applications especially in the fields of physics, engineering, and probability and statistics. In this unit, we discuss some applications of the definite integral. In particular, we deal with problems involving certain geometric quantities such as the area of a region, the volume and surface area of a solid, the length of an arc, and the center of mass of a region and of a solid. Our goals for this unit are as follows. By the end of the unit, you should be able to solve for the area of a plane region; solve for the volume of the solids of revolution using disk and cylindrical shell methods; identify when to use disk or cylindrical shell method; find centroids of plane area and solids of revolution; find the length of arc of a curve; and solve for the area of surfaces of revolution. Note: It is expected that you already know how to evaluate definite integrals. Most examples and exercise problems will only require setting up the definite integral that will solve for the measure of the object required. In case you forgot, this is how definite integrals are evaluated: RECALL: = + , then = − . MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) If is continuous on [, ] and 1 3.1 Area of a Plane Region Using vertical strips, Using horizontal strips, = = = = Area: Area: MUST REMEMBER!!! To get area of a plane region, integrate area of a strip. Use the length of the strip for the length and or for the width. Length − ( − ) − ( − ) Vertical strip: Horizontal strip: Width ILLUSTRATION: 1. Determine the area of the region bounded by = and = , Solution: When ≤ ≤ , = lies above = . Using vertical strips, the area of the described region is given by ≤≤ . = = ∫ ( − ) =∫ − = =( √ √ √ √ + ) − (− − ) = √ square units. Note: Using horizontal strips to solve the problem would be difficult. (Why?) MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) ⁄ = (− − | ⁄ 2 ILLUSTRATION: 2. Using horizontal strips, set-up the definite integral that will give the area of the region bounded by = and = , ≤ ≤ . Solution: When ≤ ≤ , = (which is equivalent to = ) lies on the right of = (which is equivalent to = ). Using horizontal strips, the area of the described region is given by = = = = ∫ ( − ) = = − + TO DO: 1. Using vertical strips, set-up the definite integral that will give the area of the region above. (Hint: Split the region into two. Solve for the points of intersection of = and = and of = and = to get the proper bounds of the integral.) MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 2. Evaluate the integrals in the previous item and in + . Compare the results. 3 TRY THIS! Use definite integral to determine the area of the following regions. Use (a) vertical strips and (b) horizontal strips. 1. 2. 3. 4. 5. 6. 7. the region bounded by = + and = , ≤ ≤ the region bounded by = √ and = the region bounded by = − and = the region bounded by = − and = − the region bounded by = − + and = − + + the region bounded by = and = , ≤ ≤ the region bounded by = and = − , ≤ ≤ For more exercises, you can refer to: Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage Learning Asia Pte. Ltd., pp. 442-445 James Stewart. (2013). Calculus, 7th Edition. Philippines: Cengage Learning Asia Pte. Ltd., pp. 349-350 For an online tutorial, follow these links: MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 4 3.2 Volume of a Solid of Revolution Volume of a cylinder = (area of a cross section) x (height) Consider the solid on the right. Let ∈ , and be the area of crosssection of the object at . Let ∆ be a certain thickness of the cross-section at . (This cross-section is perpendicular to the −axis.) Volume of a slice = Appoximate volume of the solid = (using several slices) Volume of solid = (as a Riemann sum) Volume of solid = (as a definite integral) where is an area of a cross-section. If the cross-sections that will be used are perpendicular to the −axis, the volume of a solid will be given by = ⋅ , where is an area of a cross-section and is the thickness of a slice. TO DO: Consider the solid formed by revolving the region bounded by = √, the −axis, and the line = about the −axis. To get the volume of a solid of revolution using DISKS, consider a strip of the region that is perpendicular to the axis of revolution. (Revolving this strip generates a disk.) If the strips are vertical, = ⋅ where is an area of a cross-section. Hence, = [ ] ⋅ where is the radius of a cross-section. = MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) A solid of revolution is generated by revolving a plane region about an axis of revolution that is either tangent to the region or does not pass through the region. The resulting solid has a circular cross-section: an area of ⋅ with the center on the axis of revolution. 5 If the strips are horizontal, = [ ] ⋅ where is the radius of a cross-section. ILLUSTRATION Let be the region in the first quadrant bounded by = , the −axis, and the −axis. Using disks, set-up the definite integral that will solve for the volume of the solid generated by revolving about (a) the −axis and (b) the −axis. Solution: a. Since the axis of revolution is the −axis, we need to use vertical strips. For a vertical strip, the radius of the corresponding cross-section is given by = . Noting that for the described solid, ≤ ≤ , we obtain its volume to be = ∫ [ ] ⋅ = ∫ b. Since the axis of revolution is the −axis, we need to use horizontal strips. Note that for the described region, ≤ ≤ . Hence, the function = is defined. Now, for a horizontal strip, the radius of the corresponding cross-section is given by = . Thus, the volume of the described solid is = ∫ [ ] ⋅ = ∫ TO DO: a. the line = − = b. the line = = MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) Let be the region bounded by = and the lines = and = −. Using disks, set-up the definite integral that will solve for the volume of the solid generated by revolving about 6 To get the volume of a solid of revolution (with a cavity) using WASHERS, consider a strip of the region that is perpendicular to the axis of revolution. (Revolving this strip generates a disk.) If the strips are vertical, If the strips are horizontal, = ∫ [( ) − ( ) ] ⋅ = ∫ [( ) − ( ) ] ⋅ where is the outer radius of a crosssection and is the inner radius of a cross-section. where is the outer radius of a crosssection and is the inner radius of a cross-section. ILLUSTRATION Let be the region bounded by = − + and = − + . Using washers, set-up the definite integral that will solve for the volume of the solid generated by revolving about (a) the −axis and (b) the line = −. Solution: a. Since the axis of revolution is the −axis, we need to use horizontal strips. For a horizontal strip, we need to determine the inner and outer radii of the corresponding cross-section. Note that = − + is equivalent to = ±√ − , where ≤ ≤ . However, for the described solid, ≤ ≤ . So, we choose = √ − , where ≤ ≤ . Hence, the outer radius of a crosssection is given by = √ − , where ≤ ≤ . On the other hand, the inner radius of a cross-section is given by = − since = − + is equivalent to = − . Thus, the volume of the described solid is given by = ∫ [( ) − ( ) ] ⋅ − = ∫ [(√ − ) − ( ) ] MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 7 ILLUSTRATION (cont’n.) b. Since the axis of revolution is the line = −, we need to use vertical strips. For a vertical strip, the inner and outer radii of the corresponding crosssection are given by = − + − − and = − + − − , respectively. Noting that for the described solid, ≤ ≤ , we obtain its volume to be = ∫ [( ) − ( ) ] ⋅ = [( − + − − ) − ( − + − − ) ] TO DO: Consider the region described above. Using washers, set-up the definite integral that will solve for the volume of the solid generated by revolving the region about 1. the −axis = 2. the line = = TRY THIS! Using disks or washers, set-up the definite integral that will solve for the volume of the solid of revolution generated by revolving the described region about a given axis of revolution. If it is “easy” to evaluate the integral, determine the volume of the solid. 1. the region bounded by = √, the line = , and the −axis revolved about a. the −axis b. the line = c. the line = − revolved about 3. the region bounded by = and = , ≤ ≤ revolved about a. the −axis b. the −axis 4. the region bounded by = + − and = − revolved about a. the −axis b. the line = c. the line = 5. the region bounded by = and = − revolved about a. the −axis b. the line = c. the line = MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 2. the region bounded by = and = , ≤ ≤ a. the line = b. the line = − 8 For more exercises, you can refer to: Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage Learning Asia Pte. Ltd., pp. 453-456 James Stewart. (2013). Calculus, 7th Edition. Philippines: Cengage Learning Asia Pte. Ltd., pp. 360-363 For an online tutorial, follow these links: To get the volume of a solid of revolution using CYLINDRICAL SHELLS, Volume of a cylindrical shell = ⋅ ⋅ ∆ ⋅ where = + (or the average radius) and ∆ is the thickness of the shell MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) Consider a strip of the region that is parallel to the axis of revolution. (Revolving this strip generates a shell.) 9 If the strips are vertical, If the strips are horizontal, = ∫ ⋅ ⋅ = ∫ ⋅ ⋅ where is the distance of a strip from the axis of revolution, and is the height of a strip. where is the distance of a strip fromt the axis of revolution, and is the length of a strip. ILLUSTRATION Let be the region bounded by = − + and = − + . Using cylindrical shells, set-up the definite integral that will solve for the volume of the solid generated by revolving about (a) the −axis and (b) the line = . Solution: a. Since the axis of revolution is the −axis, we need to use vertical strips. For a vertical strip, its distance from the −axis is given by = , and its height is given by = − + − − + . Noting that for the described solid, ≤ ≤ , we obtain its volume to be = ∫ ⋅ ⋅ = ∫ [ − + − − + ] b. Since the axis of revolution is the line = , we need to use horizontal strips. For a horizontal strip, we need to determine its distance from the line = and its length. It is easy to see that the distance of a horizontal strip from the line = is given by = − . Thus, the volume of the described solid is given by = ∫ ⋅ ⋅ = ∫ − (√ − − − ) MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) For the length, note that = − + is equivalent to = ±√ − , where ≤ ≤ . However, for the described solid, ≤ ≤ . So, we choose = √ − , where ≤ ≤ . On the other hand, = − + is equivalent − to = . Hence, the length of the strip is given by − = √ − − , where ≤ ≤ . 10 TO DO: Consider the region described above. Using cylindrical shells, set-up the definite integral that will solve for the volume of the solid generated by revolving the region about 1. the −axis Answer: = 2. the line = Answer: = TRY THIS! Using cylindrical shells, set-up the definite integral that will solve for the volume of the solid of revolution generated by revolving the described region about a given axis of revolution. If it is “easy” to evaluate the integral, determine the volume of the solid. 1. the region bounded by = √, the line = , and the −axis revolved about a. the −axis b. the line = c. the line = − 2. the region bounded by = and = , ≤ ≤ a. the −axis b. the line = revolved about 3. the region bounded by = and = , ≤ ≤ revolved about a. the −axis b. the −axis 4. the region bounded by = + − and = − revolved about a. the −axis b. the line = c. the line = 5. the region bounded by = and = − revolved about a. the −axis b. the line = c. the line = For more exercises, you can refer to: Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage Learning Asia Pte. Ltd., pp. 462-465 James Stewart. (2013). Calculus, 7th Edition. Philippines: Cengage Learning Asia Pte. Ltd., pp. 366-368 For an online tutorial, follow these links: MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 11 3.3 Centroid of a Plane Region Center of mass along a line (discrete case) Consider several weights distributed along a line. 0 1 2 3 4 5 6 7 8 9 10 Let be the total mass and be the moment of mass which is a product of a mass and its ̅ is the coordinate of the center of mass along the line, then directed distance from a point. If ̅= ∑= = ∑= Center of mass of a rod In a homogeneous rod, the mass is directly proportional to the length: = ⋅ where is the constant linear density and is the length of the rod. In a non-homogeneous rod, the mass is directly proportional to the length: = ⋅ where is the linear density at point on the rod. If is the linear density of a rod, then the total mass is = mass is = ⋅ and the moment of ⋅ ⋅ . TO DO: Center of mass of a plane region Consider a region on the −plane. Assumptions: the region is a lamina (the object is a “thin” sheet) and of constant area density, In effect, the center of mass will just be a function of the area of the region and its distances from the −axis and the −axis. (No “mass” in here!) In this case, the center of mass is a centroid of the plane region. MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) The length of a rod is meters and the linear density of the rod at a point meters from one end is + kg/m. Determine the center of mass of the rod. 12 Consider the region bounded by = and = , ≤ ≤ . Suppose the area density is . (Simplification: constant area density) = Total mass: = ⋅ Moments of mass: = ⋅ ⋅ − = ⋅ ⋅ − = ̅, ̅ , where Center of mass: ̅= ̅= and Using vertical strips, ⋅ [ − ] ⋅ = ⋅ ⋅ [ − ] ⋅ + = ⋅ ⋅ [ − = ⋅ In general, using horizontal or vertical strip, ⋅ ⋅ = ⋅ = ⋅ = − ⋅ ⋅ − ⋅ ⋅ , where : . ILLUSTRATION 1. Set-up the definite integrals that will give the coordinates of the centroid of the region bounded by = and = , ≤ ≤ . Solution: We use vertical strips. (Why?) When ≤ ≤ , = lies above = . Hence, we have = = = = ⋅ [ − ] ⋅ ⋅ ⋅ [ − ] ⋅ = − ⋅ − + + ⋅ [ − ] ⋅ − . ̅, ̅ =( Thus, the centroid of the described region is at , ). MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) = 13 ILLUSTRATION 2. Using horizontal strips, set-up the definite integrals that will give the coordinates of the centroid of the region bounded by = and = . Solution: The curves = and = intersect at the points , and , , and for the described region, ≤ ≤ and ≤ ≤ . Now, note that = is equivalent to = ±√. However, we choose = √ since ≤ ≤ for the described region. On the other hand, = is equivalent to = √. Hence, using horizontal strips, we have = = ⋅ [ ( √ − = = = = = = − ] ⋅ √) + ⋅ ⋅ [ √+√ ( √ − − ] ⋅ √) ⋅ ⋅ [ − ( √ − √). ⋅ ̅, ̅ =( Thus, the centroid of the described region is at , ). TRY THIS! Set-up the definite integrals that will give the coordinates of the centroid of the following regions. If it is “easy” to evaluate the integrals, determine the coordinates of the centroid of the region. 1. 2. 3. 4. 5. the region bounded by = − + , = , and the −axis the region bounded by = and = − + the region bounded by = and = , ≤ ≤ the circle defined by the equation + = , where is a positive constant the semi-circle bounded by = √ − (where is a positive constant) and the −axis Ron Larson & Bruce H. Edwards. (2016). Calculus, 10th Edition. Philippines: Cengage Learning Asia Pte. Ltd., pp. 494-496 James Stewart. (2013). Calculus, 7th Edition. Philippines: Cengage Learning Asia Pte. Ltd., pp. 585-586 For an online tutorial, follow this link: MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) For more exercises, you can refer to: 14 3.4 Centroid of a Solid of Revolution Point in the three-dimensional space: , , where : directed distance from the −plane : directed distance from the −plane : directed distance from the −plane Consider a region on the −plane revolved about a certain axis of revolution on the −plane. Let be the (constant) density of the solid of revolution. Total mass: = ⋅ Moments of mass: = ⋅ ⋅ − = ⋅ ⋅ − = ⋅ ⋅ − ̅, ̅, ̅ , where Centroid of the solid: ̅= Remark: ̅= ̅ = The centroid of a solid of revolution always lies on the axis of revolution. ̅ or ̅ is dependent on the axis of revolution. Hence, ̅ = , and either Caution! Be careful in the choice of strips: either vertical or horizontal. ILLUSTRATION 1. Let be the region in the first quadrant bounded by = , the −axis, and the −axis. Set-up the definite integrals that will give the coordinates of the centroid of the solid generated by revolving about the line = −. Solution: We first determine the volume of the solid. The inner and outer radii of the corresponding crosssection of a vertical strip are given by = − − = and = − − = + , respectively. Noting that for the described solid, ≤ ≤ , we obtain its volume to be MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) By default, ̅ = . And since the axis of revolution ̅ = −. We are left with ̅, is the line = −, then i.e., we need to determine the total mass and the moment of mass with respect to the −plane . For this problem, it is easier to use vertical strips; hence, we will do so. Equivalently, we will use washers to determine and . 15 ILLUSTRATION (cont’n.) = ∫ ⋅ [( ) − ( ) ] ⋅ = ∫ [ + − ] . Thus, = ⋅ = ∫ ⋅ ⋅ [( ) − ( ) ] ⋅ = ∫ [ + − ] . and = ⋅ − ⋅ = ∫ ⋅ ⋅ ( ⋅ [( ) − ( ) ] ⋅ ) (Note: The directed distance from the −plane is .) = ∫ [ + − ] . ̅, ̅, ̅ = ( Therefore, the centroid of the described solid is at , −, ). Remark: We can also use horizontal strips (cylindrical shells) to solve the problem. However, we should note that the directed distance from the −plane should be in terms of . In this case, the directed distance from the −plane is the average of = (which is the equation of the −axis) and = (which is the equation equivalent to = ). That is, it is . TO DO: Let be the region in the first quadrant bounded by = , the −axis, and the −axis. Set-up the definite integrals that will give the coordinates of the centroid of the solid generated by revolving about the line = . ILLUSTRATION − Let be the region bounded by = √ − and = . Using horizontal strips, set-up the definite integrals that will give the coordinates of the centroid of the solid generated by revolving about the line = . Solution: ̅ = . We are By default, ̅ = . And since the axis of revolution is the line = , then ̅, i.e., we need to determine the total mass and the moment of mass with respect left with to the −plane . For this problem, we are required to use horizontal strips. Note that the strips are perpendicular to the axis of revolution. Thus, we will use washers to determine and . MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) 2. 16 ILLUSTRATION (cont’n.) We first determine the volume of the solid. The inner and outer radii of the corresponding cross-section of a horizontal strip are given by − = − √ − and = − , respectively. Noting that for the described solid, ≤ ≤ , we obtain its volume to be = ∫ ⋅ [( ) − ( ) ] ⋅ = ∫ [( − − ) − ( − √ − ) ] Thus, = ⋅ = ∫ ⋅ ⋅ [( ) − ( ) ] ⋅ = ∫ [( − − ) − ( − √ − ) ] . and = ⋅ − ⋅ = ∫ ⋅ ⋅ ( ⋅ [( ) − ( ) ] ⋅ ) (Note: The directed distance from the −plane is .) = ∫ [( − − ) − ( − √ − ) ] . ̅, ̅, ̅ = (, Therefore, the centroid of the described solid is at , ). Remark: We can also use vertical strips (cylindrical shells) to solve the problem. However, we should note that the directed distance from the −plane should be in terms of . In this case, the directed distance from the −plane is the average of = − + where ≥ (which is the equation equivalent to = √ − ) and = − + (which is the equation equivalent to = − ). That is, it is (− +)+ −+ . − Let be the region bounded by = √ − and = . Set-up the definite integrals that will give the coordinates of the centroid of the solid generated by revolving about the line = −. MATH 27 Lecture Guide UNIT 3 (IMSP,UPLB) TO DO: 17 TRY THIS! Set-up the definite integrals that will give the coordinates of the centroid of the solid of revolution generated by revolving the described region about a given axis of revolution. If it is “easy” to evaluate the integrals, determine the coordinates of the cent...
View Full Document

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture