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Lecture Comments (3)

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Post by Steve Denton on October 22, 2012

At 21:30 in example 4, how exactly did the 1/2 cancel with the division by 2 in the next term?



1 answer

Last reply by: Jingwei Xie
Mon Apr 21, 2014 10:06 PM

Post by Steve Denton on October 22, 2012

At 14:40 or so, isn't the height of an isosceles triangle s(root3)/2 and the area then, not s^2, but s^2(root3)/4?

Revolving Solids Known Cross Sections

  • Finding volume of solid with known cross sections and bounds
    • Find area of known cross-section
    • Substitute bounding functions to variables in area equation appropriately
    • Evaluate integral of area of known cross-section
  • Knowing derivations of disk, washer, and cylindrical formulas dramatically improves understanding of known cross-sections volumes

Revolving Solids Known Cross Sections

Find the volume of a solid whose base is defined by the region bounded by y = x, y = 0, and x = 1 with the diameters of semi-circle cross-sections along the y-axis.
  • Area of a semi-circle = [1/2] πr2
  • V = [1/2] π∫01 ([x/2])2 dx
  • V = [(π)/8] [(x3)/3] |01
V = [(π)/24]
Find the volume of a solid whose base is defined by the region bounded by y = sinx and y = 0 from x = 0 to x = π with the square cross-sections.
  • A = l2
  • V = ∫0π y2 dx
  • V = ∫0π sin2 x  dx
  • V = ∫0π ([(1 − cos2x)/2]) dx
  • V = [1/2] ∫0π 1 − cos2x  dx
  • V = [1/2] (∫0π dx − ∫0π cos2x dx)
  • u = 2x
  • du = 2  dx
  • V = [1/2] (x |0π − [1/2] sin2x |0π)
V = [(π)/2]
Find the volume of a solid whose base is defined by the region bounded by y = x2 and y = x with equilateral triangle cross-sections along the y-axis.
  • Area of equilateral triangle = [(√3)/4] l2
  • Intersection points
  • x2 = x
  • x = 0, 1
  • In between 0 and 1, x ≥ x2
  • l = x − x2
  • V = [(√3)/4] ∫01 (x − x2)2 dx
  • V = [(√3)/4] ∫01 x4 − 2x3 + x2  dx
  • V = [(√3)/4] ([(x5)/5] − [(x4)/2] + [(x3)/3]) |01
  • V = [(√3)/4] ([1/5] − [1/2] + [1/3] − 0)
V = [(√3)/120]
Find the volume of a solid whose base is defined by the region bounded by y = x2 and y = x with square cross-sections along the x-axis.
  • This time, we're taking triangles parallel to the x-axis
  • We need to rewrite both of the equations in terms of x
  • x2 = y
  • x = ±√y
  • The intersection is in the first quadrant, so we only care about the positive value
  • x = √y
  • Intersects at y = 0, 1
  • A = l2
  • l = √y − y
  • V = ∫01 (√y − y)2 dy
  • V = ∫01 (y − 2y√y + y2) dy
  • V = ∫01 (y − 2 y[3/2] + y2) dy
  • V = [(y2)/2] − [(4 y[5/2])/5] + [(y3)/3] |01
  • V = [1/2] − [4/5] + [1/3]
V = [1/30]
Find the volume of a solid whose base is defined by the region bounded by y = 0, y = 2, x = 0, and x = 6 with semi-circle cross-sections along the x-axis.
  • The area bound by those lines is simply a square, with lines parallel to the axes.
  • A = [1/2] πr2
  • r = [6/2]
  • V = [(π)/2] ∫02 32 dy
  • V = [(π)/2] 9x |02
V = 9π
Find the volume of a solid whose base is defined by the region bounded by y = x2 + 1 and y = −x2 + 3 with semi-circle cross-sections along the y-axis.
  • Find their intersection points
  • x2 + 1 = −x2 + 3
  • 2x2 = 2
  • x = −1, 1
  • A = [1/2] πr2
  • r = [((−x2 + 3) − (x2 + 1))/2]
  • V = [(π)/2] ∫−11 (−x2 + 1)2  dx
  • V = [(π)/2] ∫−11 x4 − 2x2 + 1  dx
  • V = [(π)/2] ([(x5)/5] − [(2x3)/3] + x) |−11
  • V = [(π)/2] ([1/5] − [2/3] + 1 − ([(−1)/5] − [(−2)/3] − 1))
  • V = [(π)/2] ([1/5] − [2/3] + 1 + [1/5] − [2/3] + 1)
V = [(8π)/15]
Find the volume of a solid whose base is defined by the region bounded by y = cosx and y = 0 from x = 0 to x = [(π)/2] with square cross-sections along the y-axis.
  • V = ∫0[(π)/2] cos2 x  dx
  • V = ∫0[(π)/2] [(1 + cos2x)/2] dx
  • V = [1/2] (x + [1/2] sin2x) |0[(π)/2]
  • V = [1/2] ([(π)/2] + 0)
V = [(π)/4]
Find the volume of a solid whose base is defined by the region bounded by y = √{1 − x2} and y = 0 with semi-circle cross-sections along the y-axis.
  • V = [(π)/2] ∫−11 ([(√{1 − x2})/2])2 dx
  • V = [(π)/8] ∫−11 1 − x2  dx
  • We can use symmetry
  • V = [(π)/4] ∫01 1 − x2  dx
  • V = [(π)/4] (x − [(x3)/3]) |01
  • V = [(π)/4] (1 − [1/3] − 0)
V = [(π)/6]
Find the volume of a solid whose base is defined by the region bounded by x = y3 and x = y2 with square cross-sections along the x-axis.
  • V = ∫01 (y2 − y3)2 dy
  • V = ∫01 y4 − 2 y5 + y6  dy
  • V = ([(y5)/5] − [(y6)/3] + [(y7)/7]) |01
  • V = [1/5] − [1/3] + [1/7]
V = [1/105]
Find the volume of a solid whose base is defined by the region bounded by x = y3 and x = y2 with square cross-sections along the y-axis.
  • Rewrite in terms of y
  • y = 3√{x}   and  y = ±√x
  • First quadrant again, so we're only interested in the positive
  • V = ∫01 (√x − 3√{x})2 dx
  • V = ∫01 x − 2 x[5/6] + x[2/3]  dx
  • V = ([(x2)/2] − [(12 x[11/6])/11] + [(3 x[5/3])/5]) |01
  • V = [1/2] − [12/11] + [3/5]
V = [1/110]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Revolving Solids Known Cross Sections

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Revolving Solids Known Cross Sections 0:08
    • Example 1
  • Example 2 6:01
  • Example 3 11:03
  • Example 4 17:29
  • Example 5 22:19