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2.2E: Exercises

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    168408
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    Reading Questions

    1. What is the definition of a solid's cross-section?
    2. What is the general definite integral formula for the volume of a solid using the slicing method, where \(A(x)\) is the cross-sectional area at \(x\)?
    3. List the three steps in the problem-solving strategy for finding volumes by the slicing method.
    4. In Example 1.2.1, what geometric principle is used to find the side length \(s\) of the square cross-section in terms of its distance \(x\) from the apex of the pyramid?
    5. If the cross-sections of a solid are taken perpendicular to the y-axis, what variable must the area function and the integration be in terms of?
    6. According to the text, what is the mathematical definition of a cylinder? Does its base have to be a circle?

    1) Derive the formula for the volume of a sphere using the slicing method.

    2) Use the slicing method to derive the formula for the volume of a cone.

    3) Use the slicing method to derive the formula for the volume of a tetrahedron with side length \(a.\)

    Volumes by Slicing

    For exercises 4 - 8, draw a typical slice and find the volume using the slicing method for the given volume.

    4) A pyramid with height 6 units and square base of side 2 units, as pictured here.

    This figure is a pyramid with base width of 2 and height of 6 units.

    Solution:
    Here the cross-sections are squares taken perpendicular to the \(y\)-axis.
    We use the vertical cross-section of the pyramid through its center to obtain an equation relating \(x\) and \(y\).
    Here this would be the equation, \( y = 6 - 6x \). Since we need the dimensions of the square at each \(y\)-level, we solve this equation for \(x\) to get, \(x = 1 - \tfrac{y}{6}\).
    This is half the distance across the square cross-section at the \(y\)-level, so the side length of the square cross-section is, \(s = 2\left(1 - \tfrac{y}{6}\right).\)
    Thus, we have the area of a cross-section is,

    \(A(y) = \left[2\left(1 - \tfrac{y}{6}\right)\right]^2 = 4\left(1 - \tfrac{y}{6}\right)^2.\)

    \(\begin{align*} \text{Then},\quad V &= \int_0^6 4\left(1 - \tfrac{y}{6}\right)^2 \, dy \\[5pt]
    &= -24 \int_1^0 u^2 \, du, \quad \text{where} \, u = 1 - \tfrac{y}{6}, \, \text{so} \, du = -\tfrac{1}{6}\,dy, \quad \implies \quad -6\,du = dy \\[5pt]
    &= 24 \int_0^1 u^2 \, du = 24\dfrac{u^3}{3}\bigg|_0^1 \\[5pt]
    &= 8u^3\bigg|_0^1 \\[5pt]
    &= 8\left( 1^3 - 0^3 \right) \quad= \quad 8\, \text{units}^3 \end{align*}\)

    5) A pyramid with height 4 units and a rectangular base with length 2 units and width 3 units, as pictured here.

    This figure is a pyramid with base width of 2, length of 3, and height of 4 units.

    6) A tetrahedron with a base side of 4 units,as seen here.

    This figure is an equilateral triangle with side length of 4 units.

    Answer
    \(V = \frac{32}{3\sqrt{2}} = \frac{16\sqrt{2}}{3}\) units3

    7) A pyramid with height 5 units, and an isosceles triangular base with lengths of 6 units and 8 units, as seen here.

    This figure is a pyramid with a triangular base. The view is of the base. The sides of the triangle measure 6 units, 8 units, and 8 units. The height of the pyramid is 5 units.

    8) A cone of radius \( r\) and height \( h\) has a smaller cone of radius \( r/2\) and height \( h/2\) removed from the top, as seen here. The resulting solid is called a frustum.

    This figure is a 3-dimensional graph of an upside down cone. The cone is inside of a rectangular prism that represents the xyz coordinate system. the radius of the bottom of the cone is “r” and the radius of the top of the cone is labeled “r/2”.

    Answer
    \(V = \frac{7\pi}{12} hr^2\) units3

    For exercises 9 - 14, draw an outline of the solid and find the volume using the slicing method.

    9) The base is a circle of radius \( a\). The slices perpendicular to the base are squares.

    10) The base is a triangle with vertices \( (0,0),(1,0),\) and \( (0,1)\). Slices perpendicular to the \(xy\)-plane are semicircles.

    Answer

    This figure shows the x-axis and the y-axis with a line starting on the x-axis at (1,0) and ending on the y-axis at (0,1). Perpendicular to the xy-plane are 4 shaded semi-circles with their diameters beginning on the x-axis and ending on the line, decreasing in size away from the origin.

    \( V = \int_0^1 \frac{\pi(1-x)^2}{8}\, dx \quad = \quad \frac{ \pi }{24}\) units3

    11) The base is the region under the parabola \( y=1−x^2\) in the first quadrant. Slices perpendicular to the \(xy\)-plane are squares.

    12) The base is the region under the parabola \( y=1−x^2\) and above the \(x\)-axis. Slices perpendicular to the \(y\)-axis are squares.

    Answer

    This figure shows the x-axis and the y-axis in 3-dimensional perspective. On the graph above the x-axis is a parabola, which has its vertex at y=1 and x-intercepts at (-1,0) and (1,0). There are 3 square shaded regions perpendicular to the x y plane, which touch the parabola on either side, decreasing in size away from the origin.

    \( V = \int_0^1 4(1 - y)\,dy \quad = \quad 2\) units3

    13) The base is the region enclosed by \( y=x^2\) and \( y=9.\) Slices perpendicular to the \(x\)-axis are right isosceles triangles.

    14) The base is the area between \( y=x\) and \( y=x^2\). Slices perpendicular to the \(x\)-axis are semicircles.

    Answer

    This figure is a graph with the x and y axes diagonal to show 3-dimensional perspective. On the first quadrant of the graph are the curves y=x, a line, and y=x^2, a parabola. They intersect at the origin and at (1,1). Several semicircular-shaped shaded regions are perpendicular to the x y plane, which go from the parabola to the line and perpendicular to the line.

    \( V = \int_0^1 \frac{\pi}{8}\left( x - x^2 \right)^2 \, dx \quad=\quad \frac{ \pi }{240}\) units3

    For exercises 15 - 16, find the volume of the solid described.

    15) The base is the region between \( y=x\) and \( y=x^2\). Slices perpendicular to the \(x\)-axis are semicircles.

    16) The base is the region enclosed by the generic ellipse \( \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.\) Slices perpendicular to the \(x\)-axis are semicircles.

    Answer
    \(V = \frac{2ab^2 \pi }{3}\) units3

    Contributors

    Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


    This page titled 2.2E: Exercises is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Roy Simpson.

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