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Mathematics LibreTexts

6.2E: Exercises for Section 6.2

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    69809
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    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 \(a.\)

    4) Use the disk method to derive the formula for the volume of a trapezoidal cylinder.

    5) Explain when you would use the disk method versus the washer method. When are they interchangeable?

    Volumes by Slicing

    For exercises 6 - 10, draw a typical slice and find the volume using the slicing method for the given volume.

    6) 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,

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

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

    7) 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.

    8) 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

    9) 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.

    10) 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 11 - 16, draw an outline of the solid and find the volume using the slicing method.

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

    12) 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.

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

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

    14) 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.

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

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

    16) 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.

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

    Disk and Washer Method

    For exercises 17 - 24, draw the region bounded by the curves. Then, use the disk or washer method to find the volume when the region is rotated around the \(x\)-axis.

    17) \( x+y=8,\quad x=0\), and \( y=0\)

    18) \( y=2x^2,\quad x=0,\quad x=4,\) and \( y=0\)

    Answer:

    This figure is a graph in the first quadrant. It is a shaded region bounded above by the curve y=2x^2, below by the x-axis, and to the right by the vertical line x=4.

    \(\displaystyle V = \int_0^4 4\pi x^4\, dx \quad=\quad \frac{4096π}{5}\) units3

    19) \( y=e^x+1,\quad x=0,\quad x=1,\) and \( y=0\)

    20) \( y=x^4,\quad x=0\), and \( y=1\)

    Answer:

    This figure is a graph in the first quadrant. It is a shaded region bounded above by the line y=1, below by the curve y=x^4, and to the left by the y-axis.

    \(\displaystyle V = \int_0^1 \pi\left( 1^2 - \left( x^4\right)^2\right)\, dx = \int_0^1 \pi\left( 1 - x^8\right)\, dx \quad = \quad \frac{8π}{9}\) units3

    21) \( y=\sqrt{x},\quad x=0,\quad x=4,\) and \( y=0\)

    22) \( y=\sin x,\quad y=\cos x,\) and \( x=0\)

    Answer:

    This figure is a shaded region bounded above by the curve y=cos(x), below to the left by the y-axis and below to the right by y=sin(x). The shaded region is in the first quadrant.

    \(\displaystyle V = \int_0^{\pi/4} \pi \left( \cos^2 x - \sin^2 x\right) \, dx = \int_0^{\pi/4} \pi \cos 2x \, dx \quad=\quad \frac{π}{2}\) units3

    23) \( y=\dfrac{1}{x},\quad x=2\), and \( y=3\)

    24) \( x^2−y^2=9\) and \( x+y=9,\quad y=0\) and \( x=0\)

    Answer:

    This figure is a graph in the first quadrant. It is a shaded region bounded above by the line x + y=9, below by the x-axis, to the left by the y-axis, and to the left by the curve x^2-y^2=9.

    \(V = 207π\) units3

    For exercises 25 - 32, draw the region bounded by the curves. Then, find the volume when the region is rotated around the \(y\)-axis.

    25) \( y=4−\dfrac{1}{2}x,\quad x=0,\) and \( y=0\)

    26) \( y=2x^3,\quad x=0,\quad x=1,\) and \( y=0\)

    Answer:

    This figure is a graph in the first quadrant. It is a shaded region bounded above by the curve y=2x^3, below by the x-axis, and to the right by the line x=1.

    \(V = \frac{4π}{5}\) units3

    27) \( y=3x^2,\quad x=0,\) and \( y=3\)

    28) \( y=\sqrt{4−x^2},\quad y=0,\) and \( x=0\)

    Answer:

    This figure is a graph in the first quadrant. It is a quarter of a circle with center at the origin and radius of 2. It is shaded on the inside.

    \(V = \frac{16π}{3}\) units3

    29) \( y=\dfrac{1}{\sqrt{x+1}},\quad x=0\), and \( x=3\)

    30) \( x=\sec(y)\) and \( y=\dfrac{π}{4},\quad y=0\) and \( x=0\)

    Answer:

    This figure is a graph in the first quadrant. It is a shaded region bounded above by the line y=pi/4, to the right by the curve x=sec(y), below by the x-axis, and to the left by the y-axis.

    \(V = π\) units3

    31) \( y=\dfrac{1}{x+1},\quad x=0\), and \( x=2\)

    32) \( y=4−x,\quad y=x,\) and \( x=0\)

    Answer:

    This figure is a graph in the first quadrant. It is a shaded triangle bounded above by the line y=4-x, below by the line y=x, and to the left by the y-axis.

    \(V = \frac{16π}{3}\) units3

    For exercises 33 - 40, draw the region bounded by the curves. Then, find the volume when the region is rotated around the \(x\)-axis.

    33) \( y=x+2,\quad y=x+6,\quad x=0\), and \( x=5\)

    34) \( y=x^2\) and \( y=x+2\)

    Answer:

    This figure is a graph above the x-axis. It is a shaded region bounded above by the line y=x+2, and below by the parabola y=x^2.

    \(V = \frac{72π}{5}\) units3

    35) \( x^2=y^3\) and \( x^3=y^2\)

    36) \( y=4−x^2\) and \( y=2−x\)

    Answer:

    This figure is a shaded region bounded above by the curve y=4-x^2 and below by the line y=2-x.

    \(V = \frac{108π}{5}\) units3

    37) [T] \( y=\cos x,\quad y=e^{−x},\quad x=0\), and \( x=1.2927\)

    38) \( y=\sqrt{x}\) and \( y=x^2\)

    Answer:

    This figure is a graph in the first quadrant. It is a shaded region bounded above by the curve y=squareroot(x), below by the curve y=x^2.

    \(V = \frac{3π}{10}\) units3

    39) \( y=\sin x,\quad y=5\sin x,\quad x=0\) and \( x=π\)

    40) \( y=\sqrt{1+x^2}\) and \( y=\sqrt{4−x^2}\)

    Answer:

    This figure is a shaded region bounded above by the curve y=squareroot(4-x^2) and, below by the curve y=squareroot(1+x^2).

    \(V = 2\sqrt{6}π\) units3

    For exercises 41 - 45, draw the region bounded by the curves. Then, use the washer method to find the volume when the region is revolved around the \(y\)-axis.

    41) \( y=\sqrt{x},\quad x=4\), and \( y=0\)

    42) \( y=x+2,\quad y=2x−1\), and \( x=0\)

    Answer:

    This figure is a graph in the first quadrant. It is a shaded region bounded above by the line y=x+2, below by the line y=2x-1, and to the left by the y-axis.

    \(V = 9π\) units3

    43) \( y=\dfrac{3}{x}\) and \( y=x^3\)

    44) \( x=e^{2y},\quad x=y^2,\quad y=0\), and \( y=\ln(2)\)

    Answer:

    This figure is a graph in the first quadrant. It is a shaded region bounded above by the curve y=ln(2), below by the x-axis, to the left by the curve x=y^2, and to the right by the curve x=e^(2y).

    \(V = \dfrac{π}{20}(75−4\ln^5(2))\) units3

    45) \( x=\sqrt{9−y^2},\quad x=e^{−y},\quad y=0\), and \( y=3\)

    46) Yogurt containers can be shaped like frustums. Rotate the line \( y=\left(\frac{1}{m}\right)x\) around the \(y\)-axis to find the volume between \( y=a\) and \( y=b\).

    This figure has two parts. The first part is a solid cone. The base of the cone is wider than the top. It is shown in a 3-dimensional box. Underneath the cone is an image of a yogurt container with the same shape as the figure.

    Answer:
    \(V = \dfrac{m^2π}{3}(b^3−a^3)\) units3

    47) Rotate the ellipse \( \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) around the \(x\)-axis to approximate the volume of a football, as seen here.

    This figure has an oval that is approximately equal to the image of a football.

    48) Rotate the ellipse \( \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) around the \(y\)-axis to approximate the volume of a football.

    Answer:
    \(V = \frac{4a^2bπ}{3}\) units3

    49) A better approximation of the volume of a football is given by the solid that comes from rotating \( y=\sin x) around the \(x\)-axis from \( x=0\) to \( x=π\). What is the volume of this football approximation, as seen here?

    This figure has a 3-dimensional oval shape. It is inside of a box parallel to the x axis on the bottom front edge of the box. The y-axis is vertical to the solid.

    For exercises 51 - 56, find the volume of the solid described.

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

    52) 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:
    \(V = \frac{2ab^2π}{3}\) units3

    53) Bore a hole of radius a down the axis of \(a\) right cone and through the base of radius \(b\), as seen here.

    This figure is an upside down cone. It has a radius of the top as “b”, center at “a”, and height as “b”.

    54) Find the volume common to two spheres of radius \(r\) with centers that are \(2h\) apart, as shown here.

    This figure has two circles that intersect. Both circles have radius “r”. There is a line segment from one center to the other. In the middle of the intersection of the circles is point “h”. It is on the line segment.

    Answer:
    \(V = \frac{π}{12}(r+h)^2(6r−h)\) units3

    55) Find the volume of a spherical cap of height \(h\) and radius \(r\) where \(h<r\), as seen here.

    This figure a portion of a sphere. This spherical cap has radius “r” and height “h”.

    56) Find the volume of a sphere of radius \(R\) with a cap of height \(h\) removed from the top, as seen here.

    This figure is a sphere with a top portion removed. The radius of the sphere is “R”. The distance from the center to where the top portion is removed is “R-h”.

    Answer:
    \(V = \dfrac{π}{3}(h+R)(h−2R)^2\) 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.

     


    6.2E: Exercises for Section 6.2 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.