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Mathematics LibreTexts

1.1E: Exercises

  • Page ID
    18539
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    Determining Volumes by Slicing

    Exercise \(\PageIndex{1}\)

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

     
    Answer

    4324t.

    Exercise \(\PageIndex{1}\)

    Add exercises text here.

    Answer

    Add texts here. Do not delete this text first.

     

    Exercise \(\PageIndex{2}\)

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

     

    Answer

    Test \(\displaystyle \frac{1}{2} \)

    Exercise \(\PageIndex{3}\)

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

     

    Answer

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    Exercise \(\PageIndex{4}\)

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

     

    Answer

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    Exercise \(\PageIndex{5}\)

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

     

    Answer

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

    Exercise \(\PageIndex{6}\)

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

    Answer

    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,

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

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

    Exercise \(\PageIndex{7}\)

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

     

    Answer

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    Exercise \(\PageIndex{8}\)

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

     

    Answer

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

    Exercise \(\PageIndex{9}\)

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

     

    Answer

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    Exercise \(\PageIndex{10}\)

    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.

     

    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.

    Exercise \(\PageIndex{11}\)

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

     

    Answer

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    Exercise \(\PageIndex{12}\)

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

     

    Answer

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

    Exercise \(\PageIndex{13}\)

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

     

    Answer

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    Exercise \(\PageIndex{14}\)

    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

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

    Exercise \(\PageIndex{15}\)

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

     

    Answer

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    Exercise \(\PageIndex{16}\)

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

     

    Answer

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

    Exercise \(\PageIndex{17}\)

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

     

    Answer

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    Exercise \(\PageIndex{18}\)

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

     

    Answer

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

    Exercise \(\PageIndex{19}\)

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

     

    Answer

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    Exercise \(\PageIndex{20}\)

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

     

    Answer

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

    Exercise \(\PageIndex{21}\)

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

     

    Answer

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    Exercise \(\PageIndex{22}\)

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

     

    Answer

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

    Exercise \(\PageIndex{23}\)

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

     

    Answer

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    Exercise \(\PageIndex{24}\)

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

     

    Answer

    \(V = 207π\) 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.

    Exercise \(\PageIndex{25}\)

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

     

    Answer

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    Exercise \(\PageIndex{26}\)

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

     

    Answer

    \(V = \frac{4π}{5}\) units3

    Exercise \(\PageIndex{27}\)

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

     

    Answer

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    Exercise \(\PageIndex{28}\)

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

     

    Answer

    \(V = \frac{16π}{3}\) units3

    Exercise \(\PageIndex{29}\)

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

     

    Answer

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    Exercise \(\PageIndex{30}\)

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

     

    Answer

    \(V = π\) units3

    Exercise \(\PageIndex{31}\)

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

     

    Answer

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    Exercise \(\PageIndex{32}\)

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

     

    Answer

    \(V = \frac{16π}{3}\) 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.

    Exercise \(\PageIndex{33}\)

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

     

    Answer

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    Exercise \(\PageIndex{34}\)

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

     

    Answer

    \(V = \frac{72π}{5}\) units3

    Exercise \(\PageIndex{35}\)

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

     

    Answer

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    Exercise \(\PageIndex{36}\)

    \( y=4−x^2\) and \( y=2−x\)

     

    Answer

    \(V = \frac{108π}{5}\) units3

    Exercise \(\PageIndex{37}\)

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

     

    Answer

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    Exercise \(\PageIndex{38}\)

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

     

    Answer

    \(V = \frac{3π}{10}\) units3

    Exercise \(\PageIndex{39}\)

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

     

    Answer

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    Exercise \(\PageIndex{40}\)

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

     

    Answer

    \(V = 2\sqrt{6}π\) 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.

    Exercise \(\PageIndex{41}\)

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

     

    Answer

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    Exercise \(\PageIndex{42}\)

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

     

    Answer

    \(V = 9π\) units3

    Exercise \(\PageIndex{43}\)

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

     

    Answer

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    Exercise \(\PageIndex{44}\)

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

     

    Answer

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

    Exercise \(\PageIndex{45}\)

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

     

    Answer

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    Exercise \(\PageIndex{46}\)

    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 \( y=b\).

     

    Answer

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

    Exercise \(\PageIndex{47}\)

    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.

     

    Answer

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    Exercise \(\PageIndex{48}\)

    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

    Exercise \(\PageIndex{49}\)

    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?

     

    Answer

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    Exercise \(\PageIndex{50}\)

    What is the volume of the Bundt cake that comes from rotating \( y=\sin x\) around the \(y\)-axis from \( x=0\) to \( x=π\)?

     

    Answer

    \(V = 2π^2\) units3

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

    Exercise \(\PageIndex{51}\)

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

     

    Answer

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    Exercise \(\PageIndex{52}\)

    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

    Exercise \(\PageIndex{53}\)

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

     

    Answer

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    Exercise \(\PageIndex{54}\)

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

     

    Answer

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

    Exercise \(\PageIndex{55}\)

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

     

    Answer

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    Exercise \(\PageIndex{56}\)

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

     

    Answer

    \(V = \dfrac{π}{3}(h+R)(h−2R)^2\) units3