Calculus

Integration of Functions

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Volume of a Solid with a Known Cross Section

  • In this topic, we will learn how to find the volume of a solid object that has known cross sections.

    We consider solids whose cross sections are common shapes such as triangles, squares, rectangles, trapezoids, and semicircles.

    Definition: Volume of a Solid Using Integration

    Let \(S\) be a solid and suppose that the area of the cross section in the plane perpendicular to the \(x-\)axis is \(A\left( x \right)\) for \(a \le x \le b.\)

    Volume of a solid with known cross section
    Figure 1.

    Then the volume of the solid from \(x = a\) to \(x = b\) is given by the cross-section formula

    \[V = \int\limits_a^b {A\left( x \right)dx}. \]

    Similarly, if the cross section is perpendicular to the \(y-\)axis and its area is defined by the function \(A\left( y \right),\) then the volume of the solid from \(y = c\) to \(y = d\) is given by

    \[V = \int\limits_c^d {A\left( y \right)dy} .\]

    Steps for Finding the Volume of a Solid with a Known Cross Section

    1. Sketch the base of the solid and a typical cross section.
    2. Express the area of the cross section \(A\left( x \right)\) as a function of \(x.\)
    3. Determine the limits of integration.
    4. Evaluate the definite integral \[V = \int\limits_a^b {A\left( x \right)dx}.\]

  • Solved Problems

    Click a problem to see the solution.

    Example 1

    The solid has a base lying in the first quadrant of the \(xy-\)plane and bounded by the lines \(y = x,\) \(x = 1,\) \(y = 0.\) Every planar section perpendicular to the \(x-\)axis is a semicircle. Find the volume of the solid.

    Example 2

    Find the volume of a solid bounded by the elliptic paraboloid \(z = \large{\frac{{{x^2}}}{{{a^2}}}}\normalsize + \large{\frac{{{y^2}}}{{{b^2}}}}\normalsize\) and the plane \(z = 1.\)

    Example 3

    The base of a solid is bounded by the parabola \(y = 1 – {x^2}\) and the \(x-\)axis. Find the volume of the solid if the cross sections are equilateral triangles perpendicular to the \(x-\)axis.

    Example 4

    Find the volume of a regular square pyramid with the base side \(a\) and the altitude \(H.\)

    Example 5

    Find the volume of a solid if the base of the solid is the circle given by the equation \({x^2} + {y^2} = 1,\) and every perpendicular cross section is a square.

    Example 6

    Find the volume of the frustum of a cone if its bases are ellipses with the semi-axes \(A, B,\) and \(a, b\), and the altitute is equal to \(H.\)

    Example 7

    Calculate the volume of a wedge given the bottom sides \(a, b,\) the top side \(c,\) and the altitude \(H.\)

    Example 8

    Find the volume of a regular tetrahedron with the edge \(a.\)

    Example 9

    A wedge is cut out of a circular cylinder with radius \(R\) and height \(H\) by the plane passing through a diameter of the base (Figure \(10\)). Find the volume of the cylindrical wedge.

    Example 10

    The axes of two circular cylinders with the same radius \(R\) intersect at right angles. Find the volume of the solid common to both these cylinders.

    Example 1.

    The solid has a base lying in the first quadrant of the \(xy-\)plane and bounded by the lines \(y = x,\) \(x = 1,\) \(y = 0.\) Every planar section perpendicular to the \(x-\)axis is a semicircle. Find the volume of the solid.

    Solution.

    Solid whose base is a right triangle and cross sections are semicircles.
    Figure 2.

    The diameter of the semicircle at a point \(x\) is \(d=y=x.\) Hence, the area of the cross section is

    \[A(x) = \frac{{\pi {d^2}}}{8} = \frac{{\pi {x^2}}}{8}.\]

    Integration yields the following result:

    \[{V = \int\limits_0^1 {A\left( x \right)dx} }={ \int\limits_0^1 {\frac{{\pi {x^2}}}{8}dx} }={ \frac{\pi }{8}\int\limits_0^1 {{x^2}dx} }={ \frac{\pi }{8} \cdot \left. {\frac{{{x^3}}}{3}} \right|_0^1 }={ \frac{\pi }{{24}}}\]

    Example 2.

    Find the volume of a solid bounded by the elliptic paraboloid \(z = \large{\frac{{{x^2}}}{{{a^2}}}}\normalsize + \large{\frac{{{y^2}}}{{{b^2}}}}\normalsize\) and the plane \(z = 1.\)

    Solution.

    Solid bounded by the elliptic paraboloid z=x^2/a^2+y^2/b^2 and the plane z=1.
    Figure 3.

    Consider an arbitrary planar section perpendicular to the \(z-\)axis at a point \(z,\) where \(0 \lt z \le 1.\) The cross section is an ellipse defined by the equation

    \[{z = \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}},}\;\; \Rightarrow {\frac{{{x^2}}}{{{{\left( {a\sqrt z } \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {b\sqrt z } \right)}^2}}} = 1.}\]

    The area of the cross section is

    \[{A\left( z \right) = \pi \cdot \left( {a\sqrt z } \right) \cdot \left( {b\sqrt z } \right) }={ \pi abz.}\]

    Then, by the cross-section formula,

    \[{V = \int\limits_0^1 {A\left( z \right)dz} }={ \int\limits_0^1 {\pi abzdz} }={ \pi ab\int\limits_0^1 {zdz} }={ \pi ab \cdot \left. {\frac{{{z^2}}}{2}} \right|_0^1 }={ \frac{{\pi ab}}{2}}\]

    Example 3.

    The base of a solid is bounded by the parabola \(y = 1 – {x^2}\) and the \(x-\)axis. Find the volume of the solid if the cross sections are equilateral triangles perpendicular to the \(x-\)axis.
    Solid with the base bounded by the parabola y=1-x^2 and cross sections formed by equilateral triangles.
    Figure 4.

    Solution.

    The area of the equilateral triangle at a point \(x\) is given by

    \[A\left( x \right) = \frac{{{a^2}\sqrt 3 }}{4}.\]

    As the side \(a\) is equal to \(1-{x^2},\) then

    \[{A\left( x \right) = \frac{{{a^2}\sqrt 3 }}{4} }={ \frac{{\sqrt 3 }}{4}{\left( {1 – {x^2}} \right)^2}.}\]

    The parabola \(y = 1 – {x^2}\) intersects the \(x-\)axis at the points \(x=-1,\) \(x = 1.\)

    Compute the volume of the solid:

    \[{V = \int\limits_{ – 1}^1 {A\left( x \right)dx} }={ \int\limits_{ – 1}^1 {\frac{{\sqrt 3 }}{4}{{\left( {1 – {x^2}} \right)}^2}dx} }={ \frac{{\sqrt 3 }}{4}\int\limits_{ – 1}^1 {\left( {1 – 2{x^2} + {x^4}} \right)dx} }={ \frac{{\sqrt 3 }}{4}\left. {\left[ {x – \frac{{2{x^3}}}{3} + \frac{{{x^5}}}{5}} \right]} \right|_{ – 1}^1 }={ \frac{{\sqrt 3 }}{4}\left[ {\left( {1 – \frac{2}{3} + \frac{1}{5}} \right) }\right.}-{\left.{ \left( { – 1 + \frac{2}{3} – \frac{1}{5}} \right)} \right] }={ \frac{{\sqrt 3 }}{2}\left( {1 – \frac{2}{3} + \frac{1}{5}} \right) }={ \frac{{4\sqrt 3 }}{{15}}.}\]

    Example 4.

    Find the volume of a regular square pyramid with the base side \(a\) and the altitude \(H.\)

    Solution.

    Regular square pyramid with base side a and height H.
    Figure 5.

    The area of the square cross section at a point \(x\) is written in the form

    \[A\left( x \right) = {\left( {a \cdot \frac{x}{H}} \right)^2} = \frac{{{a^2}{x^2}}}{{{H^2}}}.\]

    Hence, the volume of the pyramid is given by

    \[{V = \int\limits_0^H {A\left( x \right)dx} }={ \int\limits_0^H {\frac{{{a^2}{x^2}}}{{{H^2}}}dx} }={ \frac{{{a^2}}}{{{H^2}}}\int\limits_0^H {{x^2}dx} }={ \frac{{{a^2}}}{{{H^2}}} \cdot \left. {\frac{{{x^3}}}{3}} \right|_0^H }={ \frac{{{a^2}H}}{3}}\]

    Example 5.

    Find the volume of a solid if the base of the solid is the circle given by the equation \({x^2} + {y^2} = 1,\) and every perpendicular cross section is a square.

    Solution.

    A solid whose base is the circle x^2+y^2=1 and the cross section is a square.
    Figure 6.

    An arbitrary cross section at a point \(x\) has the side \(a\) equal to

    \[a = 2y = 2\sqrt {1 – {x^2}} .\]

    Hence, the area of the cross section is

    \[A\left( x \right) = {a^2} = 4\left( {1 – {x^2}} \right).\]

    Calculate the volume of the solid:

    \[{V = \int\limits_{ – 1}^1 {A\left( x \right)dx} }={ \int\limits_{ – 1}^1 {4\left( {1 – {x^2}} \right)dx} }={ 4\left. {\left( {x – \frac{{{x^3}}}{3}} \right)} \right|_{ – 1}^1 }={ 4\left[ {\left( {1 – \frac{{{1^3}}}{3}} \right) – \left( { – 1 – \frac{{{{\left( { – 1} \right)}^3}}}{3}} \right)} \right] }={ 4\left[ {\frac{2}{3} – \left( { – \frac{2}{3}} \right)} \right] }={ \frac{{16}}{3}.}\]

    Example 6.

    Find the volume of the frustum of a cone if its bases are ellipses with the semi-axes \(A, B,\) and \(a, b\), and the altitute is equal to \(H.\)

    Solution.

    Frustum of a cone with the ellipses as bases
    Figure 7.

    The volume of the frustum of the cone is given by the integral

    \[V = \int\limits_0^H {A\left( x \right)dx} ,\]

    where \({A\left( x \right)}\) is the cross-sectional area at a point \(x.\)

    The lengths of the major and minor axes linearly change from \(a, b\) to \(A, B,\) and at the point \(x\) they are determined by the following expressions:

    \[{\text{major axis:}\;\;a + \left( {A – a} \right)\frac{x}{H}\;\;}\kern0pt{\text{minor axis:}\;\;b + \left( {B – b} \right)\frac{x}{H}}.\]

    Let’s now calculate the area of the cross section:

    \[{A\left( x \right) \text{ = }}\kern0pt{\pi \left( {a + \left( {A – a} \right)\frac{x}{H}} \right)\left( {b + \left( {B – b} \right)\frac{x}{H}} \right) }={ \pi \left[ {ab + b\left( {A – a} \right)\frac{x}{H} }\right.}+{\left.{ a\left( {B – b} \right)\frac{x}{H} }\right.}+{\left.{ \left( {A – a} \right)\left( {B – b} \right){{\left( {\frac{x}{H}} \right)}^2}} \right] }={ \pi \left[ {ab + \left( {bA + aB – 2ab} \right)\frac{x}{H} }\right.}+{\left.{ \left( {AB – aB – bA + ab} \right){{\left( {\frac{x}{H}} \right)}^2}} \right].}\]

    Then the volume is given by

    \[\require{cancel}{V = \int\limits_0^H {A\left( x \right)dx} }={ \pi \left[ {abH + \left( {bA + ab – 2ab} \right)\frac{H}{2} }\right.}+{\left.{ \left( {AB – aB – bA + ab} \right)\frac{H}{3}} \right] }={ \pi \left[ {\cancel{abH} + \frac{{bAH}}{2} }\right.}+{\left.{ \frac{{aBH}}{2} – \cancel{abH} }\right.}+{\left.{ \frac{{ABH}}{3} – \frac{{aBH}}{3} }\right.}-{\left.{ \frac{{bAH}}{3} + \frac{{abH}}{3}} \right] }={ \pi \left[ {\frac{{bAH}}{6} + \frac{{aBH}}{6} }\right.}+{\left.{ \frac{{ABH}}{3} + \frac{{abH}}{3}} \right] }={ \frac{{\pi H}}{6}\left[ {bA + aB + 2AB + 2ab} \right] }={ \frac{{\pi H}}{6}\left[ {\left( {2A + a} \right)B + \left( {A + 2a} \right)b} \right].}\]

    Example 7.

    Calculate the volume of a wedge given the bottom sides \(a, b,\) the top side \(c,\) and the altitude \(H.\)

    Solution.

    Regular wedge with the bottom sides a,b, the top side c and the height h.
    Figure 8.

    Consider an arbitrary cross section at a height \(x.\) This cross section is a rectangle with the sides \(m\) and \(n.\) It is easy to see that

    \[{m = c + \left( {a – c} \right)\frac{x}{h},\;\;}\kern0pt{n = \frac{{bx}}{h}.}\]

    Then the cross-sectional area \(A\left( x \right)\) is written as

    \[{A\left( x \right) = mn }={ \left( {c + \left( {a – c} \right)\frac{x}{h}} \right)\frac{{bx}}{h} }={ \frac{{bcx}}{h} + \frac{{ab{x^2}}}{{{h^2}}} – \frac{{bc{x^2}}}{{{h^2}}} }={ \frac{{bcx}}{h} + \frac{{b\left( {a – c} \right){x^2}}}{{{h^2}}}.}\]

    We find the volume of the wedge by integration:

    \[{V = \int\limits_0^h {A\left( x \right)dx} }={ \int\limits_0^h {\left( {\frac{{bcx}}{h} + \frac{{b\left( {a – c} \right){x^2}}}{{{h^2}}}} \right)dx} }={ \left. {\frac{{bc{x^2}}}{{2h}} + \frac{{b\left( {a – c} \right){x^3}}}{{3{h^2}}}} \right|_o^h }={ \frac{{bch}}{2} + \frac{{b\left( {a – c} \right)h}}{3} }={ \frac{{bch}}{2} + \frac{{abh}}{3} – \frac{{bch}}{3} }={ \frac{{bch}}{6} + \frac{{abh}}{3} }={ \frac{{bh}}{6}\left( {2a + c} \right).}\]

    Example 8.

    Find the volume of a regular tetrahedron with the edge \(a.\)

    Solution.

    Regular tetrahedron with the edge a
    Figure 9.

    The base of the tetrahedron is an equilateral triangle. Calculate the altitude of the base \(CE.\) By Pythagorean theorem,

    \[{{{CE}^2} = {{BC}^2} -{{EB}^2},}\;\; \Rightarrow {CE = \sqrt {{BC^2} – {EB^2}} }={ \sqrt {{a^2} – {{\left( {\frac{a}{2}} \right)}^2}} }={ \frac{{\sqrt 3 a}}{2}.}\]

    Hence

    \[CO = \frac{2}{3}CE = \frac{{\sqrt 3 a}}{3}.\]

    We express the altitude of the tetrahedron \(h\) in terms of \(a:\)

    \[{h = DO }={ \sqrt {{DC^2} – {CO^2}} }={ \sqrt {{a^2} – {{\left( {\frac{{\sqrt 3 a}}{3}} \right)}^2}} }={ \sqrt {\frac{{2{a^2}}}{3}} }={ \frac{{\sqrt 2 a}}{{\sqrt 3 }}.}\]

    The cross-sectional area \(A\left( x \right)\) is written in the form

    \[{A\left( x \right) = \frac{1}{2}{m^2}\sin 60^{\circ} }={ \frac{{\sqrt 3 {m^2}}}{4},}\]

    where \(m\) is the side of the equilateral triangle in the cross section.

    It follows from the similarity that

    \[\require{cancel}{m = \frac{{ax}}{h} }={ \frac{{\sqrt 3 \cancel{a}x}}{{\sqrt 2 \cancel{a}}} }={ \frac{{\sqrt 3 x}}{{\sqrt 2 }}.}\]

    Using integration, we find the volume:

    \[{V = \int\limits_0^h {A\left( x \right)dx} }={ \int\limits_0^h {\frac{{\sqrt 3 {m^2}}}{4}dx} }={ \int\limits_0^h {\frac{{\sqrt 3 }}{4}{{\left( {\frac{{\sqrt 3 x}}{{\sqrt 2 }}} \right)}^2}dx} }={ \int\limits_0^h {\frac{{3\sqrt 3 {x^2}}}{8}dx} }={ \frac{{3\sqrt 3 }}{8}\int\limits_0^h {{x^2}dx} }={ \frac{{3\sqrt 3 }}{8} \cdot \left. {\frac{{{x^3}}}{3}} \right|_0^h }={ \frac{{\sqrt 3 {h^3}}}{8} }={ \frac{{\sqrt 3 }}{8} \cdot {\left( {\frac{{\sqrt 2 a}}{{\sqrt 3 }}} \right)^3} }={ \frac{{2\cancel{\sqrt 3} \sqrt 2 {a^3}}}{{24\cancel{\sqrt 3} }} }={ \frac{{\sqrt 2 {a^3}}}{{12}}.}\]

    Example 9.

    A wedge is cut out of a circular cylinder with radius \(R\) and height \(H\) by the plane passing through a diameter of the base (Figure \(10\)). Find the volume of the cylindrical wedge.

    Solution.

    A wedge cut out of a cylinder of radius R and height H by the plane passing through a diameter of the base.
    Figure 10.

    A cross section of the wedge perpendicular to the \(x-\)axis is a right triangle \(ABC.\) The leg of the triangle \(AB\) is given by

    \[AB = y = \sqrt {{R^2} – {x^2}} ,\]

    and the other leg \(BC\) is expressed in the form

    \[{BC = AB \cdot \tan \alpha }={ AB \cdot \frac{H}{R} }={ \frac{H}{R}\sqrt {{R^2} – {x^2}} }\]

    Hence, the area of the cross section is written as

    \[{A\left( x \right) = \frac{{AB \cdot BC}}{2} }={ \frac{H}{R}{\left( {\sqrt {{R^2} – {x^2}} } \right)^2} }={ \frac{H}{R}\left( {{R^2} – {x^2}} \right).}\]

    Integrating yields

    \[{V = 2\int\limits_0^R {A\left( x \right)dx} }={ 2\int\limits_0^R {\frac{H}{{2R}}\left( {{R^2} – {x^2}} \right)dx} }={ \frac{H}{R}\int\limits_0^R {\left( {{R^2} – {x^2}} \right)dx} }={ \frac{H}{R}\left. {\left( {{R^2}x – \frac{{{x^3}}}{3}} \right)} \right|_0^R }={ \frac{H}{R} \cdot \frac{{2{R^3}}}{3} }={ \frac{{2{R^2}H}}{3}},\]

    so the volume of the wedge is

    \[\frac{{\frac{{2{R^2}H}}{3}}}{{\pi {R^2}H}} = \frac{2}{{3\pi }} \approx 0.212\]

    of the total volume of the cylinder. The result does not depend on \(R\) and \(H!\)

    Example 10.

    The axes of two circular cylinders with the same radius \(R\) intersect at right angles. Find the volume of the solid common to both these cylinders.

    Solution.

    The figure below shows \(\large{\frac{1}{8}}\normalsize\) of the solid of intersection.

    Two circular cylinders intersecting at right angles.
    Figure 11.

    Consider a cross section \(ABCD\) perpendicular to the \(x-\)axis at an arbitrary point \(x\). Due to symmetry, the cross section is a square with sides of length

    \[BC = AD = y = \sqrt {{R^2} – {x^2}} ,\]

    \[AB = CD = z = \sqrt {{R^2} – {x^2}}. \]

    The cross-sectional area is expressed in terms of \(x\) as follows:

    \[A\left( x \right) = {\left( {\sqrt {{R^2} – {x^2}} } \right)^2} = {R^2} – {x^2}.\]

    Then the volume of the solid common to both the cylinders (bicylinder) is given by

    \[{V = 8\int\limits_0^R {A\left( x \right)dx} }={ 8\int\limits_0^R {\left( {{R^2} – {x^2}} \right)dx} }={ 8\left. {\left( {{R^2}x – \frac{{{x^3}}}{3}} \right)} \right|_0^R }={ 8 \cdot \frac{{2{R^3}}}{3} }={ \frac{{16{R^3}}}{3}}\]