Calculus

Triple Integrals

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Calculation of Volumes Using Triple Integrals

  • The volume of a solid \(U\) in Cartesian coordinates \(xyz\) is given by

    \[V = \iiint\limits_U {dxdydz} .\]

    In cylindrical coordinates, the volume of a solid is defined by the formula

    \[V = \iiint\limits_U {\rho d\rho d\varphi dz} .\]

    In spherical coordinates, the volume of a solid is expressed as

    \[V = \iiint\limits_U {{\rho ^2}\sin \theta d\rho d\varphi d\theta } .\]


    Solved Problems

    Click a problem to see the solution.

    Example 1

    Find the volume of the cone of height \(H\) and base radius \(R\) (Figure \(1\)).

    Example 2

    Find the volume of the ball \({x^2} + {y^2} + {z^2} \le {R^2}.\)

    Example 3

    Find the volume of the tetrahedron bounded by the planes passing through the points \(A\left( {1,0,0} \right),\) \(B\left( {0,2,0} \right),\) \(C\left( {0,0,3} \right)\) and the coordinate planes \(Oxy,\) \(Oxz,\) \(Oyz\) \(\left({\text{Figure }2}\right).\)

    Example 4

    Find the volume of the tetrahedron bounded by the planes \(x + y + z = 5,\) \(x = 0,\) \(y = 0,\) \(z = 0.\) (Figure \(4\)).

    Example 5

    Find the volume of the solid formed by two paraboloids:
    \[{{z_1} = {x^2} + {y^2}\;\;\text{and}\;\;}\kern-0.3pt{{z_2} = 1 – {x^2} – {y^2}.}\]

    Example 6

    Calculate the volume of the ellipsoid
    \[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1.\]

    Example 7

    Find the volume of the solid bounded by the sphere \({x^2} + {y^2} + {z^2} = 6\) and the paraboloid \({x^2} + {y^2} = z.\)

    Example 8

    Calculate the volume of the solid bounded by the paraboloid \(z = 2 – {x^2} – {y^2}\) and the conic surface \(z = \sqrt {{x^2} + {y^2}}.\)

    Example 1.

    Find the volume of the cone of height \(H\) and base radius \(R\) (Figure \(1\)).

    Solution.

    The cone is bounded by the surface \(z = {\large\frac{H}{R}\normalsize} \sqrt {{x^2} + {y^2}} \) and the plane \(z = H\) (see Figure \(1\)).

    The cone of height H and base radius R
    Figure 1.

    Its volume in Cartesian coordinates is expressed by the formula

    \[
    {V = \iiint\limits_U {dxdydz} }
    = {\int\limits_{ – R}^R {dx} \int\limits_{ – \sqrt {{R^2} – {x^2}} }^{\sqrt {{R^2} – {x^2}} } {dy} \int\limits_{\frac{H}{R}\sqrt {{x^2} + {y^2}} }^H {dz} .}
    \]

    Calculate this integral in cylindrical coordinates that range within the limits:

    \[
    {0 \le \varphi \le 2\pi ,\;\;\;}\kern-0.3pt
    {0 \le \rho \le R,\;\;\;}\kern-0.3pt
    {\rho \le z \le H.}
    \]

    As a result, we obtain (do not forget to include the Jacobian \(\rho\)):

    \[V = \int\limits_0^R {\rho d\rho } \int\limits_0^{2\pi } {d\varphi } \int\limits_{\frac{H}{R}\rho }^H {dz} .\]

    Then the volume of the cone is

    \[ {V = \int\limits_0^R {\rho d\rho } \int\limits_0^{2\pi } {d\varphi } \int\limits_{\frac{H}{R}\rho }^H {dz} } = {2\pi \int\limits_0^R {\rho d\rho } \int\limits_{\frac{H}{R}\rho }^H {dz} } = {2\pi \int\limits_0^R {\rho d\rho } \cdot \left[ {\left. z \right|_{z = \frac{H}{R}\rho }^{z = H}} \right] } = {2\pi \int\limits_0^R {\rho \left( {H – \frac{H}{R}\rho } \right)d\rho } } = {2\pi \int\limits_0^R {\left( {H\rho – \frac{H}{R}{\rho ^2}} \right)d\rho } } = {2\pi \left[ {\left. {\left( {\frac{{{\rho ^2}H}}{2} – \frac{{{\rho ^3}H}}{{3R}}} \right)} \right|_{\rho = 0}^{\rho = R}} \right] } = {2\pi \left( {\frac{{{R^2}H}}{2} – \frac{{{R^3}H}}{{3R}}} \right) } = {\frac{{2\pi {R^2}H}}{6} } = {\frac{{\pi {R^2}H}}{3}.} \]

    Example 2.

    Find the volume of the ball \({x^2} + {y^2} + {z^2} \le {R^2}.\)

    Solution.

    We calculate the volume of the part of the ball lying in the first octant \(\left( {x \ge 0,y \ge 0,z \ge 0} \right),\) and then multiply the result by \(8.\) This yields:

    \[
    {V = 8\iiint\limits_U {dxdydz} }
    = {8\iiint\limits_{U’} {{\rho ^2}\sin \theta d\rho d\varphi d\theta } }
    = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\sin \theta d\theta } }
    = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot \left[ {\left. {\left( { – \cos \theta } \right)} \right|_0^{\large\frac{\pi }{2}\normalsize}} \right] }
    = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot}\kern0pt{ \left( { – \cos \frac{\pi }{2} + \cos 0} \right) }
    = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot 1 }
    = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^3}}}{3}} \right)} \right|_0^R} \right] }
    = {\frac{{8{R^3}}}{3}\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } }
    = {\frac{{8{R^3}}}{3} \cdot \left[ {\left. \varphi \right|_0^{\large\frac{\pi }{2}\normalsize}} \right] }
    = {\frac{{8{R^3}}}{3} \cdot \frac{\pi }{2} }
    = {\frac{{4\pi {R^3}}}{3}.}
    \]

    As a result, we get the well-known expression for the volume of the ball of radius \(R.\)

    Page 1
    Problems 1-2
    Page 2
    Problems 3-8