Calculus

Triple Integrals

Calculation of Volumes Using Triple Integrals

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

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 on problem description to see 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 \(Oxy,\) \(Oxz,\) \(Oyz\) (Figure \(2\text{).}\)

 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 \(1\)). 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:
\[
{0 \le \varphi \le 2\pi ,\;\;\;}\kern-0.3pt
{0 \le \rho \le R,\;\;\;}\kern-0.3pt
{\rho \le z \le H.}
\]

The cone of height H and base radius R

Figure 1.

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