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 or tap 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:### Example 6

Calculate the volume of the ellipsoid### 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.}

\]

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 \(R.\)