Calculation of Volumes Using Triple Integrals

Triple Integrals – Calculus 3
Triple Integrals – Calculus 3

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

Example 1.

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

Solution.

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

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:

\[0 \le \varphi \le 2\pi ,\;\; 0 \le \rho \le R,\;\; \rho \le z \le H.\]

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:

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

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

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 \(Oxy,\) \(Oxz,\) \(Oyz\) \(\left({\text{Figure }2}\right).\)

Solution.

The equation of the straight line \(AB\) in the \(xy\)-plane (Figure \(3\)) is written as \(y = 2 – 2x.\) The variable \(x\) ranges here in the interval \(0 \le x \le 1,\) and the variable \(y\) ranges in the interval \(0 \le y \le 2 – 2x.\)

Figure 2.
Figure 3.

We write the equation of the plane \(ABC\) in segment form. Since the plane \(ABC\) cuts the line segments \(1, 2,\) and \(3,\) respectively, on the \(x-,\) \(y-,\) and \(z-\)axis, then its equation can be written as

\[\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1.\]

In general case the equation of the plane \(ABC\) is written as

\[6x + 3y + 2z = 6\;\;\text{or}\;\; z = 3 – 3x – \frac{3}{2}y.\]

Hence, the limits of integration over the variable \(z\) range in the interval from \(z = 0\) to \(z = 3 – 3x – {\frac{3}{2}} y.\)

Now we can calculate the volume of the tetrahedron:

\[V = \iiint\limits_U {dxdydz}
= \int\limits_0^1 {dx} \int\limits_0^{2 – 2x} {dy} \int\limits_0^{3 – 3x – \frac{3}{2}y} {dz}
= \int\limits_0^1 {dx} \int\limits_0^{2 – 2x} {dy} \cdot \left[ {\left. z \right|_0^{3 – 3x – \frac{3}{2}y}} \right]
= \int\limits_0^1 {dx} \int\limits_0^{2 – 2x} {\left( {3 – 3x – \frac{3}{2}y} \right)dy}
= \int\limits_0^1 {dx} \cdot \Big[ {\left. {\left( {3y – 3xy – \frac{3}{4}{y^2}} \right)} \right|_{y = 0}^{y = 2 – 2x}} \Big]
= \int\limits_0^1 {\Big[ {3\left( {2 – 2x} \right) – 3x\left( {2 – 2x} \right) – \frac{3}{4}{{\left( {2 – 2x} \right)}^2}} \Big] dx}
= \int\limits_0^1 {\Big[ {6 – 6x – 6x + 6{x^2} – \frac{3}{4}\left( {4 – 8x + 4{x^2}} \right)} \Big] dx}
= \int\limits_0^1 {\left( {\color{green}{6} – \color{red}{12x} + \color{blue}{6{x^2}} – \color{green}{3} + \color{red}{6x} – \color{blue}{3{x^2}}} \right)dx}
= 3\int\limits_0^1 {\left( {\color{green}{1} – \color{red}{2x} + \color{blue}{x^2}} \right)dx}
= 3\left[ {\left. {\left( {x – {x^2} + \frac{{{x^3}}}{3}} \right)} \right|_0^1} \right]
= 3 \cdot \left( {\cancel{1} – \cancel{1^2} + \frac{{{1^3}}}{3}} \right) = 1.\]

Example 4.

Find the volume of the tetrahedron bounded by the planes

\[x + y + z = 5, x = 0, y = 0, z = 0\]

(Figure \(4\)).

Solution.

The equation of the plane \(x + y + z = 5\) can be rewritten in the form

\[z = 5 – x – y.\]

By setting \(z = 0,\) we get

\[5 – x – y = 0\;\;\text{or}\;\;y = 5 – x.\]

Hence, the region of integration \(D\) in the \(xy\)-plane is bounded by the straight line \(y = 5 – x\) as shown in Figure \(5.\)

Figure 4.
Figure 5.

Representing the triple integral as an iterated integral, we can find the volume of the tetrahedron:

\[V = \iiint\limits_U {dxdydz}
= \int\limits_0^5 {dx} \int\limits_0^{5 – x} {dy} \int\limits_0^{5 – x – y} {dz}
= \int\limits_0^5 {dx} \int\limits_0^{5 – x} {dy} \cdot \left[ {\left. z \right|_0^{5 – x – y}} \right]
= \int\limits_0^5 {dx} \int\limits_0^{5 – x} {\left( {5 – x – y} \right)dy}
= \int\limits_0^5 {dx} \left[ {\left. {\left( {5y – xy – \frac{{{y^2}}}{2}} \right)} \right|_{y = 0}^{y = 5 – x}} \right]
= \int\limits_0^5 {\left[ {5\left( {5 – x} \right) – x\left( {5 – x} \right) – \frac{{{{\left( {5 – x} \right)}^2}}}{2}} \right]dx}
= \int\limits_0^5 {\left( {25 – 5x – 5x + {x^2} – \frac{{25 – 10x + {x^2}}}{2}} \right)dx}
= \frac{1}{2}\int\limits_0^5 {\left( {25 – 10x + {x^2}} \right)dx}
= \frac{1}{2}\left[ {\left. {\left( {25x – \frac{{10{x^2}}}{2} + \frac{{{x^3}}}{3}} \right)} \right|_0^5} \right]
= \frac{1}{2}\left( {125 – 5 \cdot 25 + \frac{{125}}{3}} \right)
= \frac{{125}}{6}.\]

See more problems on Page 2.

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