Math24.net Calculus
 Home Calculus Limits and Continuity Differentiation Applications of Derivative Integration Sequences and Series Double Integrals Triple Integrals Line Integrals Surface Integrals Fourier Series Differential Equations 1st Order Equations 2nd Order Equations Nth Order Equations Systems of Equations Formulas and Tables
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 $V = \iiint\limits_U {{\rho ^2}\sin \theta d\rho d\varphi d\theta } .$
Example 1
Find the volume of the cone of height $$H$$ and base radius $$R$$ (Figure $$1$$).

Solution.
 Fig.1
The cone is bounded by the surface $$z = {\large\frac{H}{R}\normalsize} \sqrt {{x^2} + {y^2}}$$ and the plane $$z = H$$ (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: ${V = 8\iiint\limits_U {dxdydz} } = {8\iiint\limits_{U'} {{\rho ^2}\sin \theta d\rho d\varphi d\theta } } = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\sin \theta d\theta } } = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^{\large\frac{\pi }{2}\normalsize}} \right] } = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot \left( { - \cos \frac{\pi }{2} + \cos 0} \right) } = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot 1 } = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^3}}}{3}} \right)} \right|_0^R} \right] } = {\frac{{8{R^3}}}{3}\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } } = {\frac{{8{R^3}}}{3} \cdot \left[ {\left. \varphi \right|_0^{\large\frac{\pi }{2}\normalsize}} \right] } = {\frac{{8{R^3}}}{3} \cdot \frac{\pi }{2} } = {\frac{{4\pi {R^3}}}{3}.}$ As a result, we get the well-known expression for the volume of the ball of radius $$R.$$

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 $$2$$).
 Fig.2 Fig.3
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 $$0 \le y \le 2 - 2x.$$

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 $$z = 3 - 3x - {\large\frac{3}{2}\normalsize} y.$$ Now we can calculate the volume of the tetrahedron: $\require{cancel} {V = \iiint\limits_U {dxdydz} } = {\int\limits_0^1 {dx} \int\limits_0^{2 - 2x} {dy} \int\limits_0^{3 - 3x - \frac{3}{2}y} {dz} } = {\int\limits_0^1 {dx} \int\limits_0^{2 - 2x} {dy} \cdot \left[ {\left. z \right|_0^{3 - 3x - \frac{3}{2}y}} \right] } = {\int\limits_0^1 {dx} \int\limits_0^{2 - 2x} {\left( {3 - 3x - \frac{3}{2}y} \right)dy} } = {\int\limits_0^1 {dx} \cdot \left[ {\left. {\left( {3y - 3xy - \frac{3}{4}{y^2}} \right)} \right|_{y = 0}^{y = 2 - 2x}} \right] } = {\int\limits_0^1 {\left[ {3\left( {2 - 2x} \right) - 3x\left( {2 - 2x} \right) - \frac{3}{4}{{\left( {2 - 2x} \right)}^2}} \right]dx} } = {\int\limits_0^1 {\left[ {6 - 6x - 6x + 6{x^2} - \frac{3}{4}\left( {4 - 8x + 4{x^2}} \right)} \right]dx} } = {\int\limits_0^1 {\left( {\color{green}{6} - \color{red}{12x} + \color{blue}{6{x^2}} - \color{green}{3} + \color{red}{6x} - \color{blue}{3{x^2}}} \right)dx} } = {3\int\limits_0^1 {\left( {\color{green}{1} - \color{red}{2x} + \color{blue}{x^2}} \right)dx} } = {3\left[ {\left. {\left( {x - {x^2} + \frac{{{x^3}}}{3}} \right)} \right|_0^1} \right] } = {3 \cdot \left( {\cancel{1} - \cancel{1^2} + \frac{{{1^3}}}{3}} \right) = 1.}$
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 $5 - x - y = 0\;\;\text{or}\;\;y = 5 - x.$
 Fig.4 Fig.5
Hence, the region of integration $$D$$ in the $$xy$$-plane is bounded by the straight line $$y = 5 - x$$ as shown in Figure $$5.$$

Representing the triple integral as an iterated integral, we can find the volume of the tetrahedron: ${V = \iiint\limits_U {dxdydz} } = {\int\limits_0^5 {dx} \int\limits_0^{5 - x} {dy} \int\limits_0^{5 - x - y} {dz} } = {\int\limits_0^5 {dx} \int\limits_0^{5 - x} {dy} \cdot \left[ {\left. z \right|_0^{5 - x - y}} \right] } = {\int\limits_0^5 {dx} \int\limits_0^{5 - x} {\left( {5 - x - y} \right)dy} } = {\int\limits_0^5 {dx} \cdot \left[ {\left. {\left( {5y - xy - \frac{{{y^2}}}{2}} \right)} \right|_{y = 0}^{y = 5 - x}} \right] } = {\int\limits_0^5 {\left[ {5\left( {5 - x} \right) - x\left( {5 - x} \right) - \frac{{{{\left( {5 - x} \right)}^2}}}{2}} \right]dx} } = {\int\limits_0^5 {\left( {25 - 5x - 5x + {x^2} - \frac{{25 - 10x + {x^2}}}{2}} \right)dx} } = {\frac{1}{2}\int\limits_0^5 {\left( {25 - 10x + {x^2}} \right)dx} } = {\frac{1}{2}\left[ {\left. {\left( {25x - \frac{{10{x^2}}}{2} + \frac{{{x^3}}}{3}} \right)} \right|_0^5} \right] } = {\frac{1}{2}\left( {125 - 5 \cdot 25 + \frac{{125}}{3}} \right) } = {\frac{{125}}{6}.}$
Example 5
Find the volume of the solid formed by two paraboloids: ${z_1} = {x^2} + {y^2}\;\;\text{and}\;\;{z_2} = 1 - {x^2} - {y^2}.$
 Fig.6 Fig.7
Solution.
Investigate intersection of the two paraboloids (Figure $$6$$). Since $${\rho ^2} = {x^2} + {y^2},$$ the equations of the paraboloids can be written as ${z_1} = {\rho ^2}\;\;\text{and}\;\;{z_2} = 1 - {\rho ^2}.$ By setting $${z_1} = {z_2}$$ for the intersection curve, we obtain ${{\rho ^2} = 1 - {\rho ^2},}\;\; {\Rightarrow 2{\rho ^2} = 1,}\;\; {\Rightarrow {\rho ^2} = \frac{1}{2}}\;\; {\text{or}\;\;z = \frac{1}{{\sqrt 2 }} = \frac{{\sqrt 2 }}{2}.}$ For this value of $$\rho$$ (Figure $$7$$), the coordinate $$z$$ is $z = {\left( {\frac{{\sqrt 2 }}{2}} \right)^2} = \frac{1}{2}.$ The volume of the solid is expressed in terms of the triple integral as $V = \iiint\limits_U {dxdydz} .$ This integral in cylindrical coordinates becomes ${V = \iiint\limits_U {dxdydz} } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^{\frac{{\sqrt 2 }}{2}} {\rho d\rho } \int\limits_{{\rho ^2}}^{1 - {\rho ^2}} {dz} } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^{\frac{{\sqrt 2 }}{2}} {\rho d\rho } \cdot \left[ {\left. z \right|_{{\rho ^2}}^{1 - {\rho ^2}}} \right] } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^{\frac{{\sqrt 2 }}{2}} {\rho \left( {1 - {\rho ^2} - {\rho ^2}} \right)d\rho } } = {2\pi \int\limits_0^{\frac{{\sqrt 2 }}{2}} {\left( {\rho - 2{\rho ^3}} \right)d\rho } } = {2\pi \left[ {\left. {\left( {\frac{{{\rho ^2}}}{2} - \frac{{2{\rho ^4}}}{4}} \right)} \right|_0^{\frac{{\sqrt 2 }}{2}}} \right] } = {2\pi \left[ {\frac{{{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}}}{2} - \frac{{{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^4}}}{2}} \right] } = {\pi \left( {\frac{1}{2} - \frac{1}{4}} \right) } = {\frac{\pi }{4}.}$
Example 6
Calculate the volume of the ellipsoid $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1.$

Solution.
It is easier to calculate the volume of the ellipsoid using generalized spherical coordinates. Let ${x = a\rho \cos \varphi \sin \theta ,}\;\; {y = b\rho \sin \varphi \sin \theta ,}\;\; {z = c\rho \cos \theta .}$ Since the absolute value of the Jacobian for transformation of Cartesian coordinates into generalized spherical coordinates is $\left| I \right| = abc{\rho ^2}\sin\theta ,$ hence, $dxdydz = abc{\rho ^2}\sin\theta d\rho d\varphi d\theta .$ The volume of the ellipsoid is expressed through the triple integral: ${V = \iiint\limits_U {dxdydz} } = {\iiint\limits_{U'} {abc{\rho ^2}\sin\theta d\rho d\varphi d\theta } .}$ By symmetry, we can find the volume of $$\large\frac{1}{8}\normalsize$$ part of the ellipsoid lying in the first octant $$\left( {x \ge 0,y \ge 0,z \ge 0} \right)$$ and then multiply the result by $$8.$$ The generalized spherical coordinates will range within the limits: ${0 \le \rho \le 1,}\;\; {0 \le \varphi \le \frac{\pi }{2},}\;\; {0 \le \theta \le \frac{\pi }{2}.}$ Then the volume of the ellipsoid is ${V = \iiint\limits_{U'} {abc{\rho ^2}\sin\theta d\rho d\varphi d\theta } } = {8abc\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^1 {{\rho ^2}d\rho } \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\sin\theta d\theta } } = {8abc\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^1 {{\rho ^2}d\rho } \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^{\large\frac{\pi }{2}\normalsize}} \right] } = {8abc\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^1 {{\rho ^2}d\rho } \cdot \left( { - \cos \frac{\pi }{2} + \cos 0} \right) } = {8abc\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^1 {{\rho ^2}d\rho } } = {8abc\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^3}}}{3}} \right)} \right|_0^1} \right] } = {\frac{{8abc}}{3}\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } } = {\frac{{8abc}}{3} \cdot \left[ {\left. \varphi \right|_0^{\large\frac{\pi }{2}\normalsize}} \right] } = {\frac{{8abc}}{3} \cdot \frac{\pi }{2} } = {\frac{4}{3}\pi abc.}$
Example 7
Find the volume of the solid bounded by the sphere $${x^2} + {y^2} + {z^2} = 6$$ and the paraboloid $${x^2} + {y^2} = z.$$

Solution.
We first determine the curve of intersection of these surfaces. Substituting the equation of the paraboloid into the equation of the sphere, we find: ${z + {z^2} = 6,}\;\; {\Rightarrow {z^2} + z - 6 = 0,}\;\; {\Rightarrow {z_{1,2}} = \frac{{ - 1 \pm 5}}{2} = 2; - 3.}$ The second root $${z_2} = - 3$$ corresponds to intersection of the sphere with the lower shell of the paraboloid. So we do not consider this case. Thus, intersection of the solids happens at $$z = 2.$$ Obviously, the projection of the region of integration on the $$xy$$-plane is the circle (Figure $$8$$) defined by the equation $${x^2} + {y^2} = 2.$$
 Fig.8 Fig.9
The region of integration is bounded from above by the spherical surface, and from below by the paraboloid (Figure $$9$$). The volume of the solid region is expressed by the integral ${V = \iiint\limits_U {dxdydz} } = {\int\limits_{ - \sqrt 2 }^{\sqrt 2 } {dx} \int\limits_0^{\sqrt {2 - {x^2}} } {dy} \int\limits_{{x^2} + {y^2}}^{\sqrt {6 - {x^2} - {y^2}} } {dz} .}$ It is convenient to convert the integral to cylindrical coordinates: $V = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^{\sqrt 2 } {\rho d\rho } \int\limits_{{\rho ^2}}^{\sqrt {6 - {\rho ^2}} } {dz} ,$ where $${\rho ^2} = {x^2} + {y^2}$$ and the integral includes the Jacobian $$\rho.$$ As a result, we have: ${V = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^{\sqrt 2 } {\rho d\rho } \int\limits_{{\rho ^2}}^{\sqrt {6 - {\rho ^2}} } {dz} } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^{\sqrt 2 } {\rho d\rho } \cdot \left[ {\left. z \right|_{{\rho ^2}}^{\sqrt {6 - {\rho ^2}} }} \right] } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^{\sqrt 2 } {\rho \left( {\sqrt {6 - {\rho ^2}} - {\rho ^2}} \right)d\rho } } = {2\pi \int\limits_0^{\sqrt 2 } {\rho \left( {\sqrt {6 - {\rho ^2}} - {\rho ^2}} \right)d\rho } } = {\pi \int\limits_0^{\sqrt 2 } {\left( {\sqrt {6 - {\rho ^2}} - {\rho ^2}} \right)d{\rho ^2}} .}$ We change the variable: $${\rho ^2} = t$$. Here $$t = 0$$ when $$\rho = 0,$$ and, respectively, $$t = 2$$ when $$\rho = \sqrt 2.$$

Now we can calculate the volume of the solid: ${V = \pi \int\limits_0^{\sqrt 2 } {\left( {\sqrt {6 - {\rho ^2}} - {\rho ^2}} \right)d{\rho ^2}} } = {\pi \int\limits_0^2 {\left( {\sqrt {6 - t} - t} \right)dt} } = {\pi \left[ {\left. {\left( { - \frac{{2{{\left( {6 - t} \right)}^{\large\frac{3}{2}\normalsize}}}}{3} - \frac{{{t^2}}}{2}} \right)} \right|_0^2} \right] } = {\pi \left[ { - \frac{2}{3}\left( {{4^{\large\frac{3}{2}\normalsize}} - {6^{\large\frac{3}{2}\normalsize}}} \right) - 2} \right] } = {\pi \left[ {\frac{2}{3}\left( {6\sqrt 6 - 8} \right) - 2} \right] } = {\pi \left( {4\sqrt 6 - \frac{{16}}{3} - 2} \right) } = {2\pi \left( {\frac{{6\sqrt 6 - 11}}{3}} \right).}$
Example 8
Calculate the volume of the solid bounded by the paraboloid $$z = 2 - {x^2} - {y^2}$$ and the conic surface $$z = \sqrt {{x^2} + {y^2}} .$$

Solution.
First we investigate intersection of the two surfaces. By equating the coordinates $$z,$$ we get the following equation: $2 - {x^2} - {y^2} = \sqrt {{x^2} + {y^2}} .$ Let $${x^2} + {y^2} = {t^2}.$$ Then ${2 - {t^2} = t,}\;\; {\Rightarrow {t^2} + t - 2 = 0,}\;\; {\Rightarrow {t_{1,2}} = \frac{{ - 1 \pm 3}}{2} = - 2;1.}$ Only the root $$t = 1$$ has the sense in the context of the given problem, i.e. ${z = \sqrt {{x^2} + {y^2}} = 1}\;\; {\text{or}\;\;{x^2} + {y^2} = 1.}$ Thus, both the surfaces intersect at $$z = 1,$$ and the intersection is a circle (Figure $$10$$).
 Fig.10 Fig.11
The region of integration is bounded from above by the paraboloid, and from below by the cone (Figure $$11$$). To calculate the volume of the solid we use cylindrical coordinates: ${{x^2} + {y^2} = {\rho ^2},}\;\; {\sqrt {{x^2} + {y^2}} = \rho ,}\;\; {dxdydz = \rho d\rho d\varphi dz.}$ As a result, we find: ${V = \iiint\limits_U {\rho d\rho d\varphi dz} } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {\rho d\rho } \int\limits_0^{2 - {\rho ^2}} {dz} } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {\rho d\rho \cdot \left( {2 - {\rho ^2} - \rho } \right)} } = {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {\left( {2\rho - {\rho ^3} - {\rho ^2}} \right)d\rho } } = {\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {{\rho ^2} - \frac{{{\rho ^4}}}{4} - \frac{{{\rho ^3}}}{3}} \right)} \right|_0^1} \right] } = {\int\limits_0^{2\pi } {\left( {1 - \frac{1}{4} - \frac{1}{3}} \right)d\varphi } } = {\frac{5}{{12}}\int\limits_0^{2\pi } {d\varphi } } = {\frac{{5\pi }}{6}.}$