Calculation of Volumes Using Triple Integrals
Solved Problems
Example 5.
Find the volume of the solid formed by two paraboloids:
\[{z_1} = {x^2} + {y^2}\;\;\text{and}\;\;{z_2} = 1 - {x^2} - {y^2}.\]
Solution.
Investigate intersection of the two paraboloids (Figure \(6\)).
Figure 6. Figure 7. 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
\[\left| I \right| = abc{\rho ^2}\sin\theta ,\]
we have
\[dxdydz = abc{\rho ^2}\sin\theta d\rho d\varphi d\theta .\]
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 \(\frac{1}{8}\) 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
\[V = \iiint\limits_{U'} {abc{\rho ^2}\sin\theta d\rho d\varphi d\theta } = 8abc\int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^1 {{\rho ^2}d\rho } \int\limits_0^{\frac{\pi }{2}} {\sin\theta d\theta } = 8abc\int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^1 {{\rho ^2}d\rho } \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^{\frac{\pi }{2}}} \right] = 8abc\int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^1 {{\rho ^2}d\rho } \cdot \left( { - \cos \frac{\pi }{2} + \cos 0} \right) = 8abc\int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^1 {{\rho ^2}d\rho } = 8abc\int\limits_0^{\frac{\pi }{2}} {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^3}}}{3}} \right)} \right|_0^1} \right] = \frac{{8abc}}{3}\int\limits_0^{\frac{\pi }{2}} {d\varphi } = \frac{{8abc}}{3} \cdot \left[ {\left. \varphi \right|_0^{\frac{\pi }{2}}} \right] = \frac{{8abc}}{3} \cdot \frac{\pi }{2} = \frac{4}{3}\pi abc.\]
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 \({x^2} + {y^2} = 2.\)
Figure 8. The region of integration is bounded from above by the spherical surface, and from below by the paraboloid (Figure \(9\)).
Figure 9. The volume of the solid region is expressed by the integral
\[V = \iiint\limits_U {dxdydz} = \int\limits_{ - \sqrt 2 }^{\sqrt 2 } {dx} \int\limits_{-\sqrt {2 - {x^2}} }^{\sqrt {2 - {x^2}} } {dy} \int\limits_{{x^2} + {y^2}}^{\sqrt {6 - {x^2} - {y^2}} } {dz} .\]
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:
\[V = \pi \int\limits_0^{\sqrt 2 } {\left( {\sqrt {6 - {\rho ^2}} - {\rho ^2}} \right)d{\rho ^2}} = \pi \int\limits_0^2 {\left( {\sqrt {6 - t} - t} \right)dt} = \pi \left[ {\left. {\left( { - \frac{{2{{\left( {6 - t} \right)}^{\frac{3}{2}}}}}{3} - \frac{{{t^2}}}{2}} \right)} \right|_0^2} \right] = \pi \left[ { - \frac{2}{3}\left( {{4^{\frac{3}{2}}} - {6^{\frac{3}{2}}}} \right) - 2} \right] = \pi \left[ {\frac{2}{3}\left( {6\sqrt 6 - 8} \right) - 2} \right] = \pi \left( {4\sqrt 6 - \frac{{16}}{3} - 2} \right) = 2\pi \left( {\frac{{6\sqrt 6 - 11}}{3}} \right).\]
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, that is
\[z = \sqrt {{x^2} + {y^2}} = 1\;\; \text{or}\;\;{x^2} + {y^2} = 1.\]
Thus, both the surfaces intersect at \(z = 1,\) and the intersection is a circle (Figure \(10\)).
Figure 10. The region of integration is bounded from above by the paraboloid, and from below by the cone (Figure \(11\)).
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:
\[V = \iiint\limits_U {\rho d\rho d\varphi dz} = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {\rho d\rho } \int\limits_0^{2 - {\rho ^2}} {dz} = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {\rho d\rho \cdot \left( {2 - {\rho ^2} - \rho } \right)} = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {\left( {2\rho - {\rho ^3} - {\rho ^2}} \right)d\rho } = \int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {{\rho ^2} - \frac{{{\rho ^4}}}{4} - \frac{{{\rho ^3}}}{3}} \right)} \right|_0^1} \right] = \int\limits_0^{2\pi } {\left( {1 - \frac{1}{4} - \frac{1}{3}} \right)d\varphi } = \frac{5}{{12}}\int\limits_0^{2\pi } {d\varphi } = \frac{{5\pi }}{6}.\]
