Physical Applications of Surface Integrals

Solved Problems

Example 3.

Find the center of mass of the sphere \[{x^2} + {y^2} + {z^2} = {a^2}\] in the first octant, if it has constant density \({\mu_0}.\)

Solution.

Obviously that the mass of the part of the sphere lying in the first octant (Figure \(3\)) is

\[m = \frac{1}{8}\iint\limits_S {{\mu _0}dS} = \frac{{{\mu _0}}}{8}\iint\limits_S {dS} = \frac{{{\mu _0}}}{8} \cdot 4\pi {a^2} = \frac{{{\mu _0}\pi {a^2}}}{2}.\]

A portion of sphere lying in the first octant — Figure 3.

Projection of the region of integration on the xy-plane — Figure 4.

We calculate the first moment \({M_{yz}}:\)

\[{M_{yz}} = \iint\limits_S {x\mu \left( {x,y,z} \right)dS} = {\mu _0}\iint\limits_S {xdS} = {\mu _0}\iint\limits_{D\left( {x,y} \right)} x { \sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy} ,\]

where the projection \({D\left( {x,y} \right)}\) of the surface onto the \(xy\)-plane is the part of the circle that lies in the first quadrant (Figure \(4\)).

Since

\[\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\sqrt {{a^2} - {x^2} - {y^2}} = \frac{{ - x}}{{\sqrt {{a^2} - {x^2} - {y^2}} }},\]

\[\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\sqrt {{a^2} - {x^2} - {y^2}} = \frac{{ - y}}{{\sqrt {{a^2} - {x^2} - {y^2}} }},\]

then

\[\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} = \frac{a}{{\sqrt {{a^2} - {x^2} - {y^2}} }}.\]

So the first moment \({M_{yz}}\) is

\[{M_{yz}} = {\mu _0}a\int\limits_{D\left( {x,y} \right)} {\frac{{xdxdy}}{{\sqrt {{a^2} - {x^2} - {y^2}} }}} .\]

It is convenient to convert the integral into polar coordinates:

\[ {M_{yz}} = {\mu _0}a\iint\limits_{D\left( {r,\varphi } \right)} {\frac{{r\cos \varphi \cdot rdrd\varphi }}{{\sqrt {{a^2} - {r^2}} }}} = {\mu _0}a\int\limits_0^{\frac{\pi }{2}} {\cos \varphi d\varphi } \int\limits_0^a {\frac{{{r^2}dr}}{{\sqrt {{a^2} - {r^2}} }}} = {\mu _0}a \left[ {\left. {\left( { - \sin \varphi } \right)} \right|_0^{\frac{\pi }{2}}} \right] \cdot \int\limits_a^0 {\frac{{{a^2} - {r^2} - {a^2}}}{{\sqrt {{a^2} - {r^2}} }}dr} = {\mu _0}a\left[ {\int\limits_a^0 {\sqrt {{a^2} - {r^2}} dr} - {a^2}\int\limits_a^0 {\frac{{dr}}{{\sqrt {{a^2} - {r^2}} }}} } \right].\]

Calculate the first integral \({\int\limits_a^0 {\sqrt {{a^2} - {r^2}} dr} }\) in the square brackets. Make the substitution: \(r = a\sin t,\) \(dr = a\cos tdt.\) When \(r = 0,\) we have \(t = 0,\) and when \(r = a,\) we get \(t = {\frac{\pi }{2}}.\) Then the integral becomes

\[\int\limits_a^0 {\sqrt {{a^2} - {r^2}} dr} = \int\limits_{\frac{\pi }{2}}^0 {\sqrt {{a^2} - {a^2}{{\sin }^2}t} \cdot a\cos tdt} = {a^2}\int\limits_{\frac{\pi }{2}}^0 {{{\cos }^2}tdt} = {a^2}\int\limits_{\frac{\pi }{2}}^0 {\frac{{1 + \cos 2t}}{2}dt} = \frac{{{a^2}}}{2}\left[ {\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_{\frac{\pi }{2}}^0} \right] = \frac{{{a^2}}}{2} \cdot \left( { - \frac{\pi }{2}} \right) = - \frac{{\pi {a^2}}}{4}.\]

The second integral is

\[\int\limits_a^0 {\frac{{dr}}{{\sqrt {{a^2} - {r^2}} }}} = \left. {\left( {\arcsin \frac{r}{a}} \right)} \right|_a^0 = \arcsin 0 - \arcsin 1 = - \frac{\pi }{2}.\]

Thus, the first moment \({M_{yz}}\) is

\[{M_{yz}} = {\mu _0}a\left[ { - \frac{{\pi {a^2}}}{4} - {a^2}\left( { - \frac{\pi }{2}} \right)} \right] = {\mu _0}a \cdot \frac{{\pi {a^2}}}{4} = \frac{{{\mu _0}\pi {a^3}}}{4}.\]

Now we can find the coordinate \({x_C}\) of the center of mass:

\[{x_C} = \frac{{{M_{yz}}}}{m} = \frac{{\frac{{{\mu _0}\pi {a^3}}}{4}}}{{\frac{{{\mu _0}\pi {a^2}}}{2}}} = \frac{a}{2}.\]

By symmetry, we conclude that other two coordinates have the same value. Thus, the centroid of the shell is given by

\[\left( {{x_C},{y_C},{z_C}} \right) = \left( {\frac{a}{2},\frac{a}{2},\frac{a}{2}} \right).\]

Example 4.

Calculate the moment of inertia of the uniform spherical shell

\[{x^2} + {y^2} + {z^2} = 1 \left( {z \ge 0} \right)\]

with the density \({\mu_0}\) about the \(z\)-axis.

Solution.

The moment of inertia \({I_z}\) is given by the formula

\[{I_z} = \iint\limits_S {\left( {{x^2} + {y^2}} \right)\mu \left( {x,y,z} \right)dS} = {\mu _0}\iint\limits_S {\left( {{x^2} + {y^2}} \right)dS} ,\]

where the surface \(S\) is the hemisphere \({x^2} + {y^2} + {z^2} = 1\) \(\left( {z \ge 0} \right).\)

As the equation of the upper hemisphere is \(z = \sqrt {1 - {x^2} - {y^2}} ,\) the area element is

\[dS = \sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy = \frac{{dxdy}}{{\sqrt {1 - {x^2} - {y^2}} }}.\]

Then the surface integral can be expressed through the double integral as

\[ {I_z} = {\mu _0}\iint\limits_S {\left( {{x^2} + {y^2}} \right)dS} = {\mu _0}\iint\limits_{D\left( {x,y} \right)} {\frac{{{x^2} + {y^2}}}{{\sqrt {1 - {x^2} - {y^2}} }}dxdy} ,\]

where the region of integration \({D\left( {x,y} \right)}\) is the circle \({x^2} + {y^2} \le 1.\) By changing to polar coordinates, we obtain

\[ {I_z} = {\mu _0}\iint\limits_{D\left( {x,y} \right)} {\frac{{{x^2} + {y^2}}}{{\sqrt {1 - {x^2} - {y^2}} }}dxdy} = {\mu _0}\iint\limits_{D\left( {r,\varphi } \right)} {\frac{{{r^2}}}{{\sqrt {1 - {r^2}} }}rdrd\varphi } = {\mu _0}\int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {\frac{{{r^3}dr}}{{\sqrt {1 - {r^2}} }}} .\]

To calculate the latter integral, we make the substitution: \(1 - {r^2} = t,\) \(\Rightarrow - 2rdr = dt\) or \(rdr = - {\frac{{dt}}{2}}.\) When \(r = 0,\) then \(t = 1,\) and when \(r = 1,\) then on the contrary, \(t = 0.\) As a result, we get the final answer for the moment of inertia:

\[{I_z} = 2\pi {\mu _0}\int\limits_0^1 {\frac{{{r^3}dr}}{{\sqrt {1 - {r^2}} }}} = 2\pi {\mu _0}\int\limits_0^1 {\frac{{{r^2} \cdot rdr}}{{\sqrt {1 - {r^2}} }}} = 2\pi {\mu _0}\int\limits_1^0 {\frac{{\left( {1 - t} \right)\left( { - \frac{{dt}}{2}} \right)}}{{\sqrt t }}} = \pi {\mu _0}\int\limits_1^0 {\frac{{t - 1}}{{\sqrt t }}dt} = \pi {\mu _0}\left[ {\int\limits_1^0 {\sqrt t dt} - \int\limits_1^0 {\frac{{dt}}{{\sqrt t }}} } \right] = \pi {\mu _0}\left[ {\left. {\left( {\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}} - \frac{{{t^{\frac{1}{2}}}}}{{\frac{1}{2}}}} \right)} \right|_1^0} \right] = \pi {\mu _0}\left[ {\left. {\left( {\frac{2}{3}\sqrt {{t^3}} - 2\sqrt t } \right)} \right|_1^0} \right] = \frac{4}{3}\pi {\mu _0}.\]

Example 5.

Find the attractive force between the hemisphere of radius \(r\) with constant density \({\mu_0}\) centered at the origin and the point mass \(m\) placed at the origin.

Solution.

Consider a point \(M\left( {x,y,z} \right)\) of the hemisphere that belongs to the area element \(dS\) (Figure \(5\)).

A point M of the hemisphere that belongs to the area element dS — Figure 5.

We can write the attractive force \(d\mathbf{F}\left( M \right)\) between the area element \(dS\) and the point mass \(m\) in the form

\[d\mathbf{F}\left( M \right) = \frac{{G{\mu _0}mdS}}{{{r^2}}}\mathbf{e}\left( {O,M} \right),\]

where \(G\) is gravitational constant, \(\mathbf{e}\left( {O,M} \right)\) is the unit vector directed from point \(O\) to point \(M.\)

Since \(\mathbf{e}\left( {O,M} \right) = \left( {{\frac{x}{r}}, {\frac{y}{r}}, {\frac{z}{r}}} \right),\) we can write:

\[d\mathbf{F}\left( M \right) = \frac{{G{\mu _0}mdS}}{{{r^3}}}\left( {x,y,z} \right).\]

After integrating on the given hemisphere, we obtain

\[{F_x} = \frac{{G{\mu _0}m}}{{{r^3}}}\iint\limits_S {xdS} ,\;\;\ {F_y} = \frac{{G{\mu _0}m}}{{{r^3}}}\iint\limits_S {ydS} ,\;\; {F_z} = \frac{{G{\mu _0}m}}{{{r^3}}}\iint\limits_S {zdS} .\]

In spherical coordinates the equation of the hemisphere is

\[\mathbf{r}\left( {\psi ,\theta } \right) = r\cos \psi \sin \theta \cdot \mathbf{i} + r\sin \psi \sin \theta \cdot \mathbf{j} + r\cos \theta \cdot \mathbf{k},\]

where \(0 \le \psi \le 2\pi ,\) \(0 \le \theta \le {\frac{\pi }{2}}.\)

It is known that the area element for the sphere is \(dS = {r^2}\sin \theta d\psi d\theta .\) Then the force components are

\[ {F_x} = \frac{{G{\mu _0}m}}{{{r^3}}}\iint\limits_S {xdS} = \frac{{G{\mu _0}m}}{{{r^3}}}\iint\limits_{D\left( {\psi ,\theta } \right)} {r\cos \psi \sin \theta \cdot {r^2}\sin \theta d\psi d\theta } = G{\mu _0}m\int\limits_0^{2\pi } {\cos \psi d\psi } \int\limits_0^{\frac{\pi }{2}} {{\sin^2}\theta d\theta } = G{\mu _0}m \cdot \left[ {\left. {\left( {\sin \psi } \right)} \right|_0^{2\pi }} \right] \cdot \int\limits_0^{\frac{\pi }{2}} {{\sin^2}\theta d\theta } = G{\mu _0}m \cdot 0 \cdot \int\limits_0^{\frac{\pi }{2}} {{\sin^2}\theta d\theta } = 0;\]

\[ {F_y} = \frac{{G{\mu _0}m}}{{{r^3}}}\iint\limits_S {ydS} = \frac{{G{\mu _0}m}}{{{r^3}}}\iint\limits_{D\left( {\psi ,\theta } \right)} {r\sin \psi \sin \theta \cdot {r^2}\sin \theta d\psi d\theta } = G{\mu _0}m\int\limits_0^{2\pi } {\sin \psi d\psi } \int\limits_0^{\frac{\pi }{2}} {{\sin^2}\theta d\theta } = G{\mu _0}m \cdot \left[ {\left. {\left( {-\cos \psi } \right)} \right|_0^{2\pi }} \right] \cdot \int\limits_0^{\frac{\pi }{2}} {{\sin^2}\theta d\theta } = G{\mu _0}m \cdot 0 \cdot \int\limits_0^{\frac{\pi }{2}} {{\sin^2}\theta d\theta } = 0;\]

\[ {F_z} = \frac{{G{\mu _0}m}}{{{r^3}}}\iint\limits_S {zdS} = \frac{{G{\mu _0}m}}{{{r^3}}}\iint\limits_{D\left( {\psi ,\theta } \right)} {r\cos \theta \cdot {r^2}\sin \theta d\psi d\theta } = G{\mu _0}m\int\limits_0^{2\pi } {d\psi } \int\limits_0^{\frac{\pi }{2}} {\sin\theta \cos \theta d\theta } = 2\pi G{\mu _0}m \int\limits_0^{\frac{\pi }{2}} {\sin\theta d\left( {\sin \theta } \right)} = 2\pi G{\mu _0}m \cdot \left[ {\left. {\left( {\frac{{{{\sin }^2}\theta }}{2}} \right)} \right|_0^{\frac{\pi }{2}}} \right] = 2\pi G{\mu _0}m \cdot \frac{1}{2} = \pi G{\mu _0}m.\]

Note that the result \({F_x} = {F_y} = 0\) is obvious due to symmetry of the uniform surface. Therefore the resultant force \(F = {F_z} = \pi G{\mu _0}m\) is directed along the \(z\)-axis.

Example 6.

Evaluate the pressure force acting on the dam sketched in Figure \(6,\) which retains a reservoir of water of width \(W\) and depth \(H.\)

Solution.

Under condition of hydrostatic equilibrium, the gauge pressure on the dam surface depending on \(z\) is given by the formula

\[p\left( z \right) = \rho g\left( {H - z} \right),\]

where \(\rho\) is water density, \(g\) is the acceleration of gravity.

The total pressure force acting on the dam surface is

\[\mathbf{F} = \iint\limits_S {p\mathbf{n}dS} = \int\limits_0^W {\int\limits_0^H {\rho g\left( {H - z} \right) \cdot \left( { - \mathbf{i}} \right)dydz} } = \rho gW\left( { - \mathbf{i}} \right)\left[ {\left. {\left( {Hz - \frac{{{z^2}}}{2}} \right)} \right|_0^H} \right] = \frac{{\rho gW{H^2}}}{2}\left( { - \mathbf{i}} \right).\]

The vector \(\left( { - \mathbf{i}} \right)\) shows the direction of the force \(\mathbf{F}.\) The absolute value of the force is

\[\left| \mathbf{F} \right| = \frac{{\rho gW{H^2}}}{2}.\]

Example 7.

A viscous fluid is flowing along a cylindrical pipe of radius \(R\) with velocity field \[\mathbf{v} = C{e^{ - r}}\mathbf{k} \left( {\text{m} \cdot {{\text{s}}^{ - 1}}} \right),\] where \(\mathbf{k}\) is the unit vector along the axis of the pipe in the direction of the flow, \(r\) is the distance from the axis, and \(C\) is a constant (Figure \(7\)). Calculate the fluid flux through a cross section of the pipe.

Solution.

A viscous fluid flowing along a cylindrical pipe — Figure 7.

To determine the fluid flux, we should compute the surface integral

\[\Phi = \iint\limits_S {\mathbf{v} \cdot d\mathbf{S}} .\]

As the vectors \(d\mathbf{S}\) and \(\mathbf{v}\) have the same direction, we can write:

\[\Phi = \iint\limits_S {C{e^{ - r}}dS} .\]

By changing to polar coordinates, we have

\[\Phi = C\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{e^{ - r}}rdr} = 2\pi C\int\limits_0^R {{e^{ - r}}rdr} .\]

The latter integral can be found using integration by parts. Let

\[u = r,\;\;{e^{ - r}}dr = dv,\;\; \Rightarrow du = dr,\;\; v = \int {{e^{ - r}}dr} = - {e^{ - r}},\]

so that

\[\int\limits_0^R {{e^{ - r}}rdr} = \left. {\left( { - r{e^{ - r}}} \right)} \right|_0^R - \int\limits_0^R {\left( { - {e^{ - r}}} \right)dr} = \left. {\left( { - r{e^{ - r}}} \right)} \right|_0^R + \int\limits_0^R {{e^{ - r}}dr} = \left. {\left( { - r{e^{ - r}}} \right)} \right|_0^R - \left. {\left( { - {e^{ - r}}} \right)} \right|_0^R = \left. {\left[ { - {e^{ - r}}\left( {r + 1} \right)} \right]} \right|_0^R = - {e^{ - R}}\left( {R + 1} \right) + {e^0} = 1 - \left( {R + 1} \right){e^{ - R}}.\]

Thus, the fluid flux is given by

\[\Phi = 2\pi C\left[ {1 - \left( {R + 1} \right){e^{ - R}}} \right] \; \left( {{\text{m}^3} \cdot {\text{s}^{ - 1}}} \right).\]

Example 8.

Calculate the electric field of an infinite sheet with a uniform charge density \(\sigma.\)

Solution.

By symmetry the resultant electric field must have a direction normal to the plane and must have the same magnitude at all points a common distance from the plane.

We take as a Gaussian surface a cylinder of cross-sectional area \(S\) and height \(2H\) (Figure \(8\)).

The electric field of an infinite sheet with a uniform charge density sigma — Figure 8.

The electric flux is only non-zero through ends of the cylinder, so the total flux is \(\Phi = 2{\varepsilon _0}ES,\) where \(E\) is the electric field at cylinder ends. The total charge enclosed by the cylinder is \(Q = \sigma S.\) Then using Gauss' Law we find:

\[\Phi = {\varepsilon _0}\iint\limits_S {E \cdot dS} = Q,\;\; \Rightarrow 2{\varepsilon _0}ES = \sigma S\;\;\text{or}\;\; E = \frac{\sigma }{{2{\varepsilon _0}}}.\]