Calculus

Surface Integrals

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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.

The pressure force acting on the dam
Figure 6.

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:

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