Geometric Applications of Surface Integrals
Solved Problems
Example 3.
Compute the surface area of the torus with equation
\[{z^2} + {\left( {r - b} \right)^2} = {a^2} \left( {0 \le a \le b} \right)\]
in cylindrical coordinates.
Solution.
The parametric equations for the torus (see Figure \(2\)) are
\[\left\{ \begin{array}{l}
x = \left( {b + a\cos \psi } \right)\cos \varphi \\
y = \left( {b + a\cos \psi } \right)\sin\varphi \\
z = a\sin \psi
\end{array} \right..\]
Let us make sure that these equations properly desrcibe the circle in the toric section. Indeed, since \({x^2} + {y^2} = {r^2},\) then after substitution we get
\[{\left( {b + a\cos \psi } \right)^2}{\cos ^2}\varphi + {\left( {b + a\cos \psi } \right)^2}{\sin^2}\varphi = {r^2},\;\; \Rightarrow {\left( {b + a\cos \psi } \right)^2} = {r^2},\;\; \Rightarrow r = b + a\cos \psi ,\;\; \Rightarrow r - b = a\cos \psi ,\;\; \Rightarrow {\left( {r - b} \right)^2} + {z^2} = {\left( {a\cos \psi } \right)^2} + {\left( {a\sin \psi } \right)^2},\;\; \Rightarrow {\left( {r - b} \right)^2} + {z^2} = {a^2}.\]
Thus, the surface of the torus is parametrized by the position vector:
\[\mathbf{r}\left( {\varphi ,\psi } \right) = \left( {b + a\cos \psi } \right)\cos \varphi \cdot \mathbf{i}
+ \left( {b + a\cos \psi } \right)\sin\varphi \cdot \mathbf{j} + a\sin \psi \cdot \mathbf{k}.\]
The surface area is given by the formula
\[A = \iint\limits_S {dS} = \iint\limits_{D\left( {\varphi ,\psi } \right)} {\left| {\frac{{\partial \mathbf{r}}}{{\partial \varphi }} \times \frac{{\partial \mathbf{r}}}{{\partial \psi }}} \right|d\varphi d\psi } .\]
The cross product in this formula is calculated as follows:
\[\frac{{\partial \mathbf{r}}}{{\partial \varphi }} \times \frac{{\partial \mathbf{r}}}{{\partial \psi }}
= \left| {\begin{array}{*{20}{c}}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
{ - \left( {b + a\cos \psi } \right)\sin \varphi } & {\left( {b + a\cos \psi } \right)\cos\varphi } & 0\\
{ - a\sin \psi \cos \varphi } & { - a\sin \psi \sin \varphi } & {a\cos\psi }
\end{array}} \right|
= a\cos\psi \left( {b + a\cos \psi } \right)\cos\varphi \cdot \mathbf{i}
+ a\cos\psi \left( {b + a\cos \psi } \right)\sin\varphi \cdot \mathbf{j}
+ \left[ {a\sin \psi \,{{\sin }^2}\varphi \left( {b + a\cos \psi } \right)
+ a\sin \psi \,{{\cos }^2}\varphi \left( {b + a\cos \psi } \right)} \right] \cdot \mathbf{k}
= a\cos\varphi \cos \psi \left( {b + a\cos \psi } \right) \cdot \mathbf{i}
+ a\sin\varphi \cos \psi \left( {b + a\cos \psi } \right) \cdot \mathbf{j}
+ a\sin\psi \left( {b + a\cos \psi } \right) \cdot \mathbf{k}.\]
Then the absolute value of the cross product is
\[\left| {\frac{{\partial \mathbf{r}}}{{\partial \varphi }} \times \frac{{\partial \mathbf{r}}}{{\partial \psi }}} \right|
= \left[ {{a^2}{\cos^2}\varphi \,{{\cos }^2}\psi {{\left( {b + a\cos \psi } \right)}^2}} \right.
+ {a^2}{\sin^2}\varphi \,{\cos ^2}\psi {\left( {b + a\cos \psi } \right)^2}
+ {\left. {{a^2}{{\sin }^2}\psi {{\left( {b + a\cos \psi } \right)}^2}} \right]^{\frac{1}{2}}}
= a{\left[ {{{\cos }^2}\psi {{\left( {b + a\cos \psi } \right)}^2} + {{\sin }^2}\psi {{\left( {b + a\cos \psi } \right)}^2}} \right]^{\frac{1}{2}}}
= a\left( {b + a\cos \psi } \right).\]
Now we can find the surface area of the torus:
\[A = \iint\limits_S {dS} = \iint\limits_{D\left( {\varphi ,\psi } \right)} {a\left( {b + a\cos \psi } \right)d\varphi d\psi } = a\int\limits_0^{2\pi } {d\varphi } \int\limits_0^{2\pi } {\left( {b + a\cos \psi } \right)d\psi } = 2\pi a\int\limits_0^{2\pi } {\left( {b + a\cos \psi } \right)d\psi } = 2\pi a\left[ {\left. {\left( {b\psi + a\sin \psi } \right)} \right|_0^{2\pi }} \right] = 2\pi a \cdot 2\pi b = 4{\pi ^2}ab.\]
Example 4.
Find the volume of the ellipsoid \[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1.\]
Solution.
We use the following formula to calculate the volume of the ellipsoid:
\[V = \frac{1}{3}\left| {\iint\limits_S {xdydz + ydxdz + zdxdy} } \right|.\]
The surface \(S\) of the ellipsoid can be parametrized by
\[\mathbf{r}\left( {u,v} \right) = a\cos u\sin v \cdot \mathbf{i} + b\sin u\sin v \cdot \mathbf{j} + c\cos v \cdot \mathbf{k},\;\; \text{where} \;\; 0 \le u \le 2\pi ,\;\; 0 \le v \le \pi .\]
(The variables \(u, v\) correspond to spherical coordinates \(\psi\) and \(\theta\).)
According to the formula for the volume given above, we identify that the vector field here is \(\mathbf{F} = \left( {x,y,z} \right),\) so that
\[P = x = a\cos u\sin v,\;\; Q = y = b\sin u\sin v,\;\; R = z = c\cos v.\]
Since
\[\iint\limits_S {Pdydz + Qdzdx + Rdxdy}
= \iint\limits_{D\left( {u,v} \right)} {\left| {\begin{array}{*{20}{c}}
P & Q & R\\
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial z}}{{\partial u}}}\\
{\frac{{\partial x}}{{\partial v}}}&{\frac{{\partial y}}{{\partial v}}}&{\frac{{\partial z}}{{\partial v}}}
\end{array}} \right|dudv} ,\]
we obtain the following expression for the surface integral:
\[\iint\limits_S {xdydz + ydzdx + zdxdy}
= \iint\limits_{D\left( {u,v} \right)} {\left| {\begin{array}{*{20}{c}}
{a\cos u\sin v}&{b\sin u\sin v}&{c\cos v}\\
{ - a\sin u\sin v}&{b\cos u\sin v}&0\\
{a\cos u \cos v}&{b\sin u\cos v}&{ - c\sin v}
\end{array}} \right|dudv}
= \int\limits_{D\left( {u,v} \right)} {\left[ {a\cos u\sin v\left( { - bc\cos u\,{{\sin }^2}v} \right)} \right.}
- b\sin u\sin v\left( {ac\sin u\,{{\sin }^2}v} \right)
+ \left. {c\cos v\left( { - ab\,{{\sin }^2}u\sin v\cos v - ab\,{{\cos }^2}u\sin v\cos v} \right)} \right]dudv
= - abc\iint\limits_{D\left( {u,v} \right)} {\left[ {{{\sin }^3}v\left( {{{\cos }^2}u + {{\sin }^2}u} \right) + \sin v\,{{\cos }^2}v} \right]dudv}
= - abc\iint\limits_{D\left( {u,v} \right)} {\sin v\left( {{{\sin }^2}v + {{\cos }^2}v} \right)dudv}
= - abc\iint\limits_{D\left( {u,v} \right)} {\sin v dudv}
= - abc\iint\limits_0^{2\pi } {du} \int\limits_0^\pi {\sin v dv}
= - abc \cdot 2\pi \cdot \left. {\left( { - \cos v} \right)} \right|_0^\pi
= 2\pi abc \cdot \left( {\cos \pi - \cos 0} \right)
= - 4\pi abc.\]
Hence, the volume of the ellipsoid is
\[V = \left| {\frac{1}{3}\left( { - 4\pi abc} \right)} \right| = \frac{{4\pi abc}}{3}.\]