Surface integrals are used for computations of

- surface area;
- volume of a solid enclosed by a surface.

Consider these applications in more details.

### Surface Area

Suppose \(S\) be a smooth piecewise surface. The area of the surface is given by the integral

\[A = \iint\limits_S {dS} .\]

If the surface \(S\) is parameterized by the vector

\[

{\mathbf{r}\left( {u,v} \right) }

= {x\left( {u,v} \right)\mathbf{i} }+{ y\left( {u,v} \right)\mathbf{j} }+{ z\left( {u,v} \right)\mathbf{k},}

\]

then the surface area is

\[{A \text{ = }}\kern0pt{ \iint\limits_{D\left( {u,v} \right)} {\left| {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right|dudv}},\]

where \({D\left( {u,v} \right)}\) is the domain where the surface is defined.where \({D\left( {u,v} \right)}\) is the domain where the surface is defined.

If \(S\) is given explicitly by the function \({z\left( {x,y} \right)},\) the surface area is

\[{A \text{ = }}\kern0pt{ \iint\limits_{D\left( {x,y} \right)} {\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy}}\]

where \({D\left( {x,y} \right)}\) is the projection of the surface \(S\) onto the \(xy\)-plane.

### Volume of a Solid Bounded by a Closed Surface

Suppose that a solid is bounded by a smooth closed surface \(S.\) Then the volume of the solid is given by

\[{V \text{ = }}\kern0pt{ \frac{1}{3}\Big| {\iint\limits_S {xdydz + ydxdz + zdxdy} } \Big|.}\]

## Solved Problems

Click a problem to see the solution.

### Example 1

Compute the surface area of the portion of the paraboloid \(z =\) \(25 – {x^2} – {y^2}\) lying above the \(xy\)-plane.### Example 2

Find the area of a hemisphere of radius \(R.\)### 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.### Example 4

Find the volume of the ellipsoid \({\large\frac{{{x^2}}}{{{a^2}}}\normalsize} + {\large\frac{{{y^2}}}{{{b^2}}}\normalsize} \) \(+ {\large\frac{{{z^2}}}{{{c^2}}}\normalsize} \) \(= 1.\)### Example 1.

Compute the surface area of the portion of the paraboloid \(z =\) \(25 – {x^2} – {y^2}\) lying above the \(xy\)-plane.Solution.

The surface area is given by

\[

{A = \iint\limits_S {dS} \text{ = }}\kern0pt

{\iint\limits_{D\left( {x,y} \right)} {\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy} }

\]

Here we have

\[

{{\frac{{\partial z}}{{\partial x}}} = {\frac{{ – x}}{{\sqrt {25 – {x^2} – {y^2}} }}},\;\;\;}\kern0pt

{{\frac{{\partial z}}{{\partial y}}} = {\frac{{ – y}}{{\sqrt {25 – {x^2} – {y^2}} }}}}.

\]

Consequently,

\[

{A \text{ = }}\kern0pt

{\iint\limits_{D\left( {x,y} \right)} {\frac{{5dxdy}}{{\sqrt {25 – {x^2} – {y^2}} }}} .}

\]

By changing to polar coordinates, we have

\[

{A = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {\frac{{5rdr}}{{\sqrt {25 – {r^2}} }}} }

= {10\pi \int\limits_0^5 {\frac{{rdr}}{{\sqrt {25 – {r^2}} }}} }

= { – 5\pi \int\limits_0^5 {\frac{{d\left( {25 – {r^2}} \right)}}{{\sqrt {25 – {r^2}} }}} }

= { – 5\pi \left[ {\left. {\left( {\frac{{\sqrt {25 – {r^2}} }}{{\frac{1}{2}}}} \right)} \right|_0^5} \right] }

= {50\pi .}

\]