# Geometric Applications of Surface Integrals

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 or tap 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

${A = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {\frac{{5rdr}}{{\sqrt {25 – {r^2}} }}} } = {10\pi \int\limits_0^5 {\frac{{rdr}}{{\sqrt {25 – {r^2}} }}} } = { – 5\pi \int\limits_0^5 {\frac{{d\left( {25 – {r^2}} \right)}}{{\sqrt {25 – {r^2}} }}} } = { – 5\pi \left[ {\left. {\left( {\frac{{\sqrt {25 – {r^2}} }}{{\frac{1}{2}}}} \right)} \right|_0^5} \right] } = {50\pi .}$

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Problem 1
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Problems 2-4