# 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

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

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

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

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

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

$V = \frac{1}{3}\left| {\iint\limits_S {xdydz + ydxdz + zdxdy} } \right|.$

## 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 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

$A = \iint\limits_S {dS} = \iint\limits_{D\left( {x,y} \right)} {\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy}.$

Here we have

$\frac{{\partial z}}{{\partial x}} = \frac{{ - x}}{{\sqrt {25 - {x^2} - {y^2}} }},\;\; \frac{{\partial z}}{{\partial y}} = \frac{{ - y}}{{\sqrt {25 - {x^2} - {y^2}} }}.$

Consequently,

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

By changing to polar coordinates, we have

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

### Example 2.

Find the area of a hemisphere of radius $$R.$$

Solution.

Using the spherical coordinate system, we can describe the surface of the upper hemisphere as

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

where $$0 \le \psi \le 2\pi ,$$ $$0 \le \theta \le {\frac{\pi }{2}}$$ (Figure $$1$$).

Calculate the area element.

$\frac{{\partial \mathbf{r}}}{{\partial \psi }} = \frac{\partial }{{\partial \psi }}\left( {R\cos \psi \sin \theta ,R\sin \psi \sin \theta ,R\cos \theta } \right) = \left( { - R\sin \psi \sin \theta ,R\cos \psi \sin \theta ,0} \right);$
$\frac{{\partial \mathbf{r}}}{{\partial \theta }} = \frac{\partial }{{\partial \theta }}\left( {R\cos \psi \sin \theta ,R\sin \psi \sin \theta ,R\cos \theta } \right) = \left( {R\cos \psi \cos \theta ,R\sin \psi \cos \theta , - R\sin \theta } \right).$

Find the cross product of the given vectors:

$\frac{{\partial \mathbf{r}}}{{\partial \psi }} \times \frac{{\partial \mathbf{r}}}{{\partial \theta }} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ { - R\sin \psi \sin \theta } & {R\cos \psi \sin \theta } & 0\\ {R\cos \psi \cos \theta }&{R\sin \psi \cos \theta }&{ - R\sin \theta } \end{array}} \right| = - {R^2}\cos \psi \,{\sin ^2}\theta \cdot \mathbf{i} - {R^2}\sin\psi \,{\sin ^2}\theta \cdot \mathbf{j} - {R^2}\left( {{\sin^2}\psi \sin \theta \cos \theta + {{\cos }^2}\psi \sin \theta \cos \theta } \right) \cdot \mathbf{k} = - {R^2}\cos \psi \,{\sin ^2}\theta \cdot \mathbf{i} - {R^2}\sin\psi \,{\sin ^2}\theta \cdot \mathbf{j} - {R^2}\sin \theta \cos \theta \cdot \mathbf{k}.$

Hence, the area element is

$dS = \left| {\frac{{\partial \mathbf{r}}}{{\partial \psi }} \times \frac{{\partial \mathbf{r}}}{{\partial \theta }}} \right|d\psi d\theta = {R^2}\sqrt {{{\sin }^4}\theta \left( {{{\cos }^2}\psi + {{\sin }^2}\psi } \right) + {{\sin }^2}\theta \,{{\cos }^2}\theta } \,d\psi d\theta = {R^2}\sin \theta \sqrt {{{\sin }^2}\theta + {{\cos }^2}\theta } \,d\psi d\theta = {R^2}\sin \theta \,d\psi d\theta .$

Then the surface area of the hemisphere is given by

$A = \iint\limits_S {dS} = \int\limits_{D\left( {\psi ,\theta } \right)} {{R^2}\sin \theta d\psi d\theta } = {R^2}\int\limits_0^{2\pi } {d\psi } \int\limits_0^{\frac{\pi }{2}} {\sin \theta d\theta } = 2\pi {R^2} \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^{\frac{\pi }{2}}} \right] = 2\pi {R^2}\left( { - \cos \frac{\pi }{2} + \cos 0} \right) = 2\pi {R^2}.$