6.6 Surface integrals (Page 12/27)

Calculus volume 3 Page 12 / 27

Therefore, to compute a surface integral over a vector field we can use the equation

\iint_{S} F \cdot N d S = \iint_{D} (F (r (u, v)) \cdot (t_{u} \times t_{v})) d A .

Calculating a surface integral

Calculate the surface integral $\iint_{S} F \cdot N d S,$ where $F = ⟨ − y, x, 0 ⟩$ and S is the surface with parameterization $r (u, v) = ⟨ u, v^{2} - u, u + v ⟩, 0 \leq u < 3, 0 \leq v \leq 4 .$

The tangent vectors are $t_{u} = ⟨ 1, −1, 1 ⟩$ and $t_{v} = ⟨ 0, 2 v, 1 ⟩ .$ Therefore,

t_{u} \times t_{v} = ⟨ −1 - 2 v, −1, 2 v ⟩ .

By [link] ,

\begin{matrix} \iint_{S} F \cdot d S & = \int_{0}^{4} \int_{0}^{3} F (r (u, v)) \cdot (t_{u} \times t_{v}) d u d v \\ = \int_{0}^{4} \int_{0}^{3} ⟨ u - v^{2}, u, 0 ⟩ \cdot ⟨ −1 - 2 v, −1, 2 v ⟩ d u d v \\ = \int_{0}^{4} \int_{0}^{3} [(u - v^{2}) (−1 - 2 v) - u] d u d v \\ = \int_{0}^{4} \int_{0}^{3} (2 v^{3} + v^{2} - 2 u v - 2 u) d u d v \\ = {\int_{0}^{4} [2 v^{3} u + v^{2} u - v u^{2} - u^{2}]}_{0}^{3} d v \\ = \int_{0}^{4} (6 v^{3} + 3 v^{2} - 9 v - 9) d v \\ = {[\frac{3 v^{4}}{2} + v^{3} - \frac{9 v^{2}}{2} - 9 v]}_{0}^{4} \\ = 340. \end{matrix}

Therefore, the flux of F across S is 340.

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Calculate surface integral $\iint_{S} F \cdot d S,$ where $F = ⟨ 0, − z, y ⟩$ and S is the portion of the unit sphere in the first octant with outward orientation.

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Calculating mass flow rate

Let $v (x, y, z) = ⟨ 2 x, 2 y, z ⟩$ represent a velocity field (with units of meters per second) of a fluid with constant density 80 kg/m ³ . Let S be hemisphere $x^{2} + y^{2} + z^{2} = 9$ with $z \geq 0$ such that S is oriented outward. Find the mass flow rate of the fluid across S .

A parameterization of the surface is

r (ϕ, θ) = ⟨ 3 cos θ sin ϕ, 3 sin θ sin ϕ, 3 cos ϕ ⟩, 0 \leq θ \leq 2 π, 0 \leq ϕ \leq π / 2.

As in [link] , the tangent vectors are

t_{θ} ⟨ −3 sin θ sin ϕ, 3 cos θ sin ϕ, 0 ⟩ and t_{ϕ} ⟨ 3 cos θ cos ϕ, 3 sin θ cos ϕ, −3 sin ϕ ⟩,

and their cross product is

t_{ϕ} \times t_{θ} = ⟨ 9 cos θ {sin}^{2} ϕ, 9 sin θ {sin}^{2} ϕ, 9 sin ϕ cos ϕ ⟩ .

Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. Therefore we use the orientation $N = ⟨ 9 cos θ {sin}^{2} ϕ, 9 sin θ {sin}^{2} ϕ, 9 sin ϕ cos ϕ ⟩$ for the sphere.

By [link] ,

\begin{matrix} \iint_{S} ρ v \cdot d S & = 80 \int_{0}^{2 π} \int_{0}^{π / 2} v (r (ϕ, θ)) \cdot (t_{ϕ} \times t_{θ}) d ϕ d θ \\ = 80 \int_{0}^{2 π} \int_{0}^{π / 2} \\ = 80 \int_{0}^{2 π} \int_{o}^{π / 2} \begin{matrix} ⟨ 6 cos θ sin ϕ, 6 sin θ sin ϕ, 3 cos ϕ ⟩ \\ \cdot ⟨ 9 cos θ {sin}^{2} ϕ, 9 sin θ {sin}^{2} ϕ, 9 sin ϕ cos ϕ ⟩ d ϕ d θ \end{matrix} \\ = 80 \int_{0}^{2 π} \int_{0}^{π / 2} 54 {sin}^{3} ϕ + 27 {cos}^{2} ϕ sin ϕ d ϕ d θ \\ = 80 \int_{0}^{2 π} \int_{0}^{π / 2} 54 (1 - {cos}^{2} ϕ) sin ϕ + 27 {cos}^{2} ϕ sin ϕ d ϕ d θ \\ = 80 \int_{0}^{2 π} \int_{0}^{π / 2} 54 sin ϕ - 27 {cos}^{2} ϕ sin ϕ d ϕ d θ \\ = 80 \int_{0}^{2 π} {[−54 cos ϕ + 9 {cos}^{3} ϕ]}_{ϕ = 0}^{ϕ = 2 π} d θ \\ = 80 \int_{0}^{2 π} 45 d θ = 7200 π . \end{matrix}

Therefore, the mass flow rate is $7200 π kg / sec / m^{2} .$

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Let $v (x, y, z) = ⟨ x^{2} + y^{2}, z, 4 y ⟩$ m/sec represent a velocity field of a fluid with constant density 100 kg/m ³ . Let S be the half-cylinder $r (u, v) = ⟨ cos u, sin u, v ⟩, 0 \leq u \leq π, 0 \leq v \leq 2$ oriented outward. Calculate the mass flux of the fluid across S .

400 kg/sec/m

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In [link] , we computed the mass flux, which is the rate of mass flow per unit area. If we want to find the flow rate (measured in volume per time) instead, we can use flux integral $\int \int_{S} v • N d S,$ which leaves out the density. Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. Therefore, we have the following characterization of the flow rate of a fluid with velocity v across a surface S :

Flow rate of fluid across S = \int \int_{S} v • d S .

To compute the flow rate of the fluid in [link] , we simply remove the density constant, which gives a flow rate of $90 π m^{3} / sec .$

Both mass flux and flow rate are important in physics and engineering. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface.

In addition to modeling fluid flow, surface integrals can be used to model heat flow. Suppose that the temperature at point $(x, y, z)$ in an object is $T (x, y, z) .$ Then the heat flow is a vector field proportional to the negative temperature gradient in the object. To be precise, the heat flow is defined as vector field $F = − k \nabla T,$ where the constant k is the thermal conductivity of the substance from which the object is made (this constant is determined experimentally). The rate of heat flow across surface S in the object is given by the flux integral

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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