# The temperature at the point (x, y, z) in a substance with conductivity K = 7.5 is...

## Question:

The temperature at the point {eq}(x, y, z) {/eq} in a substance with conductivity {eq}K = 7.5 {/eq} is {eq}u(x,y,z)=2y^{2}+2z^{2} {/eq}. Find the rate of heat flow inward across the cylindrical surface {eq}y^{2}+z^{2}=6,0\leq x\leq 2 {/eq}.

## Applying the Divergence Theorem:

In this problem, we apply the divergence theorem to calculate the flux of a heat field through a cylinder. The axis of the cylinder is the x axis, so we modify the cylindrical coordinate system as follows:

{eq}y = r \cos \theta \\ z = r \sin \theta \\ x = x {/eq}

This makes the triple integral that appears in the divergence theorem much easier to evaluate.

## Answer and Explanation:

We wish to calculate the flux of the heat function {eq}u(x, y, z) = 2y^2 + 2z^2 {/eq} across the cylindrical surface bounded by {eq}y^2 + z^2 = 6, \: 0 \leq x \leq 2. {/eq}

We will assume that we are including the ends of the cylinder, and apply the Divergence Theorem.

The divergence of {eq}u(x, y, z) {/eq} is {eq}\mathrm{div \: u} = \displaystyle\frac{\partial}{\partial x}(0) + \frac{\partial}{\partial y}(2y^2) + \frac{\partial}{\partial z}(2z^2) = 4y + 4z. {/eq}

We can describe the cylinder in cylindrical coordinates: {eq}0 \leq \theta \leq 2\pi, \: 0 \leq r \leq \sqrt 6, \: 0 \leq x \leq 2. {/eq} Therefore the flux is

{eq}\begin{eqnarray*}\displaystyle\int_S \mathbf F \cdot \mathbf N \: dS & = & \int_0^{2\pi} \int_0^{\sqrt 6} \int_0^2 (4y + 4z) \: r \: dx \: dr \: d\theta \\ \\ & =& \int_0^{2\pi} \int_0^{\sqrt 6} \int_0^2 4r (r \cos \theta + r \sin \theta) \: dx \: dr \: d\theta \\ \\ & = & \int_0^{2\pi} \int_0^{\sqrt 6} \int_0^2 4r^2 (\cos \theta + \sin \theta) \: dx \: dr \: d\theta \\ \\ & = & \int_0^{2\pi} \int_0^{\sqrt 6} 4 x r^2 (\cos \theta + \sin \theta) \biggr|_0^2 \: dr \: d\theta \\ \\ & = & \int_0^{2\pi} \int_0^{\sqrt 6} 8 r^2 (\cos \theta + \sin \theta) \: dr \: d\theta \\ \\ & = & \int_0^{2\pi} \displaystyle\frac{8r^3}3 (\cos \theta + \sin \theta) \biggr|_0^{\sqrt 6} \: d\theta \\ \\ & = & \int_0^{2\pi} 16 \sqrt 6 (\cos \theta + \sin \theta) \: d\theta \\ \\ & = & 16 \sqrt 6 (\sin \theta - \cos \theta) \biggr|_0^{2\pi} \\ \\ & = & 0 \end{eqnarray*} {/eq}

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from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14