Answer
The rate of heat flow across the horizontal disk:
1. at $z=1$:
$ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = - 1240\pi $
2. at $z=2$:
$ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = - 2480\pi $
3. at $z=3$:
$ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = - 3720\pi $
So, the greatest rate of heat flow across the disk is at $z=3$.
Work Step by Step
We have the temperature function:
$w\left( {x,y,z} \right) = \left( {30 - {z^2}} \right)\left( {2 - \left( {{x^2} + {y^2}} \right)} \right)$
So,
$\nabla w = \left( {\frac{{\partial w}}{{\partial x}},\frac{{\partial w}}{{\partial y}},\frac{{\partial w}}{{\partial z}}} \right) = ( - 2x\left( {30 - {z^2}} \right), - 2y\left( {30 - {z^2}} \right), - 2z\left( {2 - \left( {{x^2} + {y^2}} \right)} \right)$
1. The horizontal disk at $z=1$. We choose the outward-pointing normal vector ${\bf{N}} = - {\bf{k}}$.
Evaluate:
$\nabla w\cdot{\bf{N}} = 2z\left( {2 - \left( {{x^2} + {y^2}} \right)} \right)$
At $z=1$, we get $\nabla w\cdot{\bf{N}} = 2z\left( {2 - \left( {{x^2} + {y^2}} \right)} \right)$.
Since $K=310$, the rate of heat flow across the horizontal disk at $z=1$ is
$ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = - 310\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\bf{N}}{\rm{d}}S$
$ = - 620\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {2 - \left( {{x^2} + {y^2}} \right)} \right){\rm{d}}S$
Using $x = r\cos \theta $ and $y = r\sin \theta $, we evaluate this integral in cylindrical coordinates:
$ = - 620\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 2 } \left( {2 - {r^2}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = - 620\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^{\sqrt 2 } \left( {2r - {r^3}} \right){\rm{d}}r$
$ = - 1240\pi \left( {{r^2} - \frac{1}{4}{r^4}} \right)|_0^{\sqrt 2 } = - 1240\pi \left( {2 - 1} \right) = - 1240\pi $
So, the rate of heat flow across the horizontal disk at $z=1$ is
$ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = - 1240\pi $
2. The horizontal disk at $z=2$. We choose the outward-pointing normal vector ${\bf{N}} = - {\bf{k}}$.
Evaluate:
$\nabla w\cdot{\bf{N}} = 2z\left( {2 - \left( {{x^2} + {y^2}} \right)} \right)$
At $z=2$, we get $\nabla w\cdot{\bf{N}} = 4\left( {2 - \left( {{x^2} + {y^2}} \right)} \right)$.
The rate of heat flow across the horizontal disk at $z=2$ is
$ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = - 310\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\bf{N}}{\rm{d}}S$
$ = - 1240\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {2 - \left( {{x^2} + {y^2}} \right)} \right){\rm{d}}S$
Using $x = r\cos \theta $ and $y = r\sin \theta $, we evaluate this integral in cylindrical coordinates:
$ = - 1240\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 2 } \left( {2 - {r^2}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = - 1240\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^{\sqrt 2 } \left( {2r - {r^3}} \right){\rm{d}}r$
$ = - 2480\pi \left( {{r^2} - \frac{1}{4}{r^4}} \right)|_0^{\sqrt 2 } = - 2480\pi \left( {2 - 1} \right) = - 2480\pi $
So, the rate of heat flow across the horizontal disk at $z=2$ is
$ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = - 2480\pi $
3. The horizontal disk at $z=3$. We choose the outward-pointing normal vector ${\bf{N}} = - {\bf{k}}$.
Evaluate:
$\nabla w\cdot{\bf{N}} = 2z\left( {2 - \left( {{x^2} + {y^2}} \right)} \right)$
At $z=3$, we get $\nabla w\cdot{\bf{N}} = 6\left( {2 - \left( {{x^2} + {y^2}} \right)} \right)$.
The rate of heat flow across the horizontal disk at $z=3$ is
$ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = - 310\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\bf{N}}{\rm{d}}S$
$ = - 1860\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {2 - \left( {{x^2} + {y^2}} \right)} \right){\rm{d}}S$
Using $x = r\cos \theta $ and $y = r\sin \theta $, we evaluate this integral in cylindrical coordinates:
$ = - 1860\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 2 } \left( {2 - {r^2}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = - 1860\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^{\sqrt 2 } \left( {2r - {r^3}} \right){\rm{d}}r$
$ = - 3720\pi \left( {{r^2} - \frac{1}{4}{r^4}} \right)|_0^{\sqrt 2 } = - 3720\pi \left( {2 - 1} \right) = - 3720\pi $
So, the rate of heat flow across the horizontal disk at $z=3$ is
$ - K\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \nabla w\cdot{\rm{d}}{\bf{S}} = - 3720\pi $
Comparing these results, we conclude that the greatest rate of heat flow across the disk is at $z=3$.
