Answer:
The rate of the heat flow = 1260 π
Step-by-step explanation:
From the information given :
k = 4.5
u(x,y,z) = 5y² + 5z²
Surface cylinder:
y² +z² = 5, 0 ≤ x ≤ 2
[tex]\mathtt{\overline F = \bigtriangledown u = -k(0,10y, 10z )}[/tex]
[tex]\mathtt{\overline F = -4.5(0,10y, 10z )}[/tex]
[tex]\mathtt{\overline F = (0,-45y, -45z ) \ --- (1)}[/tex]
Now parameterizing the surface by :
x = u , y = [tex]\mathtt{\sqrt{7} \ cos \ t}[/tex] , z = [tex]\mathtt{\sqrt{7} \ sin \ t}[/tex]
0 ≤ x ≤ 2 , 0 ≤ t ≤ 2π
[tex]\mathtt{{ \left. \begin{array}{1} \overline{r_y} = (1,0,0) } \\ \\ \overline{r_t} = (0, \ - \sqrt{7}\ sin \ t, \sqrt{7} \ cos \ t) \end{array} \right\} = r_u \times r_t}[/tex]
[tex]\mathtt{\overline r_u \times \overline r_t = ( -0, - \sqrt{7} \ cos \ t , - \sqrt{7} \ sin \ t) --- (2)}[/tex]
Taking integral of both equations; we have:
[tex]\mathtt{= \int ^{2}_0 \int ^{2 \pi}_{0} (0, -45y, -45 z) (0, - \sqrt{7} \ cos \ t, - \sqrt{7} \ sin \ t) \ dtdu}[/tex]
[tex]\mathtt{= \int ^{2}_0 \int ^{2 \pi}_{0} ( 45\sqrt{7} \ y\ cos \ t+ 45 \sqrt{7} \ z \ sin \ t) \ dtdu}[/tex]
[tex]\mathtt{= 45\sqrt{7}\ \int ^{2}_0 \int ^{2 \pi}_{0} (( \sqrt{7} \ cos \ t)cos \ t + (\sqrt{7} \ \ sin \ t) sin \ t) \ dtdu}[/tex]
[tex]\mathtt{= 45\times {7}\ \int ^{2}_0 \int ^{2 \pi}_{0} (1) \ dtdu}[/tex]
= 315 × (2) × (2π)
= 1260 π
