The line integral of the vector function \(u(x, y) = 2y \, \hat{i} + x \, \hat{j}\) along the straight line from (0, 0) to (2, 4) is ..........
Step 1: Parametrize the line.
The line from \((0,0)\) to \((2,4)\) can be written as \(x = t,\ y = 2t\), where \(t\) varies from 0 to 2.
Then, \(dx = dt\) and \(dy = 2dt\).
Step 2: Evaluate the line integral.
\[
\int_C \vec{u} \cdot d\vec{r} = \int_0^2 [(2y)dx + x dy]
\]
Substitute \(y = 2t,\ dx = dt,\ dy = 2dt,\ x = t\):
\[
\int_0^2 [2(2t)(1) + t(2)] dt = \int_0^2 [4t + 2t] dt = \int_0^2 6t dt
\]
\[
= [3t^2]_0^2 = 12
\]
Hence, the line integral = 12.
Step 3: Verify calculation.
All substitutions are correct, and path is linear; thus, final answer is 12.
Final Answer: 12

At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)