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
The volume integral \(\displaystyle \int_V e^{-\left(\frac{r}{R}\right)^2} \vec{\nabla} \cdot \left(\frac{\hat{r}}{r^2}\right) d^3r\), where \(V\) is the volume of a sphere of radius \(R\) centered at the origin, is equal to:
Let $f(x,y)=x^{3}-2y^{3}$. The curve along which $\nabla^{2} f = 0$ is
A curve is given by $\vec{r}(t)=t\hat{i}+t^{2}\hat{j}+t^{3}\hat{k}$. The unit vector of the tangent at $t=1$ is