A force `vecF = 2 x hati + 2 hatj + 3z^(2)hatk N` is acting on a particle .Find the work done by this force in displacing the body from `(1, 2, 3) m` to `(3,6,1)m`

Options

A. `-10J`
B. `100 J`
C. `10 J`
D. `1J`

Explanation

To find the work done by the force \(\vec{F} = 2 \hat{i} + 2 \hat{j} + 3z^2 \hat{k} \, \text{N}\) in displacing the body from the point \((1, 2, 3) \, \text{m}\) to \((3, 6, 1) \, \text{m}\), we can follow these steps:

### Step 1: Identify the displacement vector
The displacement vector \(\vec{d}\) can be calculated as:
\[ \vec{d} = (x_2 - x_1) \hat{i} + (y_2 - y_1) \hat{j} + (z_2 - z_1) \hat{k} \]
Substituting the coordinates:
\[ \vec{d} = (3 - 1) \hat{i} + (6 - 2) \hat{j} + (1 - 3) \hat{k} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k} \]

### Step 2: Set up the work done integral
The work done \(W\) by the force \(\vec{F}\) along the path of displacement \(\vec{d}\) can be calculated using the line integral:
\[ W = \int \vec{F} \cdot d\vec{s} \]
Where \(d\vec{s} = dx \hat{i} + dy \hat{j} + dz \hat{k}\).

### Step 3: Substitute the force and displacement
Substituting \(\vec{F}\) and \(d\vec{s}\) into the integral:
\[ W = \int (2x \hat{i} + 2y \hat{j} + 3z^2 \hat{k}) \cdot (dx \hat{i} + dy \hat{j} + dz \hat{k}) \]
This expands to:
\[ W = \int (2x \, dx + 2y \, dy + 3z^2 \, dz) \]

### Step 4: Determine the limits of integration
The limits for \(x\) will be from \(1\) to \(3\), for \(y\) from \(2\) to \(6\), and for \(z\) from \(3\) to \(1\).

### Step 5: Calculate each integral
1. For \(2x \, dx\):
\[ \int_{1}^{3} 2x \, dx = 2 \left[ \frac{x^2}{2} \right]_{1}^{3} = [x^2]_{1}^{3} = 9 - 1 = 8 \]

2. For \(2y \, dy\):
\[ \int_{2}^{6} 2y \, dy = 2 \left[ \frac{y^2}{2} \right]_{2}^{6} = [y^2]_{2}^{6} = 36 - 4 = 32 \]

3. For \(3z^2 \, dz\):
\[ \int_{3}^{1} 3z^2 \, dz = 3 \left[ \frac{z^3}{3} \right]_{3}^{1} = [z^3]_{3}^{1} = 1 - 27 = -26 \]

### Step 6: Combine the results
Now, we combine the results of the integrals:
\[ W = 8 + 32 - 26 = 14 \, \text{J} \]

### Step 7: Final answer
The total work done by the force is:
\[ W = 14 \, \text{J} \]