A force of `vecF=2xhati+2hatj+3z^2hatk 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. (a) `-10J`
B. (b) `100J`
C. (c) `10J`
D. (d) `1J`

Explanation

To find the work done by the force \(\vec{F} = 2x \hat{i} + 2 \hat{j} + 3z^2 \hat{k}\) when the particle is displaced from the point \((1, 2, 3)\) to \((3, 6, 1)\), we will follow these steps:

### Step 1: Define the displacement vector \(\vec{r}\)

The displacement vector \(\vec{r}\) can be calculated as:
\[ \vec{r} = \text{Final position} - \text{Initial position} = (3, 6, 1) - (1, 2, 3) = (3 - 1, 6 - 2, 1 - 3) = (2, 4, -2) \]
Thus, \(\vec{r} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k}\).

### Step 2: Express the work done \(W\)

The work done by a force during a displacement is given by the line integral:
\[ W = \int \vec{F} \cdot d\vec{r} \]
Where \(d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}\).

### Step 3: Substitute the force and displacement

We need to express the force \(\vec{F}\) in terms of the variables \(x\), \(y\), and \(z\):
\[ \vec{F} = 2x \hat{i} + 2 \hat{j} + 3z^2 \hat{k} \]

### Step 4: Set up the integral

We will integrate with respect to \(x\), \(y\), and \(z\). The limits for \(x\) will be from \(1\) to \(3\), for \(y\) from \(2\) to \(6\), and for \(z\) from \(3\) to \(1\).

The work done can be expressed as:
\[ W = \int_{1}^{3} 2x \, dx + \int_{2}^{6} 2 \, dy + \int_{3}^{1} 3z^2 \, dz \]

### Step 5: Calculate each integral

1. **First integral**:
\[ \int_{1}^{3} 2x \, dx = [x^2]_{1}^{3} = 3^2 - 1^2 = 9 - 1 = 8 \]

2. **Second integral**:
\[ \int_{2}^{6} 2 \, dy = [2y]_{2}^{6} = 2(6) - 2(2) = 12 - 4 = 8 \]

3. **Third integral**:
\[ \int_{3}^{1} 3z^2 \, dz = 3 \left[\frac{z^3}{3}\right]_{3}^{1} = [z^3]_{3}^{1} = 1^3 - 3^3 = 1 - 27 = -26 \]

### Step 6: Combine the results

Now, we combine the results of the integrals:
\[ W = 8 + 8 - 26 = 16 - 26 = -10 \text{ Joules} \]

### Final Answer

Thus, the work done by the force is:
\[ \boxed{-10 \text{ Joules}} \]