A body is displaced from origin to `(2m, 4m)` under the following two forces:
(a) `F=(2^hati + 6^hatj)N`, a constant force
(b) `F(2x^hati + 3y^(2)hatj)N`
Find work done by the given forces in both cases.

Explanation

To find the work done by the given forces in both cases, we will use the formula for work done:

\[ W = \int \mathbf{F} \cdot d\mathbf{r} \]

where \(\mathbf{F}\) is the force vector and \(d\mathbf{r}\) is the displacement vector.

### Part (a): Constant Force \(\mathbf{F} = (2\hat{i} + 6\hat{j}) \, \text{N}\)

1. **Identify the force vector and displacement vector**:
- Force: \(\mathbf{F} = 2\hat{i} + 6\hat{j}\)
- Displacement from \((0,0)\) to \((2,4)\): \(d\mathbf{r} = dx \hat{i} + dy \hat{j}\)

2. **Set up the integral for work done**:
\[ W = \int_{(0,0)}^{(2,4)} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{2} \int_{0}^{4} (2\hat{i} + 6\hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \]

3. **Calculate the dot product**:
\[ \mathbf{F} \cdot d\mathbf{r} = (2\hat{i} + 6\hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = 2dx + 6dy \]

4. **Integrate**:
\[ W = \int_{0}^{2} \int_{0}^{4} (2dx + 6dy) \]
This can be separated into two integrals:
\[ W = \int_{0}^{2} 2dx + \int_{0}^{4} 6dy \]

5. **Evaluate the integrals**:
- For the first integral:
\[ \int_{0}^{2} 2dx = 2x \bigg|_{0}^{2} = 2(2) - 2(0) = 4 \]
- For the second integral:
\[ \int_{0}^{4} 6dy = 6y \bigg|_{0}^{4} = 6(4) - 6(0) = 24 \]

6. **Add the results**:
\[ W = 4 + 24 = 28 \, \text{J} \]

### Part (b): Variable Force \(\mathbf{F} = (2x\hat{i} + 3y^2\hat{j}) \, \text{N}\)

1. **Identify the force vector and displacement vector**:
- Force: \(\mathbf{F} = 2x\hat{i} + 3y^2\hat{j}\)
- Displacement from \((0,0)\) to \((2,4)\): \(d\mathbf{r} = dx \hat{i} + dy \hat{j}\)

2. **Set up the integral for work done**:
\[ W = \int_{(0,0)}^{(2,4)} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{2} \int_{0}^{4} (2x\hat{i} + 3y^2\hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \]

3. **Calculate the dot product**:
\[ \mathbf{F} \cdot d\mathbf{r} = (2x\hat{i} + 3y^2\hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = 2x \, dx + 3y^2 \, dy \]

4. **Integrate**:
\[ W = \int_{0}^{2} \int_{0}^{4} (2x \, dx + 3y^2 \, dy) \]
This can be separated into two integrals:
\[ W = \int_{0}^{2} 2x \, dx + \int_{0}^{4} 3y^2 \, dy \]

5. **Evaluate the integrals**:
- For the first integral:
\[ \int_{0}^{2} 2x \, dx = x^2 \bigg|_{0}^{2} = 2^2 - 0^2 = 4 \]
- For the second integral:
\[ \int_{0}^{4} 3y^2 \, dy = y^3 \bigg|_{0}^{4} = 4^3 - 0^3 = 64 \]

6. **Add the results**:
\[ W = 4 + 64 = 68 \, \text{J} \]

### Final Answers:
- Work done by constant force: \(28 \, \text{J}\)
- Work done by variable force: \(68 \, \text{J}\)