The volumes of two bodies are measured to be
`V_1 = (10.2+-0.02) cm^3 and V_2 = (6.4 +- 0.01)cm^3`. Calculate sum and difference in volumes with error limits.

Explanation

To solve the problem of calculating the sum and difference of two volumes with their associated error limits, we can follow these steps:

### Step 1: Identify the Given Values
We have two volumes:
- \( V_1 = 10.2 \, \text{cm}^3 \) with an error of \( \Delta V_1 = 0.02 \, \text{cm}^3 \)
- \( V_2 = 6.4 \, \text{cm}^3 \) with an error of \( \Delta V_2 = 0.01 \, \text{cm}^3 \)

### Step 2: Calculate the Sum of the Volumes
To find the sum \( V_{\text{sum}} \):
\[ V_{\text{sum}} = V_1 + V_2 = 10.2 \, \text{cm}^3 + 6.4 \, \text{cm}^3 \]
Calculating the sum:
\[ V_{\text{sum}} = 10.2 + 6.4 = 16.6 \, \text{cm}^3 \]

### Step 3: Calculate the Error in the Sum
The error in the sum is found by adding the absolute errors:
\[ \Delta V_{\text{sum}} = \Delta V_1 + \Delta V_2 = 0.02 \, \text{cm}^3 + 0.01 \, \text{cm}^3 \]
Calculating the error:
\[ \Delta V_{\text{sum}} = 0.02 + 0.01 = 0.03 \, \text{cm}^3 \]

### Step 4: Write the Result for the Sum
Thus, the sum of the volumes with error limits is:
\[ V_{\text{sum}} = 16.6 \pm 0.03 \, \text{cm}^3 \]

### Step 5: Calculate the Difference of the Volumes
To find the difference \( V_{\text{diff}} \):
\[ V_{\text{diff}} = V_1 - V_2 = 10.2 \, \text{cm}^3 - 6.4 \, \text{cm}^3 \]
Calculating the difference:
\[ V_{\text{diff}} = 10.2 - 6.4 = 3.8 \, \text{cm}^3 \]

### Step 6: Calculate the Error in the Difference
The error in the difference is also found by adding the absolute errors:
\[ \Delta V_{\text{diff}} = \Delta V_1 + \Delta V_2 = 0.02 \, \text{cm}^3 + 0.01 \, \text{cm}^3 \]
Calculating the error:
\[ \Delta V_{\text{diff}} = 0.02 + 0.01 = 0.03 \, \text{cm}^3 \]

### Step 7: Write the Result for the Difference
Thus, the difference of the volumes with error limits is:
\[ V_{\text{diff}} = 3.8 \pm 0.03 \, \text{cm}^3 \]

### Final Results
- Sum: \( V_{\text{sum}} = 16.6 \pm 0.03 \, \text{cm}^3 \)
- Difference: \( V_{\text{diff}} = 3.8 \pm 0.03 \, \text{cm}^3 \)

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