Question

The integral of the vector \( \mathbf{A}(\rho, \varphi, z) = \frac{40}{\rho} \cos \varphi \hat{\rho} \) (standard notation for cylindrical coordinates is used) over the volume of a cylinder of height \( L \) and radius \( R_0 \), is:

• A. \( 20 \pi R_0 L ( \hat{i} + \hat{j} ) \)
• B. 0
• C. \( 40 \pi R_0 L \hat{j} \)
• D. \( 40 \pi R_0 L \hat{i} \)

Hint:

When integrating vector fields in cylindrical coordinates, remember that integrals of odd functions over symmetric intervals (like \( \cos \varphi \) over \( 0 \) to \( 2\pi \)) can result in non-zero components along certain directions.

Answer 1

The Correct Option is D

Step 1: Understanding the integral.
The vector field \( \mathbf{A}(\rho, \varphi, z) \) is given in cylindrical coordinates. The integration involves calculating the volume integral of the vector field over the cylindrical volume. Since \( \cos \varphi \) is an odd function and is being integrated over the entire angular range from \( 0 \) to \( 2\pi \), the integral will result in zero for certain components of the vector field. However, in the specific case of \( \hat{i} \), the result does not cancel, and the integral produces a non-zero value along the \( \hat{i} \)-direction.
Step 2: Conclusion.
Thus, the correct answer is option (D), and the integral evaluates to \( 40 \pi R_0 L \hat{i} \).