recategorized by
99 views
0 votes
0 votes

Consider a vector field $\vec{A}(\vec{r}).$ The closed loop line integral $\displaystyle {} \int \vec{A}\bullet\vec{dl}$ can be expressed as

  1. $\displaystyle {} \iint (\triangledown \times \vec{A}) \bullet\vec{ds}$ over the closed surface bounded by the loop
  2. $\displaystyle {} \iiint (\triangledown \bullet \vec{A}) dv$ over the closed volume bounded by the loop
  3. $\displaystyle {} \iiint (\triangledown \bullet \vec{A}) dv$ over the open volume bounded by the loop
  4. $\displaystyle {} \iiint (\triangledown \times \vec{A}) \bullet\vec{ds}$ over the open surface bounded by the loop
recategorized by

Please log in or register to answer this question.

Answer: