Let $\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }$ . Then curl F is

A)i+ j
B)i
C)0
D)j
E)k

Question

Let $\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }$ . Then curl F is

A)i+ j
B)i
C)0
D)j
E)k

Quizplus · Accepted Answer

The curl of a vector field $\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$ is given by $\nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\mathbf{k}$. For $\mathbf{F}(x, y, z) = x^2\mathbf{i} + y^2\mathbf{j} + z^2\mathbf{k}$, all partial derivatives that would contribute to the curl are zero, hence the curl of $\mathbf{F}$ is $\mathbf{0}$.

Let F(x,y,z)=x2i+y2j+z2k\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }F(x,y,z)=x2i+y2j+z2k

Let $\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }$