Curl of vector V(x, y, z) = 2x2i + 3z2j + y3k at x = y = z = 1 is A. -3i B. 3i C. 3i – 4j D. 3i – 6k

The correct answer is $\boxed{3i – 4j}$.

The curl of a vector field is a vector field that describes the circulation of a fluid around a point. It is calculated using the following formula:

$$\text{curl}(F) = \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \ F_x & F_y & F_z \end{pmatrix}$$

In this case, we have the vector field $F(x, y, z) = 2x^2i + 3z^2j + y^3k$. Substituting this into the formula for the curl, we get:

$$\text{curl}(F) = \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}

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