Question

Find the curl of the vector field \(F\). \(\mathbf{F}(x, y, z)=\arcsin y \mathbf{i}+\sqrt{1-x^{2}} \mathbf{j}+y^{2} \mathbf{k}\)

Short answer

The curl of the vector field \(\mathbf{F}\) is \(\nabla \times \mathbf{F} = 2y \mathbf{i} + \left(-\frac{x}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-y^{2}}}\right) \mathbf{k}\)

Step-by-step solution

Write out the expression for curl F

The curl of the vector field \mathbf{F} is given by: \(\nabla \times \mathbf{F} = \text{curl} \mathbf{F} = \text{det}\left\| \begin{array} {ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ {\frac{\partial}{\partial x}} & {\frac{\partial}{\partial y}} & {\frac{\partial}{\partial z}} \ \arcsin y & \sqrt{1-x^{2}} & y^{2} \end{array} \right\|\). This implies computing the determinant of a 3x3 matrix which uses partial derivatives.

Calculate the determinant

Calculate the determinant of the 3x3 matrix: \( \nabla \times \mathbf{F} = \left(\frac{\partial (y^2)}{\partial y} - \frac{\partial (\sqrt{1-x^{2}})}{\partial z}\right) \mathbf{i} - \left(\frac{\partial (y^2)}{\partial x} - \frac{\partial (\arcsin y)}{\partial z}\right) \mathbf{j} + \left(\frac{\partial (\sqrt{1-x^{2}})}{\partial x} - \frac{\partial (\arcsin y)}{\partial y}\right) \mathbf{k} \). Then plug in the appropriate values for the partial derivatives (you'll obtain zeros where the function does not depend on the variable).

Simplify the determinant

Simplify the determinant to obtain the final result: \( \nabla \times \mathbf{F} = \left(2y - 0\right) \mathbf{i} - \left(0 - 0\right) \mathbf{j} + \left(-\frac{x}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-y^{2}}}\right) \mathbf{k} \). Simplifying further yields \( \nabla \times \mathbf{F} = 2y \mathbf{i} + \left(-\frac{x}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-y^{2}}}\right) \mathbf{k}\).