How to Calculate Divergence and Curl

Step-by-Step Guide

1. 1

   Step 1: Understand what divergence is.

   Divergence is a measure of source or sink at a particular point. – In other words, how much is flowing into or out of a point.
   
   Hence, it is only defined for vector fields and outputs a scalar.
   
   Below is an example of a field with a positive divergence.
   
   The divergence is recognized by div{\displaystyle \operatorname {div} } or ∇⋅{\displaystyle \nabla \cdot }, where the dot signifies the similarity to taking a dot product.
2. 2

   Step 2: Take the dot product of the partial derivatives with the components of F{\displaystyle {\mathbf {F} }}

   This applies for vector fields F=Fxx^+Fyy^+Fzz^{\displaystyle {\mathbf {F} }=F_{x}{\mathbf {\hat {x}} }+F_{y}{\mathbf {\hat {y}} }+F_{z}{\mathbf {\hat {z}} }} defined in Cartesian coordinates only. ∇⋅F=(∂∂x,∂∂y,∂∂z)⋅(Fx,Fy,Fz)=∂Fx∂x+∂Fy∂y+∂Fz∂z{\displaystyle \nabla \cdot {\mathbf {F} }=\left({\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}}\right)\cdot (F_{x},F_{y},F_{z})={\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}} , If the vector field F{\displaystyle {\mathbf {F} }} is given in cylindrical (ρ,ϕ,z){\displaystyle (\rho ,\phi ,z)} or spherical coordinates (r,θ,ϕ){\displaystyle (r,\theta ,\phi )} (where θ{\displaystyle \theta } is the polar angle), then the divergence does not have a simple form. ∂Fx∂x+∂Fy∂y+∂Fz∂z{\displaystyle {\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}} 1ρ∂(ρFρ)∂ρ+1ρ∂Fϕ∂ϕ+∂Fz∂z{\displaystyle {\frac {1}{\rho }}{\frac {\partial (\rho F_{\rho })}{\partial \rho }}+{\frac {1}{\rho }}{\frac {\partial F_{\phi }}{\partial \phi }}+{\frac {\partial F_{z}}{\partial z}}} 1r2∂(r2Fr)∂r+1rsin⁡θ∂∂θ(Fθsin⁡θ)+1rsin⁡θ∂Fϕ∂ϕ{\displaystyle {\frac {1}{r^{2}}}{\frac {\partial (r^{2}F_{r})}{\partial r}}+{\frac {1}{r\sin \theta }}{\frac {\partial }{\partial \theta }}(F_{\theta }\sin \theta )+{\frac {1}{r\sin \theta }}{\frac {\partial F_{\phi }}{\partial \phi }}} , F=(3x2−5x2y4)x^+(xy4z2−sin⁡(2x2z3))y^+(5z2+yz)z^{\displaystyle {\mathbf {F} }=(3x^{2}-5x^{2}y^{4}){\mathbf {\hat {x}} }+(xy^{4}z^{2}-\sin(2x^{2}z^{3})){\mathbf {\hat {y}} }+(5z^{2}+yz){\mathbf {\hat {z}} }} ∇⋅F=6x−10xy4+4xy3z2+y+10z{\displaystyle \nabla \cdot {\mathbf {F} }=6x-10xy^{4}+4xy^{3}z^{2}+y+10z} As you can see, we have mapped from a vector field to a scalar field. , The curl, defined for vector fields, is, intuitively, the amount of circulation at any point.
   
   The operator outputs another vector field.
   
   A whirlpool in real life consists of water acting like a vector field with a nonzero curl.
   
   Above is an example of a field with negative curl (because it's rotating clockwise).
   
   The curl is recognized by curl{\displaystyle \operatorname {curl} } or ∇×{\displaystyle \nabla \times }, where the times symbol signifies the similarity of taking a cross product. , The curl of a function is similar to the cross product of two vectors, hence why the curl operator is denoted with a ∇×.{\displaystyle \nabla \times .} As before, this mnemonic only works if F{\displaystyle {\mathbf {F} }} is defined in Cartesian coordinates. ∇×F=|x^y^z^∂/∂x∂/∂y∂/∂zFxFyFz|{\displaystyle \nabla \times {\mathbf {F} }={\begin{vmatrix}{\mathbf {\hat {x}} }&{\mathbf {\hat {y}} }&{\mathbf {\hat {z}} }\\\partial /\partial x&\partial /\partial y&\partial /\partial z\\F_{x}&F_{y}&F_{z}\end{vmatrix}}} , Below, we do it by cofactor expansion (expansion by minors). ∇×F=(∂Fz∂y−∂Fy∂z)x^−(∂Fz∂x−∂Fx∂z)y^+(∂Fy∂x−∂Fx∂y)z^{\displaystyle \nabla \times {\mathbf {F} }=\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right){\mathbf {\hat {x}} }-\left({\frac {\partial F_{z}}{\partial x}}-{\frac {\partial F_{x}}{\partial z}}\right){\mathbf {\hat {y}} }+\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right){\mathbf {\hat {z}} }} , The curl does not have a simple form if F{\displaystyle {\mathbf {F} }} is in cylindrical or spherical coordinates. (∂Fz∂y−∂Fy∂z)x^−(∂Fz∂x−∂Fx∂z)y^+(∂Fy∂x−∂Fx∂y)z^{\displaystyle \left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right){\mathbf {\hat {x}} }-\left({\frac {\partial F_{z}}{\partial x}}-{\frac {\partial F_{x}}{\partial z}}\right){\mathbf {\hat {y}} }+\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right){\mathbf {\hat {z}} }} (1ρ∂Fz∂ϕ−∂Fϕ∂z)ρ^−(∂Fz∂ρ−∂Fρ∂z)ϕ^+1ρ(∂(ρFϕ)∂ρ−∂Fρ∂ϕ)z^{\displaystyle \left({\frac {1}{\rho }}{\frac {\partial F_{z}}{\partial \phi }}-{\frac {\partial F_{\phi }}{\partial z}}\right){\boldsymbol {\hat {\rho }}}-\left({\frac {\partial F_{z}}{\partial \rho }}-{\frac {\partial F_{\rho }}{\partial z}}\right){\boldsymbol {\hat {\phi }}}+{\frac {1}{\rho }}\left({\frac {\partial (\rho F_{\phi })}{\partial \rho }}-{\frac {\partial F_{\rho }}{\partial \phi }}\right){\mathbf {\hat {z}} }} 1rsin⁡θ(∂∂θ(Fϕsin⁡θ)−∂Fθ∂ϕ)r^−1r(∂∂r(rFϕ)−1sin⁡θ∂Fr∂ϕ)θ^+1r(∂∂r(rFθ)−∂Fr∂θ)ϕ^{\displaystyle {\begin{aligned}{\frac {1}{r\sin \theta }}\left({\frac {\partial }{\partial \theta }}(F_{\phi }\sin \theta )-{\frac {\partial F_{\theta }}{\partial \phi }}\right){\mathbf {\hat {r}} }&-{\frac {1}{r}}\left({\frac {\partial }{\partial r}}(rF_{\phi })-{\frac {1}{\sin \theta }}{\frac {\partial F_{r}}{\partial \phi }}\right){\boldsymbol {\hat {\theta }}}\\&+{\frac {1}{r}}\left({\frac {\partial }{\partial r}}(rF_{\theta })-{\frac {\partial F_{r}}{\partial \theta }}\right){\boldsymbol {\hat {\phi }}}\end{aligned}}} , F=(5x2y2−7xz3)x^+(4x−5xy−y4)y^+(xz+z2)z^{\displaystyle {\mathbf {F} }=(5x^{2}y^{2}-7xz^{3}){\mathbf {\hat {x}} }+(4x-5xy-y^{4}){\mathbf {\hat {y}} }+(xz+z^{2}){\mathbf {\hat {z}} }} , ∇×F=|x^y^z^∂/∂x∂/∂y∂/∂zFxFyFz|{\displaystyle \nabla \times {\mathbf {F} }={\begin{vmatrix}{\mathbf {\hat {x}} }&{\mathbf {\hat {y}} }&{\mathbf {\hat {z}} }\\\partial /\partial x&\partial /\partial y&\partial /\partial z\\F_{x}&F_{y}&F_{z}\end{vmatrix}}} Fx=5x2y2−7xz3{\displaystyle F_{x}=5x^{2}y^{2}-7xz^{3}} Fy=4x−5xy−y4{\displaystyle F_{y}=4x-5xy-y^{4}} Fz=xz+z2{\displaystyle F_{z}=xz+z^{2}} , (∂Fz∂y−∂Fy∂z)x^=0−0{\displaystyle \left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right){\mathbf {\hat {x}} }=0-0} (∂Fz∂x−∂Fx∂z)y^=z−(−21xz2){\displaystyle \left({\frac {\partial F_{z}}{\partial x}}-{\frac {\partial F_{x}}{\partial z}}\right){\mathbf {\hat {y}} }=z-(-21xz^{2})} (∂Fy∂x−∂Fx∂y)z^=(4−5y)−10x2y{\displaystyle \left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right){\mathbf {\hat {z}} }=(4-5y)-10x^{2}{y}} , ∇×F=−(z+21xz2)y^+(4−5y−10x2y)z^{\displaystyle \nabla \times {\mathbf {F} }=-(z+21xz^{2}){\mathbf {\hat {y}} }+(4-5y-10x^{2}y){\mathbf {\hat {z}} }} Note that we have mapped to another vector field.
3. 3

   Step 3: then sum the results.

4. 4

   Step 4: Use the formulas below as a reference.

5. 5

   Step 5: Calculate the divergence of the following function.

6. 6

   Step 6: Understand what curl is.

7. 7

   Step 7: Set up the determinant.

8. 8

   Step 8: Find the determinant of the matrix.

9. 9

   Step 9: Use the formulas below as a reference.

10. 10

    Step 10: Calculate the curl of the following function.

11. 11

    Step 11: Set up the determinant.

12. 12

    Step 12: Calculate the determinant.

13. 13

    Step 13: Arrive at the answer.

Detailed Guide

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