What is the curl and divergence here? F(x,y,z) = (x² + y² - xyz, xz² cos y, z² e^(x²+y²))
K keysar New member Joined Jul 3, 2010 Messages 2 Jul 3, 2010 #1 What is the curl and divergence here? F(x,y,z) = (x² + y² - xyz, xz² cos y, z² e^(x²+y²))
D Deleted member 4993 Guest Jul 3, 2010 #2 keysar said: What is the curl and divergence here? F(x,y,z) = (x² + y² - xyz, xz² cos y, z² e^(x²+y²)) Click to expand... What is the definition of curl and divergence? Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
keysar said: What is the curl and divergence here? F(x,y,z) = (x² + y² - xyz, xz² cos y, z² e^(x²+y²)) Click to expand... What is the definition of curl and divergence? Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Jul 5, 2010 #3 Curl is found by using the partials. Given F(x,y,z)=f(x,y,z)i+g(x,y,z)j+h(x,y,z)k\displaystyle F(x,y,z)=f(x,y,z)i+g(x,y,z)j+h(x,y,z)kF(x,y,z)=f(x,y,z)i+g(x,y,z)j+h(x,y,z)k curl F=(dhdy−dgdz)i+(dfdz−dhdx)j+(dgdx−dfdy)k\displaystyle \text{curl \;\ F}=\left(\frac{dh}{dy}-\frac{dg}{dz}\right)i+\left(\frac{df}{dz}-\frac{dh}{dx}\right)j+\left(\frac{dg}{dx}-\frac{df}{dy}\right)kcurl F=(dydh−dzdg)i+(dzdf−dxdh)j+(dxdg−dydf)k
Curl is found by using the partials. Given F(x,y,z)=f(x,y,z)i+g(x,y,z)j+h(x,y,z)k\displaystyle F(x,y,z)=f(x,y,z)i+g(x,y,z)j+h(x,y,z)kF(x,y,z)=f(x,y,z)i+g(x,y,z)j+h(x,y,z)k curl F=(dhdy−dgdz)i+(dfdz−dhdx)j+(dgdx−dfdy)k\displaystyle \text{curl \;\ F}=\left(\frac{dh}{dy}-\frac{dg}{dz}\right)i+\left(\frac{df}{dz}-\frac{dh}{dx}\right)j+\left(\frac{dg}{dx}-\frac{df}{dy}\right)kcurl F=(dydh−dzdg)i+(dzdf−dxdh)j+(dxdg−dydf)k
