- #1

greg_rack

Gold Member

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- Homework Statement
- ##\vec F=<x+y^2, y+z^2,z+x^2>##, C is the triangle with vertices (1,0,0), (0,1,0), (0,0,1). Compute ##\int_{C}^{}\vec F\cdot d \vec r##

- Relevant Equations
- Stoke's theorem

Computable form of a surface integral

From Stokes' theorem: ##\int_{C}^{}\vec F\cdot d\vec r=\iint_{S}^{}curl\vec F\cdot d\vec S=\iint_{D}^{}curl\vec F\cdot(\vec r_u \times \vec r_v)dA ##

To get to the latter surface integral, I started by parametrizing the triangular surface in ##uv## coordinates as:

$$\vec r=<1-u-v,u,v>, 0\leq u\leq 1, 0\leq v\leq 1-u$$

I then computed the curl of the vector field, the partial derivatives in ##u## and ##v## of the above parametrization and their cross product:

$$curl\vec F=<-2z, -2x, 2(x-y)>, \vec r_u \times \vec r_v=<1,1,1>$$

Now we can dot the curl with the cross product of the partial derivatives and get to a computable form of the surface integral

$$curl\vec F(\vec r(u,v))\cdot (\vec r_u \times \vec r_v)=-2(u+v)\rightarrow -2\int_{0}^{1}\int_{0}^{1-u}(u+v)dv du$$

which leads to a wrong answer. What am I missing?

To get to the latter surface integral, I started by parametrizing the triangular surface in ##uv## coordinates as:

$$\vec r=<1-u-v,u,v>, 0\leq u\leq 1, 0\leq v\leq 1-u$$

I then computed the curl of the vector field, the partial derivatives in ##u## and ##v## of the above parametrization and their cross product:

$$curl\vec F=<-2z, -2x, 2(x-y)>, \vec r_u \times \vec r_v=<1,1,1>$$

Now we can dot the curl with the cross product of the partial derivatives and get to a computable form of the surface integral

$$curl\vec F(\vec r(u,v))\cdot (\vec r_u \times \vec r_v)=-2(u+v)\rightarrow -2\int_{0}^{1}\int_{0}^{1-u}(u+v)dv du$$

which leads to a wrong answer. What am I missing?