Evaluate the surfacuse stokes's theorem to evaluate f · dr c. use a computer algebra system to verify your results. in this case, c is oriented counterclockwise as viewed from above. f(x, y, z) = 2yi + 3zj + xkc.triangle with vertices (5, 0, 0), (0, 5, 0), (0, 0, 5)e integral (x^2+y^2+z^2)ds

Since you're asked to use Stoke's theorem, I'm interpreting the question to be asking to compute the line integral

[tex]\displaystyle\int_{\mathcal C}\mathbf f(x,y,z)\cdot\mathrm d\mathbf r[/tex]

which by Stoke's theorem is equivalent to the surface integral

[tex]\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f(x,y,z)\cdot\mathrm d\mathbf S[/tex]

where [tex]\mathcal S[/tex] is the positively-oriented surface with boundary [tex]\mathcal C[/tex].

Given [tex]\mathbf f(x,y,z)=2y\,\mathbf i+3z\,\mathbf j+x\,\mathbf k[/tex], we get curl

[tex]\nabla\times\mathbf f=-3\,\mathbf i-\mathbf j-2\,\mathbf k[/tex]

We parameterize the surface [tex]\mathcal S[/tex] by

[tex]\mathbf s(u,v)=5(1-u)(1-v)\,\mathbf i+5u(1-v)\,\mathbf j+5v\,\mathbf k[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Then we take the partial derivatives of [tex]\mathbf s[/tex] and check their cross product:

[tex]\mathbf s_u\times\mathbf s_v=25(1-v)(\mathbf i+\mathbf j+\mathbf k)[/tex]

Now,

[tex]\mathrm d\mathbf S=(\mathbf s_u\times\mathbf s_v)\,\mathrm du\,\mathrm dv[/tex]

so the surface integral reduces to

[tex]\displaystyle-150\int_{v=0}^{v=1}\int_{u=0}^{u=1}(1-v)\,\mathrm du\,\mathrm dv=-75[/tex]