# Use Stokes' Theorem to evaluate int int_S CURL f * dS. F(x,y,z)=x^2y^3zi+sin(xyz)j+xyzk, S is the part of the cone y^2=x^2+z^2 that lies between the planes y = 0 and y = 2, oriented in the direction of the positive y-axis.

Use Stokes' Theorem to evaluate $\int {\int }_{S}CURLf\cdot dS$.
$F\left(x,y,z\right)={x}^{2}{y}^{3}zi+\mathrm{sin}\left(xyz\right)j+xyzk$,
S is the part of the cone ${y}^{2}={x}^{2}+{z}^{2}$ that lies between the planes y = 0 and y = 2, oriented in the direction of the positive y-axis.
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Solution:
The vector field is $F\left(x,y,z\right)={x}^{2}{y}^{3}zi+\mathrm{sin}\left(xyz\right)j+xyzk$
Find: $\int {\int }_{S}curlF·dS$
Then, the double integral becomes $\int {\int }_{S}curlF·dS={\int }_{C}F·dr$
The surface of a cone is ${y}^{2}={x}^{2}+{z}^{2}$.
At $y=2,{x}^{2}+{z}^{2}=4$.
Let the parametric equation is $r\left(t\right)=<2\mathrm{cos}t,2,2\mathrm{sin}t>\mathrm{\forall }0\le t\le 2\pi$ and its derivative is ${r}^{\prime }\left(t\right)=<-2\mathrm{sin}t,0,2\mathrm{cos}t>$.
Then, the vector field becomes
$F\left(r\left(t\right)\right)=={\left(2\mathrm{cos}t\right)}^{2}{\left(2\right)}^{3}\left(2\mathrm{sin}t\right)i+\mathrm{sin}\left(8\mathrm{sin}t\mathrm{cos}t\right)j+8\mathrm{sin}t\mathrm{cos}tk\left(64{\mathrm{cos}}^{2}t\mathrm{sin}t\right)i+\mathrm{sin}\left(8\mathrm{sin}t\mathrm{cos}t\right)j+8\mathrm{sin}t\mathrm{cos}tk$
Step 2
Obtain the integal:
${\int }_{C}F·dr={\int }_{0}^{2\pi }0<64{\mathrm{cos}}^{2}t\mathrm{sin}t,\mathrm{sin}\left(8\mathrm{sin}t\mathrm{cos}t\right),8\mathrm{sin}t\mathrm{cos}t><-2\mathrm{sin}t,0,2\mathrm{cos}t>{\int }_{0}^{2\pi }\left(-128{\mathrm{cos}}^{2}t{\mathrm{sin}}^{2}t+16\mathrm{sin}t{\mathrm{cos}}^{2}t\right)dt-32\pi$
Therefore, $\int {\int }_{S}curlF·dS=-32\pi$.