Answer:
The circulation of the field f(x) over curve C is Zero
Step-by-step explanation:
The function [tex]f(x)=(x^{2},4x,z^{2})[/tex] and curve C is ellipse of equation
[tex]16x^{2} + 4y^{2} = 3[/tex]
Theory: Stokes Theorem is given by:
[tex]I= \int \int\limits {{Curl f\cdot \hat{N }} \, dx[/tex]
Where, Curl f(x) = [tex]\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right][/tex]
Also, f(x) = (F1,F2,F3)
[tex]\hat{N} = grad(g(x))[/tex]
Using Stokes Theorem,
Surface is given by g(x) = [tex]16x^{2} + 4y^{2} - 3[/tex]
Therefore, tex]\hat{N} = grad(g(x))[/tex]
[tex]\hat{N} = grad(16x^{2} + 4y^{2} - 3)[/tex]
[tex]\hat{N} = (32x,8y,0)[/tex]
Now, [tex]f(x)=(x^{2},4x,z^{2})[/tex]
Curl f(x) = [tex]\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right][/tex]
Curl f(x) = [tex]\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\x^{2}&4x&z^{2}\end{array}\right][/tex]
Curl f(x) = (0,0,4)
Putting all values in Stokes Theorem,
[tex]I= \int \int\limits {Curl f\cdot \hat{N} } \, dx[/tex]
[tex]I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx[/tex]
[tex]I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx[/tex]
I=0
Thus, The circulation of the field f(x) over curve C is Zero
