Question

Use the surface integral in​ Stokes' Theorem to calculate the circulation of the field Bold Upper F equals x squared Bold i plus 4 x Bold j plus z squared Bold k around the curve​ C: the ellipse 16x squared plus 4 y squared equals 3 in the​ xy-plane, counterclockwise when viewed from above.

Answer 1

Answer:

The circulation of the field f(x) over curve C is Zero

Step-by-step explanation:

The function $f(x)=(x^{2},4x,z^{2})$ and curve C is ellipse of equation

$16x^{2} + 4y^{2} = 3$

Theory: Stokes Theorem is given by:

$I= \int \int\limits {{Curl f\cdot \hat{N }} \, dx$

Where, Curl f(x) = $\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right]$

Also, f(x) = (F1,F2,F3)

$\hat{N} = grad(g(x))$

Using Stokes Theorem,

Surface is given by g(x) = $16x^{2} + 4y^{2} - 3$

Therefore, tex]\hat{N} = grad(g(x))[/tex]

$\hat{N} = grad(16x^{2} + 4y^{2} - 3)$

$\hat{N} = (32x,8y,0)$

Now,  $f(x)=(x^{2},4x,z^{2})$

Curl f(x) = $\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right]$

Curl f(x) = $\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\x^{2}&4x&z^{2}\end{array}\right]$

Curl f(x) = (0,0,4)

Putting all values in Stokes Theorem,

$I= \int \int\limits {Curl f\cdot \hat{N} } \, dx$

$I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx$

$I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx$

I=0

Thus, The circulation of the field f(x) over curve C is Zero