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kupik [55]
3 years ago
10

What is the solution to the system of linear equations?

Mathematics
1 answer:
GalinKa [24]3 years ago
4 0

Answer:

(0,2)

Step-by-step explanation:

The solution would just be where these two functions cross, which is at the point (0,2) in this case.

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Use the surface integral in​ Stokes' Theorem to calculate the circulation of the field Bold Upper F equals x squared Bold i plus
Alinara [238K]

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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

3 0
3 years ago
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