Question #281717

use Stokes theorem evaluate ∮A.dr, where A=-5yi+4xj+zk and C is cirvle x²+y²=4,z=1


Expert's answer

curlA=∣ijk∂∂x∂∂y∂∂z−5y4xz∣curl \bold A=\begin{vmatrix} \bold i & \bold j & \bold k \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ -5y & 4x & z \end{vmatrix}

=9k=9\bold k

n=⟨0,0,1⟩\bold n=\langle0, 0, 1\rangle

curlA⋅n=9curl \bold A\cdot \bold n=9

∮CA⋅dr=∫∫ScurlA⋅dS∮_C\bold A\cdot d\bold r=\int\int _Scurl \bold A\cdot d\bold S

=∫∫S9dS=9∫∫SdS=\int \int_S9dS=9\int \int_SdS

The double integral in the latter formula is the area of the circle. Therefore, the integral is

∮CA⋅dr=∫∫ScurlA⋅dS∮_C\bold A\cdot d\bold r=\int\int _Scurl \bold A\cdot d\bold S

=∫∫S9dS=9∫∫SdS=9π(4)=36π=\int \int_S9dS=9\int \int_SdS=9\pi(4)=36\pi

∮CA⋅dr=36π∮_C\bold A\cdot d\bold r=36\pi


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