Question #281719

use Stokes theorem evaluate ∮A.dr, where A=y²i+x²j+z²k and C is boundary of the part of the plane x+y+z=1 like in the first octant

1
Expert's answer
2021-12-22T03:24:19-0500
curlA=ijkxyzy2x2z2curl \bold A=\begin{vmatrix} \bold i & \bold j & \bold k \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ y^2 & x^2 & z^2 \end{vmatrix}

=(2x2y)k=(2x-2y)\bold k

n=1,1,1\bold n=\langle1, 1, 1\rangle

curlAn=2x2ycurl \bold A\cdot \bold n=2x-2y

CAdr=ScurlAdS\int_C\bold A\cdot d\bold r=\int\int _Scurl \bold A\cdot d\bold S

=S(2x2y)dS=\int \int_S(2x-2y)dS

=0101y(2x2y)dxdy=\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-y}(2x-2y)dxdy

=01[x22xy]1y0dy=\displaystyle\int_{0}^1[x^2-2xy]\begin{matrix} 1-y \\ 0 \end{matrix}dy

=01((1y)22y(1y))dy=\displaystyle\int_{0}^1((1-y)^2-2y(1-y))dy

=01(12y+y22y+2y2)dy=\displaystyle\int_{0}^1(1-2y+y^2-2y+2y^2)dy

=[y32y2+y]10=0=[y^3-2y^2+y]\begin{matrix} 1 \\ 0 \end{matrix}=0

CAdr=0\int_C\bold A\cdot d\bold r=0


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