Question #210674

1) If a vector field A(x,y,z)=xz3ex - 2x2yzey+2yz4ez , calculate the curl ∇ x A at point M(1,-1,-1).


1
Expert's answer
2021-06-29T05:53:02-0400

Gives


A(x,y,z)=xz3ex^+2x2yzey^+2yz4ez^A(x,y,z)=xz^3\hat{e_x}+2x^2yz\hat{e_y}+2yz^4\hat{e_z}

PoinM(1,-1,-1)


×A=[ex^ey^ez^ddxddxddxxz32x2yz2yz4]\nabla\times A=\begin{bmatrix} \hat{e_x}&\hat{e_y} & \hat{e_z}\\ \frac{d}{dx}&\frac{d}{dx}&\frac{d}{dx}\\ xz^3&-2x^2yz&2yz^4 \end{bmatrix}

ex^(ddy(2yz4)+ddz(2x2yz))ey^(ddx(2yz4)ddx(xz3))+ez^(ddx(2x2yz)ddy(xz3))\hat{e_x}(\frac{d}{dy}(2yz^4)+\frac{d}{dz}(2x^2yz))-\hat{e_y}(\frac{d}{dx}(2yz^4)-\frac{d}{dx}(xz^3))+\hat{e_z}(\frac{d}{dx}(-2x^2yz)-\frac{d}{dy}(xz^3))

×A=ex^(2z4+2x2y)+ey^(3xz2)ez^(4xyz)\nabla\times A=\hat{e_x}(2z^4+2x^2y)+\hat{e_y}(3xz^2)-\hat{e_z}(4xyz)


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