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Let F=2zi+2yk\mathbf{F}=2 z \mathbf{i}+2 y \mathbf{k}, and let S\partial S be the intersection of the cylinder x2+y2=ayx^2+y^2=a y with the hemisphere z=a2x2y2z=\sqrt{a^2-x^2-y^2}, a>0a>0. Assuming distances in meters and force in newtons, find the work done by the force F\mathbf{F} in moving an object around S\partial S in the counterclockwise direction as viewed from above.

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Answered 2 years ago
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Using the Stokes Theorem we have that:

δSFTds=S(×F)ndS\begin{aligned} \oint_{\delta S}{\bf F}\cdot{\bf T}ds&=\iint_S(\nabla\times{\bf F})\cdot{\bf n}dS \end{aligned}

Let's find the curl of the given vector field. Since F=(2z,0,2y){\bf F}=(2z,0,2y) then:

×F=ijkxyz2z02y=(2,2,0)\begin{aligned} \nabla\times{\bf F}&=\left|\begin{array}{lcl} {\bf i} & {\bf j} & {\bf k}\\ \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z}\\ 2z & 0 & 2y \end{array}\right|\\ &=(2,2,0) \end{aligned}

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