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Ket [755]
4 years ago
13

Find the work done by the vector field F~ = x 2y~i + 1 3 x 3~j + xy~k along the curve of intersection C of the paraboloid z = y

2 − x 2 and the cylinder x 2 + y 2 = 1, oriented counterclockwise as viewed from the above.
Mathematics
1 answer:
Len [333]4 years ago
8 0

Let x=\cos t and y=\sin t. Then C can be parameterized by

\vec r(t)=\cos t\,\vec\imath+\sin t\,\vec\jmath+(\sin^2t-\cos^2t)\,\vec k

with 0\le t\le2\pi, and its derivative is

\dfrac{\mathrm d\vec r}{\mathrm dt}=-\sin t\,\vec\imath+\cos t\,\vec\jmath+4\sin t\cos t\,\vec k

Now,

\vec F(x,y,z)=x^2y\,\vec\imath+\dfrac{x^3}3\,\vec\jmath+xy\,\vec k

\implies\vec F(\vec r(t))=\cos^2t\sin t\,\vec\imath+\dfrac{\cos^3t}3\,\vec\jmath+\cos t\sin t\,\vec k

Then the work done by \vec F along C is

\displaystyle\int_C\vec F(x,y,z)\cdot\mathrm d\vec r=\int_0^{2\pi}\vec F(\vec r(t))\cdot\frac{\mathrm d\vec r}{\mathrm dt}\,\mathrm dt=\int_0^{2\pi}\left(3\cos^2t\sin^2t+\frac{\cos^4t}3\right)\,\mathrm dt=\boxed{\pi}

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