Question

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.

Answer 1

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

`(\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+\frac{x^3}3\,\vec\jmath+xy\,\vec k`

`\implies\vec F(\vec r(t))=\cos^2t\sin t\,\vec\imath+\frac{\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(\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}`