Question

Help required, work integral gives different value with different approaches, which value is the right one?

z=2-x
x^2+y^2=4
VectorField=

Answer 1

So you're looking for the work done on \(\mathbf F\) along the intersection of the surfaces?

Answer 2

Yeah. I don't know why Stokes differs from the general theorem. Neither do I know if any is right

Answer 3

Assuming that's the right interpretation, the work is given by the line integral
\[\int_C\mathbf F\cdot\mathrm d\vec{r}\]where \(C\) is the intersection of the cylinder with the plane, parameterized by
\[\vec{r}(t)=\langle 2\cos t,2\sin t,2-2\cos t\rangle\]for \(t\in[0,2\pi]\), which makes
\[\begin{align*}
\mathbf F&=\langle\cos z,x^2,2y\rangle\\[1ex]
&=\langle\cos(2-2\cos t),4\cos^2t,4\sin t\rangle
\end{align*}\]The integral is then
\[\int_0^{2\pi}\langle\cos(2-2\cos t),4\cos^2t,4\sin t\rangle\cdot\langle-2\sin t,2\cos t,2\sin t\rangle\,\mathrm dt\\[2ex]
=\int_0^{2\pi}\left(-2\sin t\cos(2-2\cos t)+8\cos^3t+8\sin^2t\right)\,\mathrm dt\]which should be easy to compute.

Answer 4

Wait, where is this cylinder coming from? I don't see it in the PDF.

Answer 5

Maybe that is what I did wrong? Since I used a parabolid? I feel so terrible I calculus 3 :(

Answer 6

If it is indeed a paraboloid you're working with, then we can get to that in a second.

Sticking with the cylinder for the moment, I'm getting the same result for the line integral above as for the surface integral.

Parameterize the surface \(S\) (the red ellipse in the plot) by
\[\mathbf x(r,t)=\langle r\cos t,r\sin t,2-r\cos t\rangle\]with \(r\in[0,2]\) and \(t\in[0,2\pi]\). Then the vector field's curl can be written as
\[\mathrm{curl}~\mathbf F=\nabla\times\mathbf F=\langle2,-\sin z,2x\rangle=\langle 2,-\sin(2-r\cos t),2r\cos t\rangle\]The work is then given by the surface integral
\[\iint_S \nabla\times \mathbf F\cdot\left(\frac{\partial\mathbf x}{\partial r}\times\frac{\partial\mathbf x}{\partial t}\right)\,\mathrm dr\,\mathrm dt=\int_0^2\int_0^{2\pi}(2r+2r^2\cos t)\,\mathrm dt\,\mathrm dr\]

Answer 7

With the paraboloid, we need to make some slight adjustments to the parameterizations.

For the line integral, we would use
\[\vec{r}(t)=\left\langle\frac{3}{2}\cos t-\frac{1}{2},\frac{3}{2}\sin t,\frac{5}{2}-\frac{3}{2}\cos t\right\rangle\]again with \(t\in[0,2\pi]\), which comes from the fact that
\[\begin{cases}x^2+y^2=z\\z=2-x\end{cases}\implies x^2+y^2=2-x\implies \left(x+\frac{1}{2}\right)^2+y^2=\left(\frac{3}{2}\right)^2\]Meanwhile, the vector field can be written as
\[\begin{align*}
\mathbf F&=\langle \cos z,x^2,y\rangle\\[1ex]
&=\left\langle\cos\left(\frac{5}{2}-\frac{3}{2}\cos t\right),\left(\frac{3}{2}\cos t-\frac{1}{2}\right)^2,3\sin t\right\rangle
\end{align*}\]Then the integral becomes
\[\int_0^{2\pi}\mathbf F\cdot\left\langle-\frac{3}{2}\sin t,\frac{3}{2}\cos t,3\cos t\right\rangle\,\mathrm dt\\[2ex]
=\frac{3}{8}\int_0^{2\pi}\left(-6+\cos t+9\cos^3t+18\sin^2t-4\sin t\cos\left(\frac{5}{2}-\frac{3}{2}\cos t\right)\right)\,\mathrm dt\]
For the surface integral, we can parameterize by
\[\mathbf x(r,t)=\left\langle r\cos t-\frac{1}{2},r\sin t,\frac{5}{2}-r\cos t\right\rangle\]with \(r\in\left[0,\dfrac{3}{2}\right]\) and \(t\in[0,2\pi]\). The curl is
\[\nabla\times \mathbf F=\left\langle2,-\sin\left(\frac{5}{2}-r\cos t\right),2r\cos t-1\right\rangle\]The integral eventually reduces to
\[\int_0^{2\pi}\int_0^{3/2}\left(r+2r^2\cos t\right)\,\mathrm dr\,\mathrm dt\]and gives the same result as the line integral.

Answer 8

Thank you so much, so how i set up the stokes theorem somehow gave the right answer? 8Pi is the work done?

Answer 9

I tend to mix up z=x^2+y^2 to be the same as x^2+y^2=Constant. But x^2+y^2=constant means the domain is a circle and the 3d plot is a cylinder with hight = z=free?
The parameters for that cylinder is always \[\sqrt{constant}*\cos (t),\sqrt{constant}*\sin(t),z\]
since the constant is the r^2?

Answer 10

Amazing, Holster you give me hope and motivation to keep on working towards my exam.¨

Apperantly i messed up the parameters since i mixed the parabolid for the cylinder, wich led to the surface version returned right answer but the direct calc returned 9/4 Pi which is right if we us the paraboloid.

you are a huge recourse to this community, keep up your good work.

Answer 11

Yup, \(8\pi\) is correct in the case of the cylinder, and \(\dfrac{9\pi}{4}\) in the case of the paraboloid.

And yes, \(z=x^2+y^2\) is a cone. If you fix \(z\), you get a circle of some radius, and if you change \(z\) to a different constant, you get a different circle. These level curves stack to form a cone through the origin. Given \(x^2+y^2=r^2\) (constant \(r\)), you have a circle of constant radius regardless of the chosen value of \(z\), so this forms a cylinder in 3D space. That parameterization is correct, though you have to make sure to account for translations of the center of each circular level curve.

Glad I could help!