Consider the following vectorfield F= Solid T is enclosed by the plane x^2+y^2+2x=0 -2

Question

Consider the following vectorfield F=<2x*z,2y*z,x^2+y^2>

Solid T is enclosed by the plane x^2+y^2+2x=0
-2<x<0, -sqrt(-x^2-2*x)<y<sqrt(-x^2-2*x), x^2+y^2<z<-2x

Evaluate the line integral F*T ds

Taking the cross product GradientF X F shows the curl is 0, therefore the line integral is 0. But in general how would I set up such an integral?

fixxer · Answer

$$\int\limits \!{\it TxF}\,{\rm d}s$$ I think the first thing i need to do is to find a parametrization for the the curve that is closing the body T.

Are the right approach maybe Greens theoreme? Thanks

fixxer · Answer

Here is a plot, how would I know the orientation of the path?

holsteremission · Answer

Here's a plot of the solid $T$. It's not clear what you mean by "curve that is closing the body $T$".

holsteremission · Answer

And from another angle to make it clearer that the solid is bounded below by the paraboloid, above by the plane, and confined within the cylinder.

fixxer · Answer

That is correct, I assume it is the outer curve that I plotted in red above. It is a problem from an earlier exam in calculus 3.

I reckon greens theorem is to be used when the curve in continious and closed.

fixxer · Answer

attachment

holsteremission · Answer

So it's the intersection of the paraboloid, plane, and cylinder? A parameterization in cylindrical coordinates might be easiest.

fixxer · Answer

Yeah tha sounds right Holster, the curve formed by intersection of the parabolid and the plane

holsteremission · Answer

Here's one parameterization.

The cylinder has equation
$$x^2+y^2+2x=0\implies (x+1)^2+y^2=1$$so we could set $x=\cos t-1$ and $y=\sin t$, with $t\in[0,2\pi]$.

Meanwhile, we have as our bounds on $z$
$$\begin{cases}
z=-2x\[1ex]x^2+y^2=z\end{cases}\implies (z-2)^2+4y^2=4\implies z=2\pm2\sqrt{1-y^2}$$and substituting $y=\sin t$ reduces this to
$$z=2\pm2\sqrt{1-\sin^2t}=2\pm2\cos t$$For reasons I have yet to figure out, we take the negative root, and the path should be parameterized by $\vec{r}(t)=\langle\cos t-1,\sin t,2-2\cos t\rangle$.

Here's a plot that confirms this works.

fixxer · Answer

I take my hat off sir, I do really struggle to make parametrizations like that.

Your work is really good :)

fixxer · Answer

Is that standard way to simplify parametrization to complete the square? Just trying to work trough your work to actually learn something :)

holsteremission · Answer

I believe it is. I've seen that approach used for finding the polar equation for a circle not centered at the origin.

holsteremission · Answer

Oh and regarding your comment about using Green's theorem, I believe you meant Stokes' theorem?

fixxer · Answer

I meant greens, to evaluate a line integral over a closed smooth simple curve. Might be wrong thought

fixxer · Answer

will you please explain how you acheived $$z=2\pm2\sqrt{1-y^2}$$ .

It is algebra and this one is doing my head :s

holsteremission · Answer

If that's the case, the theorem states that the integral of the curl $\nabla\times\mathbf F$ over the surface $\Sigma$ (the elliptical upperbound) is given by
$$\int_\Sigma \nabla \times\mathbf F\cdot\mathrm d\mathbf\Sigma=\int_C\mathbf F\cdot\mathrm d\vec{r}$$where $C$ is the red path in the plot.

I suppose you could use Green's theorem to evaluate the resulting line integral, but it's currently in three dimensions, not two. You might be able to figure out an appropriate change of coordinates but that seems like more work than necessary.

Continuing with Stokes' theorem, you have
$$\begin{align*}
\int_C\mathbf F\cdot\mathrm d\vec{r}&=\int_0^{2\pi}\langle2xz,2yz,x^2+y^2\rangle\cdot\langle x',y',z'\rangle\,\mathrm dt\[1ex]
&=\int_0^{2\pi}\left(2xzx'+2yzy'+(x^2+y^2)z'\right)\,\mathrm dt\[1ex]
&=8\int_0^{2\pi}(1-\cos t)\sin t\,\mathrm dt=\color{red}0
\end{align*}$$

fixxer · Answer

I forgot we were in space, does not greens theorem apply when we have a space curve? Stokes seems like a much easier way to go the way you put it tought :)

holsteremission · Answer

Green's theorem only works in 2D. Stokes' theorem is the generalization into higher dimensions (or at least into 3D).

And sure, here's the derivation.
$$\begin{cases}
z=-2x\[1ex]x^2+y^2=z\end{cases}\implies (z-2)^2+4y^2=4$$comes from substituting $x=-\dfrac{z}{2}$ into the second equation, then completing the square.
$$\left(-\frac{z}{2}\right)^2+y^2=z\implies \frac{z^2}{4}+y^2=z\implies z^2-4z+4y^2=0$$Completing the square gives the conclusion above.

We're converting to an off-center version of cylindrical coordinates for the parameterization with $x=\cos t-1$ and $y=\sin t$, so we get
$$(z-2)^2+4\sin^2t=4\implies (z-2)^2=4-4\sin^2t=4\cos^2t$$Take the negative square root (still wondering why negative) to get
$$z-2=-2\cos t\implies z=2-2\cos t$$

holsteremission · Answer

Looking at this again, I think I realize why we take the negative root.

We had as an upper bound to $T$ the plane $z=-2x$, which if we convert to our cylindrical coordinates gives
$$z=-2(\cos t-1)=2-2\cos t$$So I suppose the negative root reassures that the parameterization of the curve stays consistent with the boundaries of $T$.

fixxer · Answer

Thank you, you must be teaching math. Deriving the solution this quickly reflect a well educated man :)

holsteremission · Answer

Ha, I'm flattered, but I'm not a teacher. Just have a fondness for calc :P

fixxer · Answer

You should utilize your skill sets mate :) great stuff are made from mathematics.

fixxer · Answer

hi again, why isnt $$\sqrt{4-4*\sin ^2(t)}$$ a valid parameter for z?

holsteremission · Answer

Well, that choice of $z$ wouldn't be consistent with the way $T$ is defined. In the attached plot it would correspond to the blue path.

fixxer · Answer

I tried to plot it to, hence I asked.

But i was curios since $$\sqrt{4-4\sin ^2(t)}=2+2\left|\cos (t)  \right|$$

holsteremission · Answer

Minor error. That's not true, since
$$\sqrt{4-4\sin^2t}=2|\cos t|\neq\color{red}{2+~}2|\cos t|$$But it's still a valid question: Why are we allowed to use $z=2-2\cos t$ as opposed to $z=2-2|\cos t|$, or $z=2\pm2|\cos t|$?

It stems from the fact that we simultaneously have the condition $z=-2x$. With $x=\cos t-1$, and we require that $z=-2x$, it follows that $z=-2(\cos t-1)=2-2\cos t$.

This stays consistent with the choice of $y=\sin t$, because
$$\underbrace{x^2+y^2}_z\implies(\cos t-1)^2+\sin^2t=\cos^2t-2\cos t+1+\sin^2t=\underbrace{2-2\cos t}_{z(t)}$$
The takeaway is that each condition must remain unaffected after the coordinates are converted. The conditions to check are
$$\begin{cases}x^2+y^2+2x=0\[1ex]z=-2x\[1ex]z=x^2+y^2\end{cases}$$Transforming to our custom polar coordinates using the parameterization $\vec{r}(t)=\langle\cos t-1,\sin t,2-2\cos t\rangle$, you can check that each condition is an identity.

fixxer · Answer

Thanks for clarifying, I just relaised after you replied that I could substitute $$x=\cos(t)-1$$ $$y=\sin(t)$$ into either of the equations for Z and it will give -2cos(t)+2 with the right sign.

Thanks allot Holster, you has helped me so much with this problem and the surface areal problem i had here earlier.

Im so greatful :)

holsteremission · Answer

Glad I could help!