1

Let $F:\mathbb R^3 - \{0\}\to \mathbb R$ be given by $$F(x,y,z)=\frac{(0,xz,-xy)}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}.$$ Compute $\int_C F\cdot ds$ where $C$ is the unit cirlce on the plane $x+y+z=3$ with the orientation from the point $$\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}\right)$$ to $$\left(1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}\right)$$ to $$\left(1 + \frac{2}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}\right)$$ and back to $$\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}\right)$$

What I've done: By Green's theorem, the integral is $$\iint_D \operatorname{curl}(F)\cdot k\ dA$$ I got that the curl is $$\frac{(x,y,z)}{(x^2+y^2+z^2)^{3/2}}$$ so the integral is $$\int\int_D \frac{z}{(x^2+y^2+z^2)^{3/2}} \ dA$$ But I'm having difficulties with determining how to define the region $D$ to compute the last integral.

Crosby
  • 585
user557
  • 11,889

1 Answers1

0

Obviously $D$ is a disk of radius $1$ in the given plane. You just need to find it's origin. The format of the given points on the circle hints that the center is $(1,1,1)$, which you can verify.

I would do a change of variables, to some $X,Y,Z$ such that $X,Y$ are in the plane, then $Z=0$.

Andrei
  • 37,370
  • How can I verify this? And if so, and if I use the parametrization $x=1+r\cos t, y=1+r\sin t, z=3-x-y$, then the integral reduces to $$\int_0^{2\pi}\int_0^1 \frac{r-r^2(\cos t \sin t)}{(3+2r^2+2r^2\sin t \cos t)^{3/2}}drdt;$$ should I compute it in this form (if it's correct at all), or is there a way to simplify the integrand further? – user557 Jun 26 '18 at 00:45
  • Note that $D$ is the intersection between the sphere of radius $1$ and the plane. Unless the plane is horizontal ($z=const.$), the projection in the $xy$ plane is not a circle, but an ellipse. – Andrei Jun 26 '18 at 03:17
  • As for verifying that $(1,1,1)$ is the center, just plug in the given points in the equation of the sphere with center at $(1,1,1)$: $(x-1)^2+(y-1)^2+(z-1)^2=1$. Also verify that the points are in the plane – Andrei Jun 26 '18 at 03:19