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One of my practice problems asks me to compute $\int_C zdx+xdy$ where $C$ is the set of points satisfying $$x^2+y^2+z^2=4 \quad\text{and}\quad 2z+x=0$$ where $C$ is oriented counterclockwise.

If I set $z=-\frac{1}{2}x\,$ I get the ellipse $\frac{5}{8}x^2+\frac{1}{4}y^2=1.$ I'm not sure this the correct curve $C$ to parametrize. Supposing I am correct I can parametrize it to $$\vec{g}(t) = \left(\frac{2\sqrt{10}}{5}\cos t,\;2\sin t\right)$$ for $0\leq t \leq 2\pi.$ I have no idea how to proceed.

The crux of my issues lies in my not understanding what $zdx + xdy$ represents in the integrand.

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The curve is a circle, so, its projection on the $x-y$ plane is an ellipse (better $\frac{5}{16}x^2+\frac{1}{4}y^2=1$ and $x(t)=\dfrac{4\sqrt{5}}{5}\cos t$, $y(t)=2\sin t$), the one you got. Nevertheless, you almost have it: we need $\vec{g}(t) = \left(x(t),y(t),z(t)\right)$, a vector with three components, but you did the parametrization for the $x$ and the $y$ ones. We can complete it because we know that $z=-x/2$, so is, $z= -\dfrac{1}{2}\dfrac{4\sqrt{5}}{5}\cos t$

$$\vec{g}(t) = \left(\frac{4\sqrt{5}}{5}\cos t,\;2\sin t,\frac{-2\sqrt{5}}{5}\cos t\right)$$

Now, for the line integral, $\mathbb dx=x'(t)\,\mathbb dt=-\dfrac{4\sqrt{5}}{5}\sin t\,\mathbb dt$ and $\mathbb dy=y'(t)\,\mathbb dt=2\cos t\,\mathbb dt$

$$\int_C zdx+xdy=\int_0^{2\pi}\frac{-2\sqrt{5}}{5}\cos t\dfrac{-4\sqrt{5}}{5}\sin t\,\mathbb dt+\int_0^{2\pi}\frac{4\sqrt{5}}{5}\cos t\,2\cos t\mathbb dt=$$

$$=\int_0^{2\pi}\left(\dfrac{8}{5}\cos t\sin t+\frac{8\sqrt{5}}{5}\cos^2 t\right)\mathbb dt$$

Being $C$ anticlockwise as you parametrized its projection in that way.

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  • $\begingroup$ We have z parametrized that way because its the one half of x correct? We also need to parametrize the three components because we need to account for the entire three-dimensional vector field? $\endgroup$ – Trevor Mason Mar 13 at 15:31
  • $\begingroup$ Yes to both questions. I suggest you to use an online program for drawing surfaces and curves, they make some problems more intuitive. $\endgroup$ – Rafa Budría Mar 13 at 19:35

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