# Line Integral Over a Vector Field of a set of points where a sphere intersects $2z$ $+$ $x$ $=$ $0$

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.

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.

• 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? – Trevor Mason Mar 13 at 15:31
• Yes to both questions. I suggest you to use an online program for drawing surfaces and curves, they make some problems more intuitive. – Rafa Budría Mar 13 at 19:35