To evaluate a line integral along the curve of intersection of the cylinder $x^2+y^2=a^2$ and the plane $x/a+z/b=1$

Question

Let $C$ be the curve of intersection of the cylinder $x^2+y^2=a^2$ and the plane $\frac{x}{a}+\frac{z}{b}=1$; how to evaluate $$\int_C (y-z) dx+(z-x)dy+(x-y)dz?$$ I know that the parametrization for $C$ is $(a\cos t, a\sin t,b-b\cos t)$; but I can't determine the limits for $t$ as I can't visualize the intersection curve. Should I use Stoke's theorem? But then I also can't find the relevant surface.

Answer 1

You can check why the range for $t$ I provided in my comment will work by checking values of $t$ within that interval - what do you get when $t=0$? $t=\dfrac{\pi}{2}$? and so on.

The integral becomes $$\begin{align*}I&=\oint_C(y-z)\,\mathrm{d}x+(z-x)\,\mathrm{d}y+(x-y)\,\mathrm{d}z\\[1ex] &=\int_0^{2\pi}\bigg((a\sin t-b+b\cos t)(-a\sin t)+(b-b\cos t-a\cos t)(a\cos t)+(a\cos t-a\sin t)(b\sin t)\bigg)\,\mathrm{d}t\\[1ex] &=\int_0^{2\pi}\bigg(-a^2-ab+ab\cos t+ab\sin t\bigg)\,\mathrm{d}t\end{align*}$$

Answer 2

but I can't determine the limits for $t$ as I can't visulaize the intersection curve , what shape is it

The cylinder is symmetric around the $z$ axis, with radius $a$.

The plane has a normal vector $n = (1/a, 0, 1/b)$ in the $x$-$z$-plane, it is affine not needing to contain the origin. Distance from the origin is $1/\lVert n \rVert = ab/\sqrt{a^2+b^2}$.

Intersection curve seems to be an ellipse within the intersection plane. For the projection in the $y$-$z$-plane we have: $$ a^2 = y^2 + \left(a - \frac{a}{b} z\right)^2 = y^2 + \left(\frac{a(z-b)}{b}\right)^2 \iff \\ \left(\frac{y}{a}\right)^2 + \left(\frac{z-b}{b}\right) = 1 $$