$\int_c 3y dx +5x dy + \frac{2x+3}{z^2} dz$ for the intersection between two surfaces

Question

For the closed curve, $c = \{(x,y,z) \vert x^2 + y^2 - z^2 =0\} \cap \{(x,y,z) \vert (x-1)^2 + y^2 =4\}$

Find the $\int_c 3y dx +5x dy + \frac{2x+3}{x^2 + y^2} dz$

First I focused the surface $\{(x,y,z) \vert x^2 + y^2 - z^2 =0\}$ and tried to simplify like the $\int_c 3y dx +5x dy + \frac{2x+3}{z^2} dz$.

Next step is parameterizing the boundary surface $X(r,\theta)= (1+rcos\theta, rsin\theta, \pm\sqrt{1+r^2 + 2rcos\theta})$ for $0\leq r \leq 2, 0\leq \theta \leq 2\pi$

But the problem happened there are two curves. To applying the Stoke's thm, Which curve do I choose? Plus Is there are another methods without stokes thm? It is too complicated to solve by the Stokes thm. Help me.

Answer 1

Given there are two intersection curves, the question should have stated which intersection curve. It is a different matter that the line integral over either curve is same. It should have also stated orientation. Say we are interested in evaluating line integral over the intersection curve above $z = 0$ in anticlockwise direction. An easier approach in this case is direct line integral over the closed curve instead of applying Stokes' theorem. Applying Stokes' theorem would require you to consider a surface that does not cross z-axis.

At the intersection, $(x-1)^2 + y^2 = 4 \implies x^2 + y^2 = 3 + 2x = z^2$

Parametrizing the intersection curve as $r (t) = (1 + 2 \cos t, 2 \sin t, \sqrt{5+4 \cos t}), 0 \leq t \leq 2\pi$

$r'(t) = (- 2 \sin t, 2 \cos t, - \frac{2 \sin t}{\sqrt{5 + 4 \cos t}})$

$\vec F = (3y, 5x, \frac{2x+3}{x^2+y^2}) = (6 \sin t, 5 + 10 \cos t, 1)$

Then evaluate $ \displaystyle \int_0^{2\pi} \vec F \cdot r'(t) ~ dt$ and that gets an answer of $8 \pi$. The work can be simplified using the fact that integral of $\cos t$ and $\cos 2t$ over $(0, 2\pi)$ evaluates to zero.

Answer 2

$x^2 + y^2 - z^2 = 0$ represents a cone with axis along the $z$ axis, and $(x-1)^2 + y^2 = 4 $ is a right circular cylinder with axis parallel to the $z$ axis and passing through $(1, 0)$ and a radius of $2$.

Let $x = 1 + 2 \cos \theta , y = 2 \sin \theta $, then this point lies on the cylinder, the intersection with the cone results in

$z^2 = x^2 + y^2 = 5 + 4 \cos \theta $

Take $z$ to be positive, then $z = \sqrt{ 5 + 4 \cos \theta } $

Differentials:

$dx = -2 \sin \theta d\theta $

$dy = 2 \cos \theta d\theta $

$dz = \dfrac{- 2 \sin \theta d \theta }{ \sqrt{5 + 4 \cos \theta }} $

$\displaystyle \int_C 3 y dx + 5 x dy + \dfrac{2x+3}{x^2 + y^2} dz $

Becomes

$\displaystyle \int_0^{2 \pi} (-6 \sin^2 \theta + 10(1 + 2 \cos \theta) \cos \theta + \dfrac{ 2(1 + 2 \cos \theta) + 3 }{5 + 4 \cos \theta } \dfrac{- 2 \sin \theta}{ \sqrt{5 + 4 \cos \theta }}) d\theta $

which simplifies to

$\displaystyle \int_0^{2 \pi} (-12 \sin^2 \theta + 20 \cos^2 \theta + 10 \cos \theta + \dfrac{- 2 \sin \theta}{ \sqrt{5 + 4 \cos \theta }}) d\theta = \dfrac{1}{2} (-12 + 20)(2 \pi) = \boxed{8 \pi}$