<s>[INST] $\int_c 3y dx +5x dy + \frac{2x+3}{z^2} dz$ for the intersection between two surfaces 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.
 [/INST] $x^2 + y^2 - z^2 = 0$ 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
$\displaystyle \int_0^{2 \pi} (-12 \sin^2 \theta + 20 \cos^2 \theta + 10 \cos \theta + \dfrac{- 2 \sin \theta}{ \sqrt{5 + 4 \cos \theta }}) d\theta = \dfrac{1}{2} (-12 + 20)(2 \pi) = \boxed{8 \pi}$
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