Calculate the path integral $\int x^3dx + y^3dy + z^3dz$ over the curve $\{z=x^2+y^2,\, y+z=3\}$

Question

I've plotted the curve. I suppose I need to parametrize it, e.g. $\{x = r\cos{\alpha},\,y = r\sin{\alpha}-0.5,\,z = 3.5 - r\sin{\alpha}\}$, but substituting the values of $x,y,z$ into the integral yields long cubic formulas. Maybe it's possible to apply Stokes Theorem here, but I haven't found much information on how to do it with parametrized curves. Is there any simpler way to compute this integral? ($r = \sqrt{13}/2$ is a constant)

Answer 1

The curl of the vector field $({x^3,y^3,z^3})$ is zero, so the vector field is conservative. Path integrals over closed curves in conservative fields are 0, so the answer is 0.

Answer 2

$$x^2+y^2=3-y \implies x^2+\left(y+\frac12\right)^2=\frac{13}4$$

Let $x(t)=\frac{\sqrt{13}}2\cos(t)$ and $y(t)=\frac{\sqrt{13}}2\sin(t)-\frac12$, so it follows that $z(t)=3-y(t)=\frac72-\frac{\sqrt{13}}2\sin(t)$. You get the curve of intersection for $0\le t\le2\pi$.

Denote this parameterization of the path $C$ by $\vec r(t)$; that is,

$$\vec r(t)=x(t)\,\vec\imath+y(t)\,\vec\jmath+z(t)\,\vec k$$

with derivative

$$\frac{\mathrm d\vec r(t)}{\mathrm dt}=x'(t)\,\vec\imath+y'(t)\,\vec\jmath+z'(t)\,\vec k$$

Denoting the given inherent vector field by $\vec F(x,y,z)$, the line integral over $C$ is then

$$\int_C\vec F(x,y,z)\cdot\mathrm d\vec r=\int_C\vec F(\vec r(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt=\int_0^{2\pi}g(t)\,\mathrm dt$$

where $g(t)$ is short for

$$-\frac{211\sqrt{13}}8\cos(t)+\frac{39\sqrt{13}}8\cos^3(t)+\frac{4069}{32}\cos(t)\sin(t)-\frac{507}{32}\cos^3(t)\sin(t)-\frac{117\sqrt{13}}8\cos(t)\sin^2(t)+\frac{507}{32}\cos(t)\sin^3(t)$$

and whose antiderivative is easy enough to find. But this method is overkill, and you should use shortcuts (Stokes' theorem, checking that $\vec F$ is conservative) when you can.