Evaluate $\int_C (z^2,xz,2xy)\cdot dr$ where $C$ is the intersection of the surface $z=1-y^2, z\ge0$, and $2x+3z=6$ oriented anti-clockworkwise.

Question

Evaluate the line integral of $\int_C F \cdot \,dr : F(x,y,z)=(z^2,xz,2xy)$ and $C$ is the curve obtained by the intersection of the surface $z=1-y^2$, $z \ge 0$, with the plane $2x+3z=6$, oriented anti clockwork wise.

Answer 1

Let's parameterize the curve:

$$y=t$$ $$z=1-t^2$$ $$x=3-\frac{3}{2}z=3-\frac{3}{2}(1-t^2)=\frac{3}{2}t^2+\frac{3}{2}$$

This makes:

$$dr=(dx,dy,dz)=(3t,1,-2t)\ dt$$

$$z^2\ dx=3t(1-t^2)^2\ dt=3t(t^4-2t^2+1)\ dt=(3t^5-6t^3+3t)\ dt$$

$$xz\ dy=\left(\frac{3}{2}t^2+\frac{3}{2}\right)(1-t^2)\ dt=\left(-\frac{3}{2}t^4+\frac{3}{2}\right)\ dt$$

$$2xy\ dz=-4t^2\left(\frac{3}{2}t^2+\frac{3}{2}\right)\ dt=(-6t^4-6t^2)\ dt$$

$$F\cdot dr=(z^2,xz,2xy)\cdot(dx,dy,dz)=\left[(3t^5-6t^3+3t)+\left(-\frac{3}{2}t^4+\frac{3}{2}\right)+(-6t^4-6t^2)\right]\ dt$$ $$=\left(3t^5-\frac{15}{2}t^4-6t^3-6t^2+3t+\frac{3}{2}\right)\ dt$$

For $z\geq 0$ we have $|t|\leq 1$.

Therefore, assuming anticlockwise is to the observer looking from above, then $t$ starts at $1$ and goes to $-1$. This would give:

$$\int_CF\cdot dr=\int_1^{-1}\left(3t^5-\frac{15}{2}t^4-6t^3-6t^2+3t+\frac{3}{2}\right)\ dt$$ $$=\left(\frac{t^6}{2}-\frac{3}{2}t^5-\frac{3}{2}t^4-2t^3+\frac{3}{2}t^2+\frac{3}{2}t\right)\bigg|_1^{-1}=\left(-\frac{3}{2}t^5-2t^3+\frac{3}{2}t\right)\bigg|_1^{-1}=2-(-2)=4$$