Answer:
1:2 - one dog to two cats.
If it's a square all sides are 9 feet...9x9=81
With
![\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20r%28t%29%3D4t%5C%2C%5Cvec%5Cimath%2B6t%5C%2C%5Cvec%5Cjmath-t%5E2%5C%2C%5Cvec%20k)
we have
![\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cvec%20r%3D%284%5C%2C%5Cvec%5Cimath%2B6%5C%2C%5Cvec%5Cjmath-2t%5C%2C%5Cvec%20k%29%5C%2C%5Cmathrm%20dt)
The vector field evaluated over this parameterization is
![\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20f%28x%2Cy%2Cz%29%3D%5Cvec%20f%28x%28t%29%2Cy%28t%29%2Cz%28t%29%29%3D4t%5C%2C%5Cvec%5Cimath%2Bt%5E2%5C%2C%5Cvec%5Cjmath%2B6t%5C%2C%5Cvec%20k)
so the line integral is
![\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B-1%7D%5E1%284t%5C%2C%5Cvec%5Cimath%2Bt%5E2%5C%2C%5Cvec%5Cjmath%2B6t%5C%2C%5Cvec%20k%29%5Ccdot%284%5C%2C%5Cvec%5Cimath%2B6%5C%2C%5Cvec%5Cjmath-2t%5C%2C%5Cvec%20k%29%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_%7B-1%7D%5E1%2816t%2B6t%5E2-12t%5E2%29%5C%2C%5Cmathrm%20dt%3D-4)
A circle with center
and radius
has equation
![(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x = 16](https://tex.z-dn.net/?f=%28x-3%29%5E2%2By%5E2%3D25%20%5Ciff%20x%5E2%20%2B%20y%5E2-%206%20x%20%20%3D%2016)
If we substitute
in this equation, we have
![x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0](https://tex.z-dn.net/?f=x%5E2%2B%282x%2Bk%29%5E2-6x%3D16%20%5Ciff%205x%5E2%2B%284k-6%29x%2Bk%5E2-16%3D0)
This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:
![\Delta=b^2-4ac=(4k-6)^2-4\cdot 5\cdot (k^2-16)>0](https://tex.z-dn.net/?f=%5CDelta%3Db%5E2-4ac%3D%284k-6%29%5E2-4%5Ccdot%205%5Ccdot%20%28k%5E2-16%29%3E0)
The expression simplifies to
![-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89](https://tex.z-dn.net/?f=-4%20%28k%5E2%20%2B%2012%20k%20-%2089%29%3E0%20%5Ciff%20k%5E2%20%2B%2012%20k%20-%2089%3C0)
The solutions to the associated equation are
![k^2 + 12 k - 89=0 \iff k=-6\pm 5\sqrt{5}](https://tex.z-dn.net/?f=k%5E2%20%2B%2012%20k%20-%2089%3D0%20%5Ciff%20k%3D-6%5Cpm%205%5Csqrt%7B5%7D)
So, the parabola is negative between the two solutions:
![-6-5\sqrt{5}](https://tex.z-dn.net/?f=-6-5%5Csqrt%7B5%7D%3Ck%3C-6%2B5%5Csqrt%7B5%7D)