# Thread: Curve of intersection of surfaces

1. ## Curve of intersection of surfaces

The problem statement, all variables and given/known data

See first figure attached

The attempt at a solution

I was able to sketch the two curves individually to get an idea of what I'm looking at, but I still can't really visualize how the two curves would intersect each other in the first octant.

Is this crucial to answering this question?

If I can write parametric equations for x, y and z then I could express the curve of intersection in that form. Then I'd simply have to look at values of t for which x=0, y=0 and z=0, right? Those would be the endpoints of the curve when they are exiting the first octant.

Any ideas/suggestions/tips?

Thanks again!

2. You have the two surfaces defined by $x^2+ \frac{z^2}{4}= 1$, a cylinder, and $z= \sqrt{4x^2+ y^2}$, the "upper half" of the cone given by $z^2= 4x^2+ y^2$.

If you replace the $z^2$ in the first equation by $4x^2+ y^2$, you get $x^2+ x^2+ \frac{y^2}{4}= 1$ or $2x^2+ \frac{y^2}{4}= 1$, an ellipse with major semi-axis along the y axis with length 2 and minor semi-axis along the x-axis, of length $\sqrt{2}{2}$. That is the equation of the projection into the xy-plane.

I don't think the problem actually requires you to find parametric equations for the curve, but if you want to, a standard parameterization of the ellipse $2x^2+ \frac{y^2}{4}= 1$ would be $x= \frac{\sqrt{2}}{2}cos(t)$, $y= 2 sin(t)$. Then $z= \sqrt{4x^2+ y^2}= \sqrt{2cos^2(t)+ 4 sin^2(t)}= \sqrt{2+ 2sin^2(t)}$.

The boundaries of the first octant are the coordinate planes, x= 0, y= 0, and z= 0. If z= 0, then the first equation becomes $x^2= 1$ and since we are in the first octant, x= 1. But then the second equation becomes $0= \sqrt{4+ y^2}$ which is impossible. The curve of intersection does not cross the xz-plane.

If x= 0, then the first equation becomes $\frac{z^2}{4}= 1$ so $z^2= 4$ and, since we are in the first octant, z= 2. then the second equation becomes $2= \sqrt{y^2}$ so $y= 2$. The endpoint in the yz-plane is (0, 2, 2).

If y= 0, then the second equation becomes $z= \sqrt{4x^2}$ or $z= 2x$. Then the first equation becomes $x^2+ x^2= 2x^2= 1$ so that $x= \sqrt{2}{2}$, $z= \sqrt{2}$ and the endpoint in the xz-plane is $(\sqrt{2}{2}, 0, \sqrt{2})$.