# Thread: Am I doing this right?

1. ## Am I doing this right?

Find a vector function that represents the curve of intersection of the two surfaces.

The cylinder x^2 + y^2 = 4 and the surface z = xy.

What I do is make z = 0, then x^2 + y^2 = 4 is a circle and
xy = 0.

Then I make these equations equal each other.

x^2 + y^2 - 4 = xy and simplify

x^2 + y^2 - xy - 4 = 0

But now I can't seem to complete the square.

Any help would be appreciated.

2. Originally Posted by Undefdisfigure
Find a vector function that represents the curve of intersection of the two surfaces.

The cylinder x^2 + y^2 = 4 and the surface z = xy.

What I do is make z = 0, then x^2 + y^2 = 4 is a circle and
xy = 0.

Then I make these equations equal each other.

x^2 + y^2 - 4 = xy and simplify

x^2 + y^2 - xy - 4 = 0

But now I can't seem to complete the square.

Any help would be appreciated.
You can finish this by rotating the coordinate system about the z axis by some angle to eliminate the xy term.

However, I'm confused about something. Why are you specifically setting z = 0?

-Dan

3. I'm setting z = 0 to simplify the expression so I can complete the square. It appears that this is not the way to go. Could you write out in mathematical terms the rotation about the z - axis here for me please. I guess when you deal with surfaces its not the same as curves.

4. Originally Posted by Undefdisfigure
Find a vector function that represents the curve of intersection of the two surfaces.

The cylinder x^2 + y^2 = 4 and the surface z = xy.

What I do is make z = 0, then x^2 + y^2 = 4 is a circle and
xy = 0.

Then I make these equations equal each other.

x^2 + y^2 - 4 = xy and simplify

x^2 + y^2 - xy - 4 = 0

But now I can't seem to complete the square.

Any help would be appreciated.
It's easier than you think:

The curve is the set of points that satisfy $x^2 + y^2 = 4$ and $z = xy$. Using parametric equations:

To satisfy $x^2 + y^2 = 4$, $x = 2 \cos t\,$ and $\,y = 2 \sin t\,$ (note: clearly other choices are possible).

Therefore, to satisfy $z = xy$ it is required that $\,z = (2 \cos t) \, (2 \sin t) = 2 \sin(2 t)$.

Note that $-\infty < t < \infty$.

Therefore the vector equation of the curve is:

$r = (\,2 \cos t\,) i + (\,2 \sin t\,) j + (\,2 \sin(2t)\,) k$, where $-\infty < t < \infty$.

5. Originally Posted by Undefdisfigure
x^2 + y^2 - xy - 4 = 0
Let
$x = x^{\prime}~cos(\theta) - y^{\prime}~sin(\theta)$
and
$y = x^{\prime}~sin(\theta) + y^{\prime}~cos(\theta)$

(This represents a rotational change of coordinates.)

Then
$x^2 + y^2 - xy - 4 = 0$
becomes
$(x^{\prime}~cos(\theta) - y^{\prime}~sin(\theta))^2 + (x^{\prime}~sin(\theta) + y^{\prime}~cos(\theta))^2 -$ $(x^{\prime}~cos(\theta) - y^{\prime}~sin(\theta))(x^{\prime}~sin(\theta) + y^{\prime}~cos(\theta)) - 4 = 0$

Now expand this and find the angle $\theta$ that makes the $x^{\prime}y^{\prime}$ coefficient 0. (It's a bit of work and more than I wish to type out at the moment.)

-Dan