Question

the length of the curve of intersection of the cylinder (4x^2) + y^2 = 4 and the plane x + y + z = 2

I'm so lost. Any help is appreciated!

Answer 1

Suppose some point (x,y,z) is on the curve of inersection. Then we can find a point

(x+dx, y+dy, z+dz)

that is infinitessimally close on the curve of intersection. We have:

d[4x^2 + y^2] = 0

d[x+y+z] = 0

----->

8x dx + 2y dy = 0 (1)

dx + dy + dz = 0 (2)

From (1):

dy = -4x/y dx

From (2)

dz = -(dx + dy) = (4x/y - 1)dx

Length element of curve ds follows from Pythagoras's formula:

(ds)^2 = (dx)^2 + (dy)^2 + (dz)^2 =

(express everything in terms of dx) =


[1 + 16 x^2/y^2 + (4x/y - 1)^2] (dx)^2

[2 + 32 x^2/y^2 -8x/y] (dx)^2

So, we have:

ds = sqrt[2 + 32 x^2/y^2 -8x/y]
dx

y is a known function of x (dtermined by the equation of the cylinder). So, we can obtain the curve length by integrating over x from x = -1 to x = 1. You will then get half of the length (if you take one solution for y the range from x = -1 to 1 will move you along one half pof the curve).

To simplify the integration you can use a trig substitution.

Answer 2

The cylinder can be rewritten
x^2 + (y/2)^2 = 1, with z any value.

The curve will be the intersection of that elliptical cylinder, parallel to the z axis, with a plane that is inclined 45 degrees to x, y and z axes. My guess is that the intersection curve will also be an ellipse, with major and minor axes larger than those of the cylinder by a factor sqrt 2.

The major and minor axis lengths for the elliptical cylinder are 2 and 1. The length of an ellipse is pi*(major axis)*(minor axis). If the intersecing plane were perpendicular to the z axis, the answer (the length of the line of intersection) would be 2 pi. For the inclined plane, it is 4 pi.