Question

Find, correct to four decimal places, the length of the curve of intersection of the cylinder 16x2 + y2 = 16 and the plane x + y + z = 1

Answer 1

Let the curve C be the intersection of the cylinder  

$16x^2+y^2=16$

and the plane

$x+y+z=1$

The projection of C on to the x-y plane is the ellipse

$16x^2+y^2=16$

To see clearly that this is an ellipse, le us divide through by 16, to get

$\frac{x^2}{1}+ \frac{y^2}{16}=1$

or  

$\frac{x^2}{1^2}+ \frac{y^2}{4^2}=1$,

We can write the following parametric equations,

$x=cos(t), y=4sin(t)$

for  

$0\le t \le 2\pi$

Since C lies on the plane,

$x+y+z=1$

it must satisfy its equation.

Let us make z the subject first,  

$z=1-x-y$

This implies that,

$z=1-sin(t)-4cos(t)$

We can now write the vector equation of C, to obtain,

$r(t)=(cos(t),4sin(t),1-cos(t)-4sin(t))$

The length of the curve of the intersection of the cylinder and the plane is now given by,

$\int\limits^{2\pi}_0 {|r'(t)|} \, dt$

But  

$r'(t)=(-sin(t),4cos(t),sin(t)-4cos(t))$

$|r'(t)|=\sqrt{(-sin(t))^2+(4cos(t))^2+(sin(t)-4cos(t))}$

$\int\limits^{2\pi}_0 {\sqrt{2sin^2(t)+32cos(t)-8sin(t)cos(t)} }\, dt=24.08778184$

Therefore the length of the curve of the intersection  intersection of the cylinder and the plane is 24.0878 units correct to four decimal places.