Multivariable calculus problem

[born2chill]

I am not sure where to start with this problem....Any help I receive would be greatly appreciated:


Compute the surface area of the portion of the cone x^2 + y^2 = z^2 which lies between the planes z=0 and x + 2z = 3.

[HallsofIvy]

This is a non-trivial problem. (I think I'll add it to my Calculus III test!!)

The first thing I would do is determine the area in the xy-plane involved here: find the equation of intersection of the cone and project down to the xy-plane.

The plane is x+ 2z= 3 or z= 3/2+ x/2. Putting that into the equation of the cone, we come to the equation of an ellipse (that should be no surprise):
\frac{(x-1)^2}{4}+\frac{y^2}{3}= 1

Parametric equations for that ellipse are x= 2cos(θ)-1, y= √(3)sin(θ) and we can cover the entire (filled) ellipse by taking x= 2rcos(θ)- 1,
y= √(3)rsin(θ) with r running from 0 to 1, θ running from 0 to 2π.

The cone can be expressed in terms of r and &theta as
x= 2rcos(θ)- 1, y= √(3)rsin(θ),z= √(4r2cos2(θ)+ 3r2sin2(θ)) or, in terms of the "position vector" as X= (2rcos(θ)- 1)i+ √(3)rsin(θ)j+ √(4r2cos2(θ)+ 3r2sin2(θ))k.

To find the "differential of surface area" with respect to r and θ differentiate that with respect to each variable and find the cross product of those two vectors.
The differential of surface area is the length of that cross product time dr dθ

Integrate with r going from 0 to 1, &theta from 0 to 2π