# Multivariable calculus problem

Multivariable calculus problem involving surface area

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.

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## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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(&theta;)-1, y= &radic;(3)sin(&theta;) and we can cover the entire (filled) ellipse by taking x= 2rcos(&theta;)- 1,
y= &radic;(3)rsin(&theta;) with r running from 0 to 1, &theta; running from 0 to 2&pi;.

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

To find the "differential of surface area" with respect to r and &theta; 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 d&theta;

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