Thread: Surface Integral Qu

Hi, I am new to this forum and would be grateful if anyone could help me in this surface integral question.

Find the area of the surface of the cone z^2=3(x^2+y^2) cut out by the paraboloid z=x^2+y^2.

The answer is 6pi but I'm not sure how exactly to get this.
Here is my working

We are trying to find (int z) dS

z=sqrt(3x^2+3y^2). So dz/dx=(3x)/(sqrt(3x^2+3y^2) and dz/dy=(3y)/sqrt(3x^2+3y^2). Using the formula dS= sqrt((dz/dx)^2+(dz/dy)^2+1), we are left with

Double integral sqrt(3x^2+3y^2)*2 dxdy. In polar coordinates this gives

Double integral sqrt(3r^2)*2r drdtheta. The radius is sqrt3, so the first integral is from 0 to sqrt3.

The problem is that I thought the second integral is from 0 to 2pi. But this gives a final answer of 12pi. So the second integral must be from 0 to pi.

Can anyone explain why this would be the case, or have I made a mistake somewhere else?

Originally Posted by Jtown

Hi, I am new to this forum and would be grateful if anyone could help me in this surface integral question.

Find the area of the surface of the cone z^2=3(x^2+y^2) cut out by the paraboloid z=x^2+y^2.

The answer is 6pi but I'm not sure how exactly to get this.
Here is my working

We are trying to find (int z) dS

z=sqrt(3x^2+3y^2). So dz/dx=(3x)/(sqrt(3x^2+3y^2) and dz/dy=(3y)/sqrt(3x^2+3y^2). Using the formula dS= sqrt((dz/dx)^2+(dz/dy)^2+1), we are left with

Double integral sqrt(3x^2+3y^2)*2 dxdy. In polar coordinates this gives

Double integral sqrt(3r^2)*2r drdtheta. The radius is sqrt3, so the first integral is from 0 to sqrt3.

The problem is that I thought the second integral is from 0 to 2pi. But this gives a final answer of 12pi. So the second integral must be from 0 to pi.

Can anyone explain why this would be the case, or have I made a mistake somewhere else?

The surface area is given by , NOT . Therefore:

Surface area

where the region of integration in the xy-plane is bounded by the circle .