# I Parameterize an offset ellipse and calculate the surface area

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1. Mar 16, 2016

### Thales Costa

I'm given that:

S is the surface z =√(x² + y²) and (x − 2)² + 4y² ≤ 1

I tried parametrizing it using polar coordinates setting
x = 2 + rcos(θ)
y = 2rsin(θ)
0≤θ≤2π, 0≤r≤1
But I'm not getting the ellipse that the original equation for the domain describes
So far I've tried dividing everything by 4 and also tried the method of completing the square, but no success.

I'm supposed to calculate the surface area of S. But without the parametric equations, calculating the normal vector is impossible.

EDIT: Messing with the equations on Wolfram I got the following:

x = 2 + cos(u)
y = (1/2)sin(u)
0≤ u ≤2π

But when I multiply the cosine and sine by r and make r vary from 0 to 1, the parametric plot changes to something completely different

Last edited: Mar 16, 2016
2. Mar 17, 2016

### andrewkirk

Are you sure you need to parametrize? What is the shape of the unconstrained surface? Does the angle between the normal to the surface and the $z$ axis change over the surface? If not, can you simplify the problem and solve it without parametrization?

3. Mar 17, 2016

### Thales Costa

So I didn't need to parametrize the surface.

It was a matter of just simply calculating dS, which is the cross-product between dz/dx and dz/dy. dS = √2 dA

Then integrating over the surface ∫∫√2⋅dA I would get that the surface area I was looking for is √2 times the Area of the ellipse.

I was going in the harder direction trying to figure out the integral that would give me the area of the ellipse. And this I still don't know how.

4. Mar 17, 2016

### andrewkirk

If you change the inequality to an equality that gives you the equation of the ellipse. The area won't change if you translate it in the x direction, so translate it so that the first term becomes just $x^2$. Now you have an ellipse centred on the origin. If you just integrate the area under the branch in one of the four quadrants and multiply that by four, there's your area of the ellipse.

5. Mar 17, 2016

### Thales Costa

This is what I figured. I took me a while to understand that I was integrating a vector field over a surface and that the position of the surface could be shifted to the origin without changing the result.

So integrating the field over x² +4y² = 1 would be the same as integrating it over the original surface. Is that correct?

6. Mar 17, 2016

### andrewkirk

It would if you also shifted the vector field in the X direction by -2.
But you don't need to worry about that. Above you concluded that
Once you've reasoned your way to there, you can forget about the vector field and just calculate the area of the ellipse.