# Surface Area of a Multivariable function

1. Jun 1, 2012

### mariya259

I don't know how to calculate the surface area after setting everything up. I have tried both MAPLE 15 program and wolfram alpha, but I can't find the answer.

I have the function:
f(x,y)= x*(y^2)*e^-((x^2+y^2)/4)

The form I found for surface are was the square root of (sum of squares of partials with respect to x and y).
The work I have so far
∫∫√(e^(-(x^2+y^2)/4))*(y^4-(3x^2*y^4)+(4y^2*x^2)+((x^4*y^4)/4)+((x^2*y^6)/4)+1 dx dy
I have to integrate from -3 to 3 for both x and y.
I have tried putting it into MAPLE 15, but it won't solve the function for a value.
What should I do? Is there any way to simplify this algebraically so that it is easier to solve? Is it even possible to solve it by hand?

2. Jun 1, 2012

### tt2348

try spherical/cylindrical coordinates

3. Jun 1, 2012

### mariya259

What would the limits of integration be though? It is not shaped anything like a cylinder or sphere.

4. Jun 1, 2012

### Ray Vickson

I doubt that there is a closed-form formula for the area, so try a numerical approach. If you call dA your integrand, the following works for me in Maple 11:
Jx:=evalf(Int(dA,y=-3..3));
Area:=evalf(Int(Jx,x=-3..3));
Area := 44.43229369

RGV

5. Jun 1, 2012

### algebrat

[STRIKE]Isn't it an odd function in the appropriate sense? You do not have to do the integration, you can realize the value "by inspection".[/STRIKE]

Ray Vickson has pointed out below that I made a (large) oversight.

Last edited: Jun 2, 2012
6. Jun 1, 2012

### Ray Vickson

I think you have misunderstood the problem. For a graph of the form $z = f(x,y)$ the surface area S of the graph over a region $A \subset R^2$ in (x,y)-space is
$$S = \int\int_{A} \sqrt{1 + f_x^2 + f_y^2} \: dx \, dy,$$
where $f_x = \partial f/\partial x,\; f_y = \partial f/\partial y.$ For $$f = x y^2 \exp{\left(-\frac{x^2 + y^2}{4}\right)}$$ we have
$$f_x = \frac{1}{2}y^2 (x^2-2) \exp{\left(-\frac{x^2 + y^2}{4}\right)}\\ f_y = \frac{1}{2}x y (y^2-4) \exp{\left(-\frac{x^2 + y^2}{4}\right)}.$$
Do you really think the double integral S can be evaluated by inspection?

RGV

7. Jun 2, 2012

### algebrat

Yes, that was my mistake, thank you Ray!

8. Jun 2, 2012

### dimension10

Try parametrising it in some other way.

9. Jun 2, 2012

### algebrat

[STRIKE]So dimension10, are you thinking some thing like polar coords., r goes from 0 to 3 cos theta?[/STRIKE]

Oh yeah sorry, I'm not reading very carefully.

Last edited: Jun 2, 2012
10. Jun 2, 2012

### dimension10

[Post cleared]

Last edited: Jun 2, 2012
11. Jun 2, 2012

### algebrat

never mind, deleted

12. Jun 2, 2012

### mariya259

how would i parametrize this function?

13. Jun 2, 2012

### dimension10

I'm not sure but I would suggest a way through which you can use some algebraic/trigonometric identities OR reverse chain rule/reverse product rule.

14. Jun 3, 2012

### Ray Vickson

No matter what you do the function will not have a closed-form double integral. The only reason to bother parametrizing is to make it easier to fit the shape of the region $A \subset R^2$ over which you want to take the surface area. So, if A is rectangular with sides parallel to the x and y axes, just use the original parametrization. If A is a circular disc, you could switch to polar coordinates. If A is an elliptical disc, re-scale x and y to make it a circle, then switch to polar coordinates. In every case you will be forced to get a numerical answer.

RGV

15. Jun 3, 2012

### mariya259

16. Jun 3, 2012

### tt2348

with your region, in wolfram alpha, I think you need to put the 1 outside the parenthesis in the square root

17. Jun 3, 2012

### Ray Vickson

18. Jun 3, 2012

### tt2348

observe some of the symmetries, even along y, odd along x. so your surface area would basically be 4x (integral over region of [0,3]^2)

19. Jun 3, 2012