## Surface Area of a Multivariable function

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

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## Surface Area of a Multivariable function

 Quote by 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?
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
 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". Ray Vickson has pointed out below that I made a (large) oversight.

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 Quote by algebrat 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".
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
 Yes, that was my mistake, thank you Ray!

 Quote by 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?
Try parametrising it in some other way.
 So dimension10, are you thinking some thing like polar coords., r goes from 0 to 3 cos theta? Oh yeah sorry, I'm not reading very carefully.
 [Post cleared]
 never mind, deleted
 how would i parametrize this function?

 Quote by mariya259 how would i parametrize this function?
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.

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 Quote by mariya259 how would i parametrize this function?
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
 I think I would have to keep the original though, because here is how the region looks like http://www.wolframalpha.com/input/?i...2F4%29%2B1+%29
 with your region, in wolfram alpha, I think you need to put the 1 outside the parenthesis in the square root

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 Quote by mariya259 I think I would have to keep the original though, because here is how the region looks like http://www.wolframalpha.com/input/?i...2F4%29%2B1+%29
I think you have the wrong integrand: your integrand should be of the form
$$\sqrt{1 + e^{-(x^2+y^2)/2} \cdot \text{some function of x and y}},$$ but that is not what you have.

RGV

 Tags integration, surface area