Recent content by mariya259

Homework Statement When is electric potential negative and when is it positive? Homework Equations V=Es The Attempt at a Solution I first thought that when you are moving in the same direction as the electric field the electric potential will be a positive value. However I...

Yes, you are right. here is what it looks like now. http://www.wolframalpha.com/input/?i=sqrt%28%28e%5E%28-%28x%5E2%2By%5E2%29%2F2%29%29*%28y%5E4-%283x%5E2*y%5E4%29%2B%284y%5E2*x%5E2%29%2B%28%28x%5E4*y%5E4%29%2F4%29%2B%28%28x%5E2*y%5E6%29%2F4%29%29%2B1+%29+graph

I think I would have to keep the original though, because here is how the region looks like http://www.wolframalpha.com/input/?i=sqrt%28%28e%5E%28-%28x%5E2%2By%5E2%29%2F2%29%29*%28y%5E4-%283x%5E2*y%5E4%29%2B%284y%5E2*x%5E2%29%2B%28%28x%5E4*y%5E4%29%2F4%29%2B%28%28x%5E2*y%5E6%29%2F4%29%2B1+%29

how would i parametrize this function?

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

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...

I understand you can use sum of squares instead of the sqrt of everything, I don't understand why you can do that. Wouldn't the square root change the answer of where the maximum is?

Alright. To find the partials and critical points can I just use the function with the sum of squares or do I need to take the function of the square root(sum of squares)?

I took the sum of squares. From what I understand now I need to set it equal to 0 and solve for critical points?

So only this part is what I need to do? e^(-(x^2+y^2)/4)*((y^2)-.5(x^2)(y^2)))^2 + (e^(-(x^2+y^2)/4)*(2yx-.5x(y^3)))^2

Do you mean taking the square root of each of the partials squared to find the maginitude: sqrt((e^(-(x^2+y^2)/4)*((y^2)-.5(x^2)(y^2)))^2 + (e^(-(x^2+y^2)/4)*(2yx-.5x(y^3)))^2) ?

When I solved for the critical point, I got (x,0) as one of the points. looking at the graph I saw that is a min and max. But I am confused about how a whole line of points could be considered a saddle point? Or would (0,0) be the saddle point in this case, because if you put in (0,0) as a...

The whole question is to find the average value of f(x,y)= x*(y^2)*e^-((x^2+y^2)/4), with x and y from -3 to 3 In my textbook the formula given to find it is (1/Area)*∫∫f(x,y) I took the double integral and found that the answer is 0. The area would be 36, since both x and y are from -3...

If you look at the function along the y axis, when x is less than 0, the function is a local maximum and when x is more than 0 the function is a local minimum. This is looking only at the points on y=0. It looks a bit like a cylinder cut in half. I used the second derivative test and found the...

I have the function: f(x,y)= x*(y^2)*e^-((x^2+y^2)/4), with x and y from -3 to 3 I took the integral of this function and got 0 as my answer. I need to find the average value, which is 1/area multiplied by the double integral. Since the double integral is 0, would the average value also be...