Integrating to find average value

[mariya259]

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 equal to 0?
How would I solve this?

 

[DryRun]

What is the whole question?

 

[SammyS]

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 equal to 0?
How would I solve this?

Yes, that integral is zero.

If this is a problem you have been given to work on, Please, give us the entire problem, word for word, as it was given to you.

I see that sharks beat me to the punch !

 

[mariya259]

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 to 3.
1/36 * 0 = 0
Can the average value really be 0 though?

 

[SammyS]

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 to 3.
1/36 * 0 = 0
Can the average value really be 0 though?

Yes! Why not ?

 

[algebrat]

Notice it is an odd function when viewed from the x-axis. Try graphing it in wolfram alpha.