1. ## Iterated Integrals

Hey Everyone,

Suppose I have this function f(x,y) = 8xy

I want to evaluate the double integral of this function over the region enclosed by the curve x^2 + y^2 + xy = 3/4

I am having a little trouble constructing the integrands. I believe that -1 < x < 1, but I am not positive about what the range of y is.

Thanks again!

Hey Everyone,

Suppose I have this function f(x,y) = 8xy

I want to evaluate the double integral of this function over the region enclosed by the curve x^2 + y^2 + xy = 3/4

I am having a little trouble constructing the integrands. I believe that -1 < x < 1, but I am not positive about what the range of y is.

Thanks again!
you need to rotate your curve to get rid off the term $\displaystyle xy$ and get a standard form of an ellipse. the substitution $\displaystyle x=u+v, \ y=u-v$ will do that for you.

you can also directly transform your curve to a circle instead of a (standard) ellipse and then a circle. the substitution $\displaystyle x=u+\sqrt{3}v, \ y=u-\sqrt{3}v$ will do the job.

Hey Everyone,

Suppose I have this function f(x,y) = 8xy

I want to evaluate the double integral of this function over the region enclosed by the curve x^2 + y^2 + xy = 3/4

I am having a little trouble constructing the integrands. I believe that -1 < x < 1, but I am not positive about what the range of y is.

Thanks again!
How about just solving it as is where is first as in $\displaystyle 2\int_0^1 \int_{g_1(y)}^{g_2(y)} 8 xy\,dxdy$ where $\displaystyle g(y)=\frac{-y\pm \sqrt{3-3y^2}}{2}$? Pretty sure that's correct but haven't worked through it yet. If it's not, try and fix it please.

Hey Everyone,

Suppose I have this function f(x,y) = 8xy

I want to evaluate the double integral of this function over the region enclosed by the curve x^2 + y^2 + xy = 3/4
Put $\displaystyle x=\frac{u-\sqrt{3}v}{\sqrt{6}},\,y=\frac{u+\sqrt{3}v}{\sqrt{ 6}}$ and your region will be transfered to $\displaystyle u^{2}+v^{2}\le \frac{3}{2}$ and then apply the usual polar transformation.

5. Thanks everyone.

I tried doing it all three ways and I either cannot finish the integration or I get a negative answer, which I do not think is correct.

If anyone does calculate this and gets a plausible answer, please let me know!

Thanks again

6. UPDATE!

I tried using a change of variables and then solving along with jacobian value and got an answer of 8/sqrt(3). Whether this is right or wrong, it is anybody's guess.

I am almost positive that it should be possible by find the range of x and y and solving the integral, without making any substitutions.

I found it to be -1<x<1 and -x/2 - sqrt(3-3x^2)/2 < y < -x/2 + sqrt(3-3x^2)/2.

However, when I try to solve this...i get -sqrt(3)pi/2...

Any ideas? or do you see any errors i may have made?
Thanks

7. Since 8xy is negative and positive (in your region) you cannot claim that
$\displaystyle \int\int 8xy dxdy>0$, so a negative answer is not impossible.

8. Originally Posted by matheagle
Since 8xy is negative and positive (in your region) you cannot claim that
$\displaystyle \int\int 8xy dxdy>0$, so a negative answer is not impossible.
Thanks! I had a feeling that did not make sense. However, I don't think I made any integration mistakes so that means the region i am integrating on is incorrect...but then i am pretty sure that is right also. I guess it is time to do an extensive check of all my calculations