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I am having trouble with determining the integral bounds in change of variable problems.

The problem:

Consider the transformation given by $\space x = u - \sqrt{(\frac13)}\cdot v \space$ and $\space y = u + \sqrt{(\frac13)}\cdot v$ $- g(u,v) \space$ and $h(u,v)$ respectively

$R = {(x,y)\in R^2 : x^2 −xy+y^2 \le1}$ and $S = {(u,v)∈ R^2 : (u - \sqrt{(\frac13)}\cdot v, u + \sqrt{(\frac13)\cdot v)} \in R}$

I understand that we take the integral of $f(g(u,v), h(u,v))\cdot\text{Jacobian} - \text{Jacobian}$ evaluates to $\frac{2}{\sqrt{(\frac13)}}$

This is where I get stuck. I know we can find the bounds for $x$ and $y$ from $x^2 −xy+y^2 \le1$ but I'm unsure how to convert that to $u,v$ variables.

In the problem I have to show this \begin{equation} \iint_Rx^2-xy+y^2\ dA=\frac{\pi}{\sqrt{3}} \end{equation}

And also have to draw the transformation from $(x,y)$ to $(u,v)$

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  • $\begingroup$ My question is what is your region R? What is your area bounded by? $\endgroup$ Commented Jun 18, 2020 at 19:52
  • $\begingroup$ The $x^2-xy+y^2$ is the multivariable function you are integrating. $\endgroup$ Commented Jun 18, 2020 at 19:54
  • $\begingroup$ Is the problem Type I or Type 2, is it $dydx$, or $dxdy$ $\endgroup$ Commented Jun 18, 2020 at 19:55
  • $\begingroup$ To me at least your problem is missing the initial region, whether it's type 1 or type 2, the point of the transformation, your determinant on how you got your Jacobian, your partials. Things that would help solve the problem. $\endgroup$ Commented Jun 18, 2020 at 20:03
  • $\begingroup$ if $x=u-\frac{1}{\sqrt{3}}v$, and $y=u+\frac{1}{\sqrt{3}}v$, then you Jacobian is setup as follows. \begin{bmatrix}\frac{\delta x}{\delta u}&\frac{\delta x}{\delta y}\\ \frac{\delta y}{\delta u} & \frac{\delta y}{\delta v}\end{bmatrix} $\endgroup$ Commented Jun 18, 2020 at 20:25

1 Answer 1

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Consider the assignment of the variables $u(x, y) = \dfrac{x + y} 2$ and $v(x, y) = \dfrac{\sqrt 3(y - x)} 2.$ We have that $$u^2 + v^2 = \frac{x^2 + 2xy + y^2} 4 + \frac{3(y^2 - 2xy + x^2)} 4 = x^2 - xy + y^2.$$ Consequently, we obtain a transformation $G(x, y) = (u(x, y), v(x, y))$ with Jacobian $$\operatorname{Jac}(G) = \det \begin{pmatrix} \frac 1 2 & -\frac{\sqrt 3} 2 \\ \frac 1 2 & \phantom -\frac{\sqrt 3} 2 \end{pmatrix} = \frac{\sqrt 3}{2}.$$ Observe that the region in the $uv$-plane that is mapped onto by $G$ is given by $0 \leq u^2 + v^2 \leq 1,$ i.e., the disk of radius $1$ centered at the origin. Using polar coordinates, this region is $\{(r, \theta) \,|\, 0 \leq r \leq 1 \text{ and } 0 \leq \theta \leq 2 \pi\}.$ By the Change of Variables Formula, therefore, we have that $$\iint_R (x^2 - xy + y^2) \, dA = \frac 2 {\sqrt 3} \iint_U (u^2 + v^2) \, dA = \frac 2 {\sqrt 3} \int_0^{2 \pi} \int_0^1 r^3 \, dr \, d \theta = \frac{\pi}{\sqrt 3}.$$

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