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How to calculate $$\int\int_E e^{5x^2+2xy+y^2}dA,$$ where $E=\{(x,y)\mid 5x^2+2xy+y^2\leq 1\}$? I know I have to use the change-of-variable formula by first finding the change-of-variable function $\Psi$. However, what's the function should I use? And how to deal with the annoying $xy$-term?

Eric
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    $5x^2+ 2xy+ y^2= 4x^2+ x^2+ 2xy+ y^2= 4x^2+ (x+ y)^2$. Let z= x+ y and that becomes $4x^2+ z^2\le 1$, an ellipse, and its interior, in xz-space. – user247327 Jan 31 '18 at 14:13

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Assume that the quadratic form $q(x,y)=ax^2+2bxy+cy^2$, associated to the matrix $Q=\begin{pmatrix}a & b \\ b & c \end{pmatrix}$ is positive definite, i.e. $a>0$ and $ac>b^2$ (Sylvester's criterion). Let $$E_q=\{(x,y)\in\mathbb{R}^2: q(x,y)\leq 1\}.$$ By the spectral theorem, $$ \iint_{E_q}e^{q(x,y)}\,dx\,dy = \iint_{\lambda_1 x^2+\lambda_2 y^2\leq 1}e^{\lambda_1 x^2+\lambda_2 y^2}\,dx\,dy $$ where $\lambda_1,\lambda_2$ are the eigenvalues of $Q$. By straightforward substitutions, the RHS equals $$ \frac{1}{\sqrt{\lambda_1 \lambda_2}}\iint_{X^2+Y^2\leq 1}e^{X^2+Y^2}\,dX\,dY=\frac{1}{\sqrt{\det Q}}\int_{0}^{2\pi}\int_{0}^{1}\rho e^{\rho^2}\,d\rho\,d\theta=\color{red}{\frac{\pi}{\sqrt{ac-b^2}}}. $$ Can you see what happens by considering $a=5$ and $b=c=1$?

Jack D'Aurizio
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  • Thanks. Can you elaborate more on the "straightforward substitutions" part? – Eric Feb 01 '18 at 15:14
  • Rescale and switch to polar coordinates. – Jack D'Aurizio Feb 01 '18 at 15:52
  • Thank you. I think I understand the process. Use the spectrum theorem suggest a way on how to choose the "right" transformation. And after applying such transformation, the region consequently changed. A slightly generalized question: how about the integrand is not bad, but the integral region is very bad, like $\int_{\Bbb R^2}e^{-(4x^2+4xy+5y^2)}d(x,y)$? This time, is not like the original one that the region is also a somewhat "correspoding ellipse" (so the change-of-variable formula works fine.) This is a problem came from an degree exam years ago. – Eric Feb 01 '18 at 16:02
  • And by the way, the eigenvalues of the quadratic form in this post is https://imgur.com/a/GFohi, which looks very ugly, including the orthonormal eigenvectors. It seems hard to compute, is there anything wrong? – Eric Feb 01 '18 at 16:15
  • The point is that if the integration domain is sufficiently symmetric, you do not need to compute the Jordan decomposition of Q, since Q is conjugated to a diagonal matrix through an orthogonal transformation an the product of the eigenvalues equals the determinant of Q. – Jack D'Aurizio Feb 01 '18 at 16:40
  • Hmm.. not sure I understand. I think I should post a new question, can you help me there? Thanks. – Eric Feb 01 '18 at 16:55