0
$\begingroup$

Let S be the region satisfying $ 3x^2 + 2xy + y^2 \leq\ 1$ in the real plane $R^{2}$. Then compute the value of the double integral

$$\int\int_S\ e^{(3x^2+2xy+y^2)}dxdy$$

i learned double integral at polar coordinates. but i didn't solve this problem. Please give me hint or answer.

$\endgroup$
  • $\begingroup$ You can use something similar to polar coordinates but transformed in such a way that the circle becomes the ellipse $\endgroup$ – Paul Castle Jul 11 '16 at 6:42
2
$\begingroup$

$$3x^2 +2xy +y^2 = 3(x + y/3)^2 + \frac{2 y^2}{3}$$ So try the substitution $u = \sqrt{3} x + y/\sqrt{3}, v = \sqrt{2/3} y$

This should transform it into an integral over the unit circle. Then you can use polar coordinates.

$\endgroup$
0
$\begingroup$

Referring to another answer and transforming the ellipse into:

$$(2+\sqrt{2})X^2+(2-\sqrt{2})Y^2=1$$

Let $(X,Y)=\displaystyle \left( \frac{u\cos v}{\sqrt{2+\sqrt{2}}}, \frac{u\sin v}{\sqrt{2-\sqrt{2}}} \right)$, then $dX \, dY=\displaystyle \frac{u\, du \, dv}{\sqrt{2}}$

Now $$\iint_{(2+\sqrt{2})X^2+(2-\sqrt{2})Y^2<1} e^{(2+\sqrt{2})X^2+(2-\sqrt{2})Y^2} dX \, dY =\int_{0}^{2\pi} \int_{0}^{1} \frac{e^{u^2}}{\sqrt{2}} u\, du \, dv$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.