Question

Evaluate the double integral of (e^-(x^2+(y-x)^2+y^2))dA with the bounds -inf 11 years ago

Answer 1

To clarify: the integral is
\[\large\int_{-\infty}^\infty\int_{-\infty}^\infty \exp\bigg[-\left(x^2+(y-x)^2+y^2\right)\bigg]~dx~dy~~?\]

Answer 2

You must be away or something. I'll assume the integral is as described above.

The region \(x\in(-\infty,\infty),y\in(-\infty,\infty)\) is the entire \(x-y\) plane, so when transforming to polar coordinates, you have \(r\in(0,\infty)\) and \(\theta\in(0,2\pi)\).

In polar coordinates, you have
\[\begin{cases}x=r\cos\theta\\y=r\sin\theta\\dx~dy=r~dr~d\theta\end{cases}\]
So the integral itself, post-transformation, is equivalent to
\[\large\int_0^{2\pi}\int_0^\infty\exp\bigg[-\left(r^2+(r\sin\theta-r\cos\theta)^2\right)\bigg]~r~dr~d\theta\\
\large\int_0^{2\pi}\int_0^\infty\exp\bigg[-r^2-r^2(\sin\theta-\cos\theta)^2\bigg]~r~dr~d\theta\]
Can you take it from here?

Answer 3

Yeah, thanks. For some reason, the professor recommended we utilize a jacobian, but I just don't see how it's practical with this problem. I guess I'll find out later. Again, thanks for the help.

Answer 4

Well the Jacobian is still used here.
\[f(r,\theta)=r\cos\theta\\
g(r,\theta)=r\sin\theta\]
\[J=\begin{vmatrix}\frac{\partial f}{\partial r}&\frac{\partial f}{\partial\theta}\\
\frac{\partial g}{\partial r}&\frac{\partial g}{\partial\theta}\end{vmatrix}=\begin{vmatrix}\cos\theta&-r\sin\theta\\
\sin\theta&r\cos\theta\end{vmatrix}=r\cos^2\theta+r\sin^2\theta=r\]
Generally, the Jacobian is used in any change of variables, and for the polar conversion you get:
\[\int\int_Rf(x,y)~dx~dy~~\iff~~\int\int_Sf(r,\theta)J~dr~d\theta\]
where \(J=r\).