I have made an attempt on a problem from an old exam.
I'm not sure if my method is correct or not, as it differs from the teacher's solution and I'm unsure of the theory. Does my solution lack any important
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
As I understand it, your solution is not correct.
The reason is that you cannot write your double integral as a product of 2 simple integrals as you did, since the integral with respect to $x$ depends on $y$. When you say "no problem" for this integral (with respect to $x$), you are forgetting that it depends on $x$, so that you cannot simply ignore it when looking at the convergence of your double integral.
What you can do (and I presume what your teacher does), is first compute the integral with respect to $x$ (this will give you something that depends on $y$), and then integrate what you get with respect to $y$.
Edit. As suggested by David, you can also integrate first with respect to $y$ and then with respect to $x$.
-
-
$\begingroup$ @DavidH If you mean "much easier to integrate first with respect to $x$", I agree; except for the "much". $\endgroup$– EtienneApr 19, 2014 at 10:28
-
$\begingroup$ but the teacher seems to change the region info $[1,\infty)^2$ which has double the area of the one I drew in my sketch. $\endgroup$– jacobApr 20, 2014 at 8:12
-
$\begingroup$ @Etienne oops, you're right, my comment didn't make much sense. I must have been tired. I meant to say it's easier to integrate wrt y first. $\endgroup$– David HApr 20, 2014 at 20:31