Reverse Integral and Evaluation: A Trick Question?

[NINHARDCOREFAN]

One question on it was reverse the following integral and then evaluate it:

$$\int_{0}^{\pi}\int_{x^2}^{\pi^2} x^3y dydx$$

So I got the following reversed integral:
$$\int_{0}^{\pi^2}\int_{-\sqrt{y}}^{\sqrt{y}} x^3y dxdy$$

When you first evaluate in respect to x you get 0, am I doing something wrong?

 

[snoble]

minor mistake really... the second integral should be
$$\int_0^{\pi^2}\int_0^{\sqrt{y}} x^3 y \partial x \partial y$$
I really doubt you would lose a lot of marks

unless I've made a mistake which is possible

 

[Theelectricchild]

Yeah, as always, drawing out your region really helps. This is an interesting part of calculus where you can actually visualize things... unlike many other math courses.

 

[NINHARDCOREFAN]

Damn it, I put -sqrt(y) because in the book, the same problem (with x^2 in the lower bound of the first integral) like it put -sqrt(y) in the lower bound for it's solution. I would get lot of points taken off... it's a 15 point problem, the professors are harsh, i would get taken off -10 or something like that

 

[NINHARDCOREFAN]

The book problem...

The book problem was:
$$\int_{-2}^{2}\int_{x^2}^{4} x^2y dydx$$

Their answer was:
$$\int_{0}^{4}\int_{-\sqrt{y}}^{\sqrt{y}} x^2y dydx$$

Why was there a -sqrt(y) in this problem?

 

[Theelectricchild]

The reason for that comes from the fact that part of your domain for the region extends into the dreaded negative x.

 

[Theelectricchild]

and might i ask, did you mean to interchange the differential elements in the second integrand?

And also, NIN rocks!

 

[NINHARDCOREFAN]

Yes, I meant to do that. Damn it! I should have changed there right then because the problem would have never come out to zero. Oh well. Thanks.