# Just got done with a test

1. Apr 13, 2005

### 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?

2. Apr 13, 2005

### 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

3. Apr 13, 2005

### 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.

4. Apr 13, 2005

### 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

5. Apr 13, 2005

### 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?

6. Apr 13, 2005

### Theelectricchild

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

7. Apr 13, 2005

### Theelectricchild

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

And also, NIN rocks!!!

Last edited: Apr 13, 2005
8. Apr 13, 2005

### 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.

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