Thread: Evaluate the following double integrals

1. Evaluate the following double integrals

I swear, our Prof. must get shot at at least once a year for some of the crazy assignments he gives out. At least one student per year in his class must go insane. I digress, however.

These two double integrals (the material for which I'm trying to catch up to) are a lot more difficult than they look (or I made a mistake somewhere). I have the final answers due to an online website, but it doesn't give processes for definite integrals (which I'm trying to work out right now).

a) $\int_0^4 \int_{\sqrt{y}}^2 \frac{1}{1+x^3} dxdy$
Ans. $\approx 0.732408$

b) $\int_0^1 \int_x^1 \frac{x}{1+y^6} dydx$
Ans. $\approx 0.1309$

I'll put up work on this a bit later when I have more, but any help that could be provided would be appreciated.

2. Originally Posted by Runty
I swear, our Prof. must get shot at at least once a year for some of the crazy assignments he gives out. At least one student per year in his class must go insane. I digress, however.

These two double integrals (the material for which I'm trying to catch up to) are a lot more difficult than they look (or I made a mistake somewhere). I have the final answers due to an online website, but it doesn't give processes for definite integrals (which I'm trying to work out right now).

a) $\int_0^4 \int_{\sqrt{y}}^2 \frac{1}{1+x^3} dxdy$
Ans. $\approx 0.732408$

b) $\int_0^1 \int_x^1 \frac{x}{1+y^6} dydx$
Ans. $\approx 0.1309$

I'll put up work on this a bit later when I have more, but any help that could be provided would be appreciated.
The trick with these is to reverse the order of integration. Start by drawing a diagram of the region over which the integral occurs. For (a), the region described by $\sqrt y\leqslant x\leqslant 2,\ 0\leqslant y\leqslant 4$ is the region under the parabola $y=x^2$ between x=0 and x=2. So it could also be described as the region $0\leqslant y \leqslant x^2,\ 0\leqslant x\leqslant 2$.

That means that when you switch the order of integration, the integral becomes $\displaystyle\int_0^2\!\!\int_0^{x^2}\!\!\!\frac1{ 1+x^3}\,dydx$, which you should find much easier to do. (I make it $\frac13\ln9$.)

I haven't looked at (b), but I think that it will almost certainly respond to the same technique.

3. Wow, and I was doing that one the hard way. That just shows how out of this sort of thing I am. I definitely picked the wrong degree; I'm better at number crunching and data entry, not critical thinking.

EDIT: Would the region for b) be as follows for reversing the order of integration?
$0\leq x\leq y,\ 0\leq y\leq 1$

4. Originally Posted by Runty
EDIT: Would the region for b) be as follows for reversing the order of integration?
$0\leq x\leq y,\ 0\leq y\leq 1$
Correct. In problems such as these, I always find it helpful to draw the region enclosed.