Calculate the following double integral

[Hannisch]

Homework Statement

Calculate the double integral:

\iint\limits_D x^{5}y^{6}dxdy
where D = {(x,y): x⁹ ≤ y ≤ x^1/9}

Homework Equations

The Attempt at a Solution

I didn't think this problem would be too hard, but it seems I'm really not good with double integrals.

Anyway, I first tried to find the right interval. The curve y=x⁹ is below y=x^1/9 in two intervals, first of all 0 ≤ x ≤ 1 and -∞ ≤ x ≤ -1. I was wondering about this for some time, but then I thought that I probably needed an enclosed D, so I discarded the second interval.

When x=0, y=0, and when x=1, y=1 in both curves.

\iint\limits_D x^{5}y^{6}dxdy = \int ^{1}_{0} x^5 dx \int ^{1}_{0} y^6 dy

I tried that, because that was something they explained to us during the lecture and well.. it seemed to apply. And those two integrals are really very easy:

\int ^{1}_{0} x^5 dx \int ^{1}_{0} y^6 dy = \left[ \frac{1}{6} x^6 \right] ^{1}_{0} \left[ \frac{1}{7} y^7 \right]^{1}_{0} = ( \frac{1}{6})( \frac{1}{7}) = \frac{1}{42}

And this is wrong.

Help?

 

[Mark44]

The two curves intersect at (-1, -1), (0, 0), and (1, 1). When 0 <= x <= 1, x^(1/9) >= x^9, and when -1 <= x <= 0, x^9 >= x^(1/9). So that means you'll need to split your double integral into a pair of iterated integrals, one for each of the two intervals.
Hope that helps.

 

[flatmaster]

You must integrate over y first because the limits of the y integral contain x values. After the y integral, you do the x integral. You should end up with an expression with only x's.

 

[Mark44]

Also, I don't think there's a justification for splitting up the iterated integrals as you did. Flatmaster's advice of integrating with respect to y first, and then with respect to x, is good.

 

[Hannisch]

Oh thank you! Talk about change in thinking.

And anyway, I didn't use that second interval, because it didn't fit with the area that I was given, since it stated that x^(1/9) needed to be >= x^9. It gave me the right answer, so I'm very happy now!