I must compute the double integral

$$\iint_D x^6y^6 dx dy$$


$$D = \left\{ (x,y) : x^{2}\le y\le x^{1/8} \right\}$$

Functions $x^2=x^{1/8}$ are going to be equal for $0$ and $1$. The region looks as follows.

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So, I have

$$\int_0^1 \left[ \cfrac{x^6y^7}{7} \right]_{x^2}^{x^{1/8}}$$

But that gives me

$$\left[ \cfrac{x^{55/8}-x^{20}}{7} \right]_{0}^{1} = 0$$

I am missing a big chunk of the theory. But, what?


2 Answers 2


So, let's recap what you've done so far. You've represented your double integral as an iterated integral in the following order: $$\iint_D x^6y^6\,dx\,dy=\int_0^1\int_{x^2}^{x^{1/8}} x^6y^6\,dy\,dx=\cdots$$ Then you evaluated the inside integral: $$\cdots=\int_0^1\left[\frac{x^6y^7}{7}\right]_{y=x^2}^{y=x^{1/8}}\,dx=\int_0^1\left[\frac{x^{55/8}-x^{20}}{7}\right]\,dx.$$ So far everything is good! But then your mistake is that it's too early to plug in the limits of integration $0$ and $1$ for $x$ into the expression in the brackets as you did. You must integrate the remaining integral first (i.e. find its antiderivative), and only then do you plug in the numbers.

  • $\begingroup$ I'm sorry I do not understand. $\endgroup$
    – Dovendyr
    Jul 3, 2018 at 8:25
  • $\begingroup$ What do you mean by ''You must integrate the remaining integral first (i.e. find its antiderivative)''? Is there a hidden derivative in $\frac{x^n}{n}$? $\endgroup$
    – Dovendyr
    Jul 3, 2018 at 12:39
  • $\begingroup$ @Dovendyr: A double integral requires integrating twice. You have only integrated once so far. So you do need to do integrate one more time. $\endgroup$
    – zipirovich
    Jul 3, 2018 at 13:26
  • $\begingroup$ But that's what I do, and then I have: $$\int_0^1\left[\frac{x^{55/8}-x^{20}}{7}\right]\,dx$$ = $$\left[\frac{1^{55/8}-1^{20}}{7}\right]$$ $\endgroup$
    – Dovendyr
    Jul 3, 2018 at 15:13
  • 1
    $\begingroup$ @Dovendyr: No, that's not true. Here's a different example. Suppose you're given the following question: $\int_0^1 x^2\,dx=?$ What is the correct solution: "$\int_0^1 x^2\,dx=1^2-0^1$" or "$\int_0^1 x^2\,dx=\left.\frac{x^3}{3}\right|_0^1=\frac{1^3}{3}-\frac{0^3}{3}$"? $\endgroup$
    – zipirovich
    Jul 3, 2018 at 16:10

Edit 1: zipirovich's answer is considerably simpler, since you can just switch the order of integration like that, but if you feel like going through a pointless exercise in re-expressing the bounds of an integral, read on!

With that order of integration and those bounds, the integral you would need to evaluate is $$\int_{y=x^2}^{x^{1/8}} \int_{x=0}^1 x^6y^6 dx dy,$$ but that makes no sense, since the functions of $x$ in the outermost integral mean your final answer would be a function of $x$ instead of a number. Since it's a number you're after, that clearly won't do. You're going to have to re-express the boundaries as functions of $y$ and then do the integral like so: $$\int_{y=0}^1 \int_{x=y^8}^{\sqrt{y}} x^6y^6dxdy$$ Can you take it from here?

Edit 2: We do the inside integral first, to get $$\int_0^1\left[\frac{x^7y^6}{7}\right]_{y^8}^{\sqrt{y}}dy$$ Plugging in the bounds on the integral we just did, we end up with $$\int_0^1\left[\frac{\left(y^{1/2}\right)^7\left(y^6\right)-\left(y^8\right)^7\left(y^6\right)}{7}\right]dy=\int_0^1\left[\frac{\left(y^{7/2}\right)\left(y^6\right)-\left(y^{56}\right)\left(y^6\right)}{7}\right]dy$$ $$=\int_0^1\frac{y^{19/2}-y^{62}}{7}dy$$ This is a much simpler integral we're left with, and evaluating this one to arrive at the final answer is what zipirovich was alluding to. All you have to do from here is find the antiderivative of $\frac{y^{19/2}-y^{62}}{7}$ and plug in $0$ and $1$ in the right places, and you'll be all done.

  • $\begingroup$ It's interesting and I tried, and I come to $\dfrac {y^{19/2}-y^{64}}{7}$. But since I don't understand the part ''You must integrate the remaining integral first (i.e. find its antiderivative)'' in @zipirovich answer I can't go further. $\endgroup$
    – Dovendyr
    Jul 3, 2018 at 8:29
  • $\begingroup$ @Dovendyr Did my edit clear things up? $\endgroup$ Jul 3, 2018 at 15:33
  • $\begingroup$ Yes, thank you! $\endgroup$
    – Dovendyr
    Jul 5, 2018 at 6:04

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