Evaluate the double integral $\iint_\Omega (x^4 +y^2)dxdy$

Where $\Omega$ is a bounded region between $y=x^2$ and $y=x^3$

I have found by the points of intersection to be $(0,0)$ and $(1,1)$ making my limits then:

$\int_0^1 \int_0^1(x^4 +y^2)dydx$

followed it through and got 8/15 as my answer which I know is wrong. If someone could show me how to set up the double integral correctly that would be great, as I feel as if my limits are incorrect

  • 1
    $\begingroup$ The bounds for $y$ should be $x^3$ and $x^2$ : $\int_0^1\int_{x^3}^{x^2}(x^4+y^2)\,dydx$. $\endgroup$ – Guest Dec 8 '15 at 21:17

You are integrating over the following region: img http://puu.sh/lOlhY/3932ae7112.png
Let's take your order of integration as $\,dy\,dx$. In that case: $x\in[0, 1]$ and $y\in[x^3,x^2]$ This is because you imagine drawing a pillar from the bottom function to the top function. In this case, the bottom curve (black in the graph) is $y=x^3$ and the top curve is $y=x^2$ (orange). You can check this pretty easily by seeing that $x^3$ is always less than $y=x^2$ as long as $0<x<1$.
Your integral is now: $$\int_0^1\int_{x^3}^{x^2}(x^4+y^2)\,dy\,dx$$ The way you were doing it, you are integrating over the box $x\in[0, 1] \cup y\in[0, 1]$.

  • $\begingroup$ This is a great help to visualize it, Ill endeavour to sketch more in the future I think. $\endgroup$ – Clovers Dec 8 '15 at 21:34
  • $\begingroup$ My advice would be to always sketch the region, and do the pillar method going from the lower function to the upper function (or in the case of the $\,dx\,dy$ order, the from the left function to the right function). Cheers! $\endgroup$ – Eli Berkowitz Dec 9 '15 at 0:36


The region bounded by $$ 0\le x\le 1 \quad \land \quad 0\le y\le 1 $$ is a square.

The limits of your region are: $$ 0\le x\le 1 \quad \land \quad x^3\le y\le x^2 $$

can you do from this?

  • $\begingroup$ So, making my limits x^3 and x^2, integrating wrt to y first, I get 9/280? $\endgroup$ – Clovers Dec 8 '15 at 21:33
  • $\begingroup$ Evaluating the iterated integral you with these limits you get $\int_0^1\left(x^4y+\frac{y^3}{3}\right)\Big|_{y=x^3}^{y=x^2}\,dx$ $=\int_0^1\left(x^6+\frac{x^6}{3}-x^7-\frac{x^9}{3}\right)\,dx$ $\endgroup$ – Erik Olesen Dec 8 '15 at 22:09

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