0
$\begingroup$

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

$\endgroup$
  • 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
2
$\begingroup$

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 $x\in[0, 1] \cup y\in[0, 1]$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

Hint:

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.