# Double integration to find the area of a bounded region

• May 7th 2011, 09:00 PM
Glitch
Double integration to find the area of a bounded region
The question

Use double integration to find the area bounded by $y = x^3$ and $y = x^2$

My attempt
I know that finding the area of a bounded region is given by:

$\iint_\Omega {1 \ dx\ dy}$

So I drew the graph and decided to attempt it by taking two integrals:

$\int_0^x\int_{x^2}^{x^3} {1 \ dy\ dx}$ +
$\int_{-x}^0\int_{x^3}^{x^2} {1 \ dy\ dx}$

But upon solving these I get $\frac{-2x^3}{3}$ rather than the solution $\frac{1}{12}$

I'm fairly sure I'm attempting this wrong. :/

Any help would be most appreciated!
• May 7th 2011, 11:41 PM
HappyJoe
The reason why you get an answer that varies with x, rather than just an plain number, is that your limits on the outer integrals vary with x. The limits should be from 0 to 1 (and similar).
• May 8th 2011, 12:02 AM
Glitch
Quote:

Originally Posted by HappyJoe
The reason why you get an answer that varies with x, rather than just an plain number, is that your limits on the outer integrals vary with x. The limits should be from 0 to 1 (and similar).

Ahh I see. So how do I choose the limits? The functions are unbounded on the x-axis, so do I just choose an arbitrary bound? Thanks.
• May 8th 2011, 02:03 AM
Chris L T521
Quote:

Originally Posted by Glitch
Ahh I see. So how do I choose the limits? The functions are unbounded on the x-axis, so do I just choose an arbitrary bound? Thanks.

Pretend this isn't a double integral we're evaluating; instead, think of this as a problem from single variable calculus. Suppose we wanted to find the area between the curves $y=x^2$ and $y=x^3$. What would you do first? Find the points of intersection! Clearly, they are the points (0,0) an (1,1).

If you integrate with respect to x (treating it as a type I region), then the integral for area would be $\int_0^1 x^2-x^3\,dx$, BUT...as a double integral, this is just simply $\int_0^{1}\int_{x^3}^{x^2}1\,dy\,dx$.

Analogously, if you integrate with respect to y (treating it as a type II region), then the integral for area would be $\int_0^{1}\sqrt[3]{y}-\sqrt{y}\,dy$, BUT...as a double integral, this is just simply $\int_0^{1}\int_{\sqrt{y}}^{\sqrt[3]{y}}1\,dx\,dy$.

In this case, the first integral is easier to evaluate. I leave it for you to come up with the desired result.

I hope this makes sense.
• May 8th 2011, 02:13 AM
Glitch
Thanks Chris. For some reason I didn't think there was an intersection other than at the origin. >_< That'd explain my confusion! Thanks!