Evaluating Double Integrals

• Apr 15th 2008, 04:22 AM
Warrick2236
Evaluating Double Integrals
Calculate the double integrals:

I)
IntInt R (x-y^2)dA

where R is the region bounded by the curves y = x^2 and y = x^3

II)
IntInt D x^3 sin(y^3)dydx

where D is the region bounded by the parabola y = x^2 and the two straight lines x = 0 and y = 1

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I will be forever grateful for all the help I can get.
• Apr 15th 2008, 04:52 AM
mr fantastic
Quote:

Originally Posted by Warrick2236
Calculate the double integrals:

I)
IntInt R (x-y^2)dA

where R is the region bounded by the curves y = x^2 and y = x^3

II)
IntInt D x^3 sin(y^3)dydx

where D is the region bounded by the parabola y = x^2 and the two straight lines x = 0 and y = 1

-------------

I will be forever grateful for all the help I can get.

Have you drawn the region of integration in each case?

I) $= \int_{x = 0}^{x = 1} \int_{y = x^3}^{y = x^2} x - y^2 \, dy \, dx$.

II) $= \int_{x = 0}^{x = 1} x^3 \left( \int_{y = x^2}^{y = 1} \sin(y^3) \, dy \right) \, dx$.
• Apr 15th 2008, 04:56 AM
mr fantastic
Quote:

Originally Posted by mr fantastic
[snip]
II) $= \int_{x = 0}^{x = 1} x^3 \left( \int_{y = x^2}^{y = 1} \sin(y^3) \, dy \right) \, dx$.

But since the y-integral is a tad tricky you might struggle with this order of integration (Rofl)

So I'll let you consider how to reverse the order of integration ..... (You're again advised to first draw the region of integration).
• Apr 15th 2008, 09:10 AM
Krizalid
Quote:

Originally Posted by mr fantastic
(You're again advised to first draw the region of integration).

Well actually one can even play with inequalities:

In the $dy\,dx$ order our region is bounded by $0\le x\le1,\,x^2\le y\le1,$ so $0\le x^2\le1$ and from here $0\le x^2\le y\le1$ and the limits for $dx\,dy$ order will be $0\le y\le1,\,0\le x\le\sqrt y.$

Of course, this trick not always works, or may be hard sometimes if you only want to play with inequalities. In fact is useful for quick stuff.
• Apr 16th 2008, 07:55 AM
Warrick2236
Thanks for showing the regions of the integration, but im still having some troubles solving both integration. And im not even sure if I get the correct answer.

Any help would be highly appreciated!
• Apr 16th 2008, 01:46 PM
mr fantastic
Quote:

Originally Posted by Warrick2236
Thanks for showing the regions of the integration, but im still having some troubles solving both integration. And im not even sure if I get the correct answer.

Any help would be highly appreciated!

The hard work has been done. All you have to do is integrate.

It would be best if you showed your working so that additional help can be targeted to your exact troubles.