Changing the order of integration

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November 19th 2008, 06:50 AM
thelostchild

Changing the order of integration

Hey I'm having some trouble with multiple integrals if anyone would care to explain (I think I'm just not using the correct region!)

Quote:

Evaluate
$\int_{x=0}^2\int_{y=\frac{x}{2}}^1 2xy^2 dy dx$

first directly then changing the order of integration

I did the first part and got 0.8 as the answer

just for the second I'm having trouble

I think the surface being integrated over the the triangle bounded by the lines y=x/2 x=0 and y=1 but I may be wrong which gives

$\int_{y=0}^1\int_{x=2y}^2 2xy^2 dx dy$

but the value of that doesnt agree, can anyone explain how to do the limits correctly?

cheers

Simon
November 19th 2008, 08:19 AM
Peritus

$<br /> <br /> \int\limits_0^1 {\int\limits_0^{x = 2y} {2xy^2 dxdy} } <br /> <br />$
November 19th 2008, 12:53 PM
Mathstud28

Quote:

Originally Posted by thelostchild

Hey I'm having some trouble with multiple integrals if anyone would care to explain (I think I'm just not using the correct region!)

I did the first part and got 0.8 as the answer

just for the second I'm having trouble

I think the surface being integrated over the the triangle bounded by the lines y=x/2 x=0 and y=1 but I may be wrong which gives

$\int_{y=0}^1\int_{x=2y}^2 2xy^2 dx dy$

but the value of that doesnt agree, can anyone explain how to do the limits correctly?

cheers

Simon

$0\leqslant{x}\leqslant{2}$ and $\frac{x}{2}\leqslant{y}\leqslant{1}\implies{x\leqs lant{2y}\leqslant{2}}$ . Stringing these two inequalites together gives

$0\leqslant{x}\leqslant{2y}\leqslant{2}$

From there Peritus's answer should be more apparent.
November 19th 2008, 01:27 PM
Krizalid

Quote:

Originally Posted by Mathstud28

$0\leqslant{x}\leqslant{2}$ and $\frac{x}{2}\leqslant{y}\leqslant{1}\implies{x\leqs lant{2y}\leqslant{2}}$ . Stringing these two inequalites together gives

$0\leqslant{x}\leqslant{2y}\leqslant{2}$

From there Peritus's answer should be more apparent.

But be careful 'cause, this method not always works. :)
November 19th 2008, 03:25 PM
Mathstud28

Quote:

Originally Posted by Krizalid

But be careful 'cause, this method not always works. :)

Really? I have never encountered a case where it hasn't. Would you mind giving me an example of one please?
November 19th 2008, 03:28 PM
Krizalid

Try it with $\int_0^1\int_{x^3}^{\sqrt[3]x}dy\,dx.$
November 19th 2008, 03:31 PM
Mathstud28

Quote:

Originally Posted by Krizalid

Try it with $\int_0^1\int_{x^3}^{\sqrt[3]x}dy\,dx.$

Thank you, I will report back later when I have had time to look at it.
November 20th 2008, 09:30 AM
Moo

Also be careful that reversing integration order is submitted to the condition that $\int \int \left|f(x,y)\right| ~ dx ~ dy$ is a finite value.
See Fubini's theorem - Wikipedia, the free encyclopedia for further information (advanced calculus imo)