$\displaystyle \iint xy dy dx$

I don't know how to text in the limits but the interior are $\displaystyle \sqrt{x} $ and $\displaystyle x$, and the exterior are 0 and 2.

What do I do with this please?

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- Jun 9th 2011, 08:26 PMTeaWithoutMussoliniRepeated Integral in x and y?
$\displaystyle \iint xy dy dx$

I don't know how to text in the limits but the interior are $\displaystyle \sqrt{x} $ and $\displaystyle x$, and the exterior are 0 and 2.

What do I do with this please? - Jun 9th 2011, 08:35 PMchisigma
- Jun 9th 2011, 08:36 PMTeaWithoutMussolini
Those are the values but no it is two integrals, then xy dy dx. Its all on one side.

- Jun 9th 2011, 08:51 PMchisigma
All right!... the integral is...

$\displaystyle \int_{0}^{2} \int_{\sqrt{x}}^{x} x\ y\ dx\ dy$ (1)

... and first step is to separate the variables splitting the (1) into two integrals...

$\displaystyle \int_{0}^{2} x\ dx\ \int_{\sqrt{x}}^{x} y\ dy$ (2)

... and at this point You don't have difficulties to proceed...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jun 9th 2011, 08:53 PMTeaWithoutMussolini
Can you just DO that? Without any hassle? Just move them out? Does it then matter what order you take them in?

I still don't get it? You will get a number, from the first one, you will get a term in x, what do you do with it in the second? - Jun 9th 2011, 09:02 PMchisigma
Yes... without any hassle!... You have to consider that the integral is...

$\displaystyle \int_{0}^{2} x\ g(x)\ dx$ (1)

... where...

$\displaystyle g(x)=\int_{\sqrt{x}}^{x} y\ dy$ (2)

... so that the order of integration is : first integrate in y and after in x...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jun 9th 2011, 09:08 PMTeaWithoutMussolini
Product? I made the y integral $\displaystyle \frac{1}{2}x^2 - \frac{1}{2}x$. Is this ok? Do I then multiply both of those by x, integrate wrt x and solve between 0 and 2?

Well I make the answer $\displaystyle \frac{2}{3}$ but I am probably wrong. - Jun 9th 2011, 09:18 PMchisigma
All right!...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jun 9th 2011, 09:25 PMTeaWithoutMussolini
Its right isn't it!

Well I have an exam in 3 hours and I just learned the syllabus in one night. Let's see if I can pass this! - Jun 10th 2011, 03:59 AMHallsofIvy
This is completely correct. However, what you wrote before:

$\displaystyle \int_0^2 x dx \int_{\sqrt{x}}^x y dy$

which would be likely to be interpreted as the product of two integrals:

$\displaystyle \left(\int_0^2 x dx\right)\left(\int_{\sqrt{x}}^x y dy\right)$

which is equal to $\displaystyle (1/2)(4- 0)(1/2)(x^2- x)= x^2- x$

A better notation would be

$\displaystyle \int_{x=0}^2 \left(\int_{y= \sqrt{x}}^x ydy\right) xdx$ - Jun 10th 2011, 05:13 PMchisigma
A substantial detail: if the requested integral is...

$\displaystyle \int \int_{A} x\ y\ dx\ dy$ (1)

... where A is the 'red area' in the figure...

http://digilander.libero.it/luposabatini/MHF117.bmp

... then the integral must be splitted in two as...

$\displaystyle \int \int_{A} x\ y\ dx\ dy = \int_{0}^{1} x\ dx (\int_{x}^{\sqrt{x}} y\ dy) + \int_{1}^{2} x\ dx (\int_{\sqrt{x}}^{x} y\ dy) = \frac{3}{4}$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jun 10th 2011, 05:39 PMQuacky
- Jun 10th 2011, 10:31 PMProve It
While what ChiSigma said is correct, it's only valid to separate the variables if the variables can be separated, as they can be in this case. A better way in my opinion, is to perform the innermost integration of the ENTIRE function, then the outer integration of the entire function. This will work in ALL cases.

So in your case...

$\displaystyle \displaystyle \begin{align*} \int_0^2{\int_{\sqrt{x}}^x{x\,y\,dy\,dx}} &= \int_0^2{\left(\int_{\sqrt{x}}^x{x\,y\,dy}\right) \,dx} \\ &= \int_0^2{\left[\frac{x\,y^2}{2}\right]_{\sqrt{x}}^x\,dx} \\ &= \int_0^2{\left[\frac{x(x)^2}{2} - \frac{x(\sqrt{x})^2}{2}\right]\,dx} \\ &= \int_0^2{\frac{x^3}{2} - \frac{x^2}{2}\,dx} \\ &= \left[\frac{x^4}{8}- \frac{x^3}{6}\right]_0^2 \\ &= \left(\frac{2^4}{8} - \frac{2^3}{6}\right) - \left(\frac{0^4}{8} - \frac{0^3}{6}\right) \\ &= \frac{2}{3}\end{align*} $