$\displaystyle \iint xy dy dx$
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?
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
$\displaystyle \chi$ $\displaystyle \sigma$
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
$\displaystyle \chi$ $\displaystyle \sigma$
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
$\displaystyle \int_{x=0}^2 \left(\int_{y= \sqrt{x}}^x ydy\right) xdx$
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...
... 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
$\displaystyle \chi$ $\displaystyle \sigma$
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*} $