# Math Help - Repeated Integral in x and y?

1. ## Repeated Integral in x and y?

$\iint xy dy dx$

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

What do I do with this please?

2. Originally Posted by TeaWithoutMussolini
$\\iint xy dy dx$
I don't know how to text in the limits but the interior are $\sqrt{x}$ and $x$, and the exterior are 0 and 2.

What do I do with this please?
$\int_{0}^{2} x\ dx \int_{\sqrt{x}}^{x} y\ dy$ (1)

... is the integral You want to compute?...

Kind regards

$\chi$ $\sigma$

3. Those are the values but no it is two integrals, then xy dy dx. Its all on one side.

4. All right!... the integral is...

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

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

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

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

Kind regards

$\chi$ $\sigma$

5. 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?

6. Yes... without any hassle!... You have to consider that the integral is...

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

... where...

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

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

Kind regards

$\chi$ $\sigma$

7. Product? I made the y integral $\frac{1}{2}x^2 - \frac{1}{2}x$. 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 $\frac{2}{3}$ but I am probably wrong.

8. All right!...

Kind regards

$\chi$ $\sigma$

9. 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!

10. Originally Posted by chisigma
Yes... without any hassle!... You have to consider that the integral is...

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

... where...

$g(x)=\int_{\sqrt{x}}^{x} y\ dy$ (2)
This is completely correct. However, what you wrote before:
$\int_0^2 x dx \int_{\sqrt{x}}^x y dy$
which would be likely to be interpreted as the product of two integrals:
$\left(\int_0^2 x dx\right)\left(\int_{\sqrt{x}}^x y dy\right)$
which is equal to $(1/2)(4- 0)(1/2)(x^2- x)= x^2- x$

A better notation would be
$\int_{x=0}^2 \left(\int_{y= \sqrt{x}}^x ydy\right) xdx$

11. A substantial detail: if the requested integral is...

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

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

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

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

Kind regards

$\chi$ $\sigma$

12. Originally Posted by TeaWithoutMussolini
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!
Forgive the tangent, but is this your usual technique? Is this average? If it is, I can't wait until September.

13. Originally Posted by TeaWithoutMussolini
$\iint xy dy dx$

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

What do I do with this please?
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.

\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*}