0
$\begingroup$

I am having trouble discerning the difference between a double integral and an iterated integral. I have seen alternate notation to the effect of: \begin{align*} \iint\limits_R f(x,y) \ \ dR = \int_a^b \int_c^d f(x,y) \ \ dx \ dy = \int_a^b \left(\int_y^{y^2} f(x,y) \ dx\right) \ dy \end{align*} My question is: what is the difference between these two types of integrals, and do the above integrals I've written represent one or both of these?

Thanks.

$\endgroup$
  • 1
    $\begingroup$ a double integral is a more compact way of describing an integral. $R$ describes the region being integrated over, and $dA$ = $dxdy$ $\endgroup$ – Aniruddh Venkatesan Apr 5 at 1:01
1
$\begingroup$

The double integral refers to the actual abstract operation that takes in the function $f$ and the region $R$ and returns the integral as its output. In this case, $R$ can be any region in the plane that you may want to integrate over.

The iterated integral, on the other hand, is literally an integral inside of another integral, computed as one integral, and then the integral of the integral. You see it has constant bounds for $x$ and $y$, actual numbers for which the answer can be expressly calculated.

The connection between them is that in order to computer a double integral, it must be first transformed into an iterated integral, by writing the region $R$ as some $x$ and $y$ bounds.

$\endgroup$
1
$\begingroup$

If a, b, c, d are all constants, then the region, R is a rectangle $R=[c,d]\times[a,b]$, so you would be integrating $f(x,y)$ over this rectangle.

The region described for the right side integral is a type 2 region where $c=h_1(y)$ and $d=h_2(y)$, in other words x is a function of y and a and b are horizontal lines satisfying $y=a$ and $y=b$. So the region is whatever is enclosed by the two horizontal lines and the two curves.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.