Double integrals and area?

Question

What does $\displaystyle \iint_U \, dx \,dy$ describe, where $U$, say, is some region in, say, the $(x,y)$ plane?

Is it the net area of $U$ in the $(x,y)$ plane?

How about $\displaystyle \iint_U f \, dx \,dy$, for some (appropriate?) $f$. Is the net area enclosed by $U$ in $f$?

I'm confused by so many different types of integrals in multivariable calculus. Thanks.

Answer 1

Yes and yes.

For a subset $U$ of $\Bbb R^2$ we have $$\text{Area of $U$}=\iint_U dA$$ where $dA=dxdy$. This is because you're just integrating at height $1$ above all of $U$, so you get $1$ times the area of $U$.

Now if we throw in an integrable function $f:U\to\Bbb R$ then $$\text{Area below $f$ on $U$}=\iint_U f ~dA$$ This is because you're integrating over $U$ at a height of $f(x,y)$, which changes with $x$ and $y$.

Note: I use "area" under $f$, but this is really a 3D volume.

Answer 2

Just as a definite integral of the form $\int f dx$ is an area under $y=f(x)$, $\int f dx dy$ is a volume under $z=f(x,\,y)$. The special case $f=1$ gives your first integral.

Answer 3

In fact, integration and Measure Theory are quite synonymous, which can be explained in very simple terms. We can say that the measure of a set can be calculated as follows: $$ \mu(A) = \int_A 1 dx$$

For instance, suppose we wanted to find the distance between two locations $a$ and $b$, we can represent this as $A = [a,b]$ then we would expect the measure (distance) of this set to be its length, $b-a$.
$$\mu(A)= \int_{a}^{b} 1~dx$$ $$~~~~~~~ = b - a$$ Now say a rectangle has one side with length $a$ and another of length $b$. We would expect the area to be $ab$. Note from the example before, the measure of $[a,b]$ is some $x\in \mathbb{R}$. So why not say one side of the rectangle is just some set $[c,d]$ and the other is another set $[e,f]$?

Then if $B= [c,d]\times[e,f]$, the area of the rectangle is \begin{align*} \mu(B) &= \int_{c}^{d} \int_{e}^{f} 1~dxdy\\ &= (d-c)\times(f-e) \\ \end{align*}

Hopefully this gives you an intuition of integration from another perspective? To get the area of the rectangle we are finding the measure (integral) in the $x$-plane first, and then in the $y$-plane second, which covers all of the set. Note triple integrals are used to calculate the volume of a $3$D object and so on.