Double integrals over rectangles

The double integral $\displaystyle\iint_R f(x,y)\,dA$ is the signed volume between the rectangle $R:\ a \le x \le b,\ c \le y \le d$ and the surface $z = f(x,y)$. We have met it in two forms.

Its definition, as the limit of double Riemann sums over $R$ as the grid is refined:

And a way to compute it, as an iterated integral in either order:

That was stated, not explained. Why does integrating twice give the volume, whichever variable goes first?

Slicing the solid into slabs answers it.

The first view cuts the whole solid; the second shows a single slab.

The volume of all the slabs is the sum of the individual slab volumes. As the slabs thin they fill the solid, so this sum approaches the signed volume. It is a Riemann sum, and the limit of the sum is written as an integral:

So that integral is the signed volume. Informally, it adds up slab volumes $A(x)\,dx$ from $x = a$ to $x = b$, with the slabs now infinitely thin.

And since the area of each slab's face is $A(x) = \displaystyle\int_c^d f(x,y)\,dy$, it is the iterated integral written above.

Cutting across $y$ instead uses the cross-sections at fixed $y$, and the same argument gives the other order.

This is what makes a double integral computable: two ordinary integrals, evaluated one after the other with single-variable tools, in whichever order is easier. The two integral signs become two ordinary integrals, and $dA$ becomes $dy\,dx$.

Fubini's theorem turns a double integral over a rectangle into an iterated integral. Before anything is integrated, that iterated integral has to be written down, and the only information needed is the four limits of integration that $R$ supplies.

The picture shows where each limit comes from and which integral sign it belongs on, in either order of integration.

Each pair stays with its own variable: $a$ and $b$ sit on the integral written with $dx$, and $c$ and $d$ on the integral written with $dy$, whichever of the two is inner. By Fubini's theorem, for $f$ continuous on $R$, the order is a free choice:

Write $\displaystyle\iint_R \left(5 - xy\right)\,dA$ over $R:\,-1\le x\le 3,\ 2\le y\le 6$ as an iterated integral, in both orders.

Solution

The rectangle supplies two pairs of limits of integration: $-1$ and $3$ bound $x$, and $2$ and $6$ bound $y$.

Taking the order $dy\,dx$, we put the pair belonging to $y$ on the inner integral, the one written with $dy$, and leave the pair belonging to $x$ on the outer:

For the other order we move each pair to the other position, still keeping it with its own variable, so $-1$ and $3$ now sit on the inner integral and $2$ and $6$ on the outer:

Since $5 - xy$ is continuous on $R$, Fubini's theorem sets these two iterated integrals equal, and each of them gives the signed volume $\displaystyle\iint_R \left(5 - xy\right)\,dA$.

Once $\displaystyle\iint_R f(x,y)\,dA$ is written as an iterated integral, evaluating it is evaluating that iterated integral: inside out, one variable at a time.

Because the limits of integration are all constants, the inner integral always produces a function of one variable, and the outer integral evaluates that to a number. Either order gives the same number, so choose whichever inner integral is easier.

Evaluate $\displaystyle\iint_R \left(100 - 6x^2y\right)\,dA$ over $R:\,0\le x\le 2,\ -1\le y\le 1$.

Solution

Reading the limits of integration off $R$ gives $0$ and $2$ for $x$, and $-1$ and $1$ for $y$. We take the order $dy\,dx$, which puts the pair belonging to $y$ on the inner integral:

The inner integral is taken with $x$ held fixed, so $6x^2$ is a constant multiple of $y$ and the antiderivative of $y$ is $\dfrac{y^2}{2}$:

Substituting the limits $y = 1$ and $y = -1$:

The two $x^2$ terms cancel, so the inner integral is the constant $200$. That is the integrand of the outer integral, taken between $0$ and $2$:

So $\displaystyle\iint_R \left(100 - 6x^2y\right)\,dA = 400$.