Double integrals as volume or area

Double integrals as volume

You are probably familiar that in one-variable calculus, the integral ∫ abf(x)dx for positive f(x) can be interpreted as the area under the curve f(x) over the interval [a,b].

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The integral is the area between the curve f(x) and the x-axis.

In the same way, the double integral ∫∫Df(x,y)dA of positive f(x,y) can be interpreted as the volume under the surface z = f(x,y) over the region D. Imagine that the blue object below is the surface z = f(x,y) floating above the xy-plane. The double integral ∫∫Df(x,y)dA can be interpreted as the volume between the surface z = f(x,y) and the xy-plane, i.e, the ”cylinder” above the region D.

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You can also see this from the Riemann sum approximating the integral

∑ i,jf(xij,yij)ΔxΔy
Each term in the Riemann sum is the volume of a thin box with base Δx × Δy and height f(xij,yij).

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Hence, the total Riemann sum approximates the volume under the surface by the volume of a bunch of these thin boxes. In the limit as Δx, Δy → 0, we obtain the total volume under the surface over the region D, i.e., ∫∫Df(x,y)dA.

If f(x,y) > g(x,y) can you see how the double integral

∫∫D(f(x,y) - g(x,y))dA
is the volume between the surface z = f(x,y) and the surface z = g(x,y)?

Double integrals as area

In one variable calculus, what happens if you integrate the function f(x) = 1 over the interval [a,b]? You can calculate that

∫ abf(x)dx = ∫ ab1dx = x| ab = b - a.
The integral of the function f(x) = 1 is just the length of the interval [a,b]. It also happens to be the area of the rectangle of height 1 and length (b-a), but we can interpret it as the length of the interval [a,b].

We can do the same trick for double integrals. We can integrate the function f(x,y) = 1 over the region D. We could interpret the integral as the volume under the function f(x,y) = 1 over D. But the integral of f(x,y) = 1 is also the area of the region D. This can be a nifty way of calculating the area of the region D. If we let A be the area of the region D, we can write this as

A = ∫ ∫DdA.

Click here for an example of using the double integral to compute area.