Double Integrals as Volume
Math Insight
You are probably familiar that in one-variable calculus, the integral ∫baf(x)dx for
positive f(x) can be interpreted as the area under the curve f(x) over the
interval [a,b].
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.
You can also see this from the Riemann sum approximating the integral
Each term in the Riemann sum is the volume of a thin box with base Δx×Δy and
height f(xij,yij).
Source URL: http://mathinsight.org/double_integral_volume
Saylor URL: http://www.saylor.org/courses/ma103/
Attributed to: [Duane Q. Nykamp]
www.saylor.org
Page 1 of 5
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.
The below image illustrates how the volume of the boxes approaches the area
under the surface as Δx and Δy decrease to zero.
Double integral Riemann sum. The volume of the small boxes illustrates a Riemann sum
approximating the volume under the graph of z=f(x,y) over the region D, i.e., the double
integral ∬DfdA for f(x,y)=cos2x+sin2y and D defined by 0≤x≤2 and 0≤y≤1. The volume of the
boxes is
where xi is the midpoint of the ith interval along the x-axis and yj is the midpoint of the jth interval
along the y-axis. The purple line of the cyan slider shows the volume estimated by the volume of
the boxes, and the blue line of the cyan slider shows the actual volume underneath the surface.
As Δx and Δy approach zero, the purple line approaches the blue line, illustrating how the
estimated volume approaches the actual volume.
If f(x,y)>g(x,y) can you see how the double integral
is the volume between the surface z=f(x,y) and the surface z=g(x,y)?
Source URL: http://mathinsight.org/double_integral_volume
Saylor URL: http://www.saylor.org/courses/ma103/
Attributed to: [Duane Q. Nykamp]
www.saylor.org
Page 2 of 5
The below examples assume you know how to write double integrals as interated
integrals1.
Example 1
Calculate the volume under the surface z=3+x2−2y over the region D defined
by 0≤x≤1 and −x≤y≤x.
Solution: The volume V is the double integral of 3+x2−2y over D.
Example 2
Calculate the volume between the surfaces z=f(x,y)=3−x2−y2 and the
surface z=g(x,y)=2x2+2y2.
Solution: At the origin, f is above g, since f(0,0)=3 and g(0,0)=0. As one moves
away from the origin, f gets large and negative, while g gets large and positive.
Therefore, f and g must cross somewhere, and the finite volume
between f and g is the integral of f−g over the region where f>g.
We can calculate where the surfaces cross by setting f=g. This occurs when
The curve where the surfaces cross is the unit circle, and the region D over
which we must integrate f−g is the unit disk x2+y2≤1.
The volume V is
Source URL: http://mathinsight.org/double_integral_volume
Saylor URL: http://www.saylor.org/courses/ma103/
Attributed to: [Duane Q. Nykamp]
www.saylor.org
Page 3 of 5
This integral is very simple to calculate if you know how to change
variables2 to polar coordinates3. (If you don't yet know how to do this, you can
still calculate the integral if you are good at doing integrals, but it gets pretty ugly.
We'll show the procedure below.)
In polar coordinates, the unit disk is the region 0≤r≤1, 0≤θ≤2π. Including the extra
factor of r from the change of variables, the integral becomes
If you don't know how to change variables to polar coordinates, you can do the
integration over disk by writing it as −1≤x≤1,
. As you can
see below, you'll have to do a trigonometric substitution x=sinθ as well as other
trigonometric identities to evaluate the integral. With this trigonometric
substitution, the range of θ is −π/2≤≤π/2 so that x=sinθ ranges from −1≤x≤1.
Rather than explain every step, we'll just show some of the intermediate
calculations. You won't want to try to follow this anyway unless you are already
skilled at integration.
Here's the calculation:
Source URL: http://mathinsight.org/double_integral_volume
Saylor URL: http://www.saylor.org/courses/ma103/
Attributed to: [Duane Q. Nykamp]
www.saylor.org
Page 4 of 5
Notes and Links:
1. http://mathinsight.org/double_integral_iterated
2. http://mathinsight.org/double_integral_change_variables_introduction
3. http://mathinsight.org/polar_coordinates
Source URL: http://mathinsight.org/double_integral_volume
Saylor URL: http://www.saylor.org/courses/ma103/
Attributed to: [Duane Q. Nykamp]
www.saylor.org
Page 5 of 5