In single variable calculus, you learned how to calculate area between curves and volumes of revolution. Now that you know how to evaluate double integrals, you can use that knowledge to calculate the same areas and also volumes of more irregular shapes that lie above a plane, usually the xyplane.
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Actually, you already know how to calculate area and volume with a double integral that is already set up. You can think of the limits of integration as defining the area in the plane. To find that area, the integrand is just one. For volume, the integrand is the height of the volume above that area. So on this page, we show you how to set up the double integral to calculate an area and a volume.
In your classes and in many of the videos you will watch, the instructor will often generate a 3D plot of the volume you are trying to calculate. These are interesting and can give you some perspective on what is going on but we have found them unnecessary when actually setting up the integral. The only plotting you really need to do is 2D in the base plane, i.e. the plane where the shadow of the volume falls. This is easily done with Geogebra or other graphing utilities and can usually also be done by hand. This 2D plot is a necessary part of setting up your double integral since you use it not only to determine your limits of integration but also in what order you integrate.
Whether you are calculating an area or a volume, you start out by describing the area in the plane. Just like you did when calculating volumes of revolution using either the washer or cylinder method, you need to plot the region in the base plane and describe it mathematically. This panel explains how to do that.
To describe an area in the xyplane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot (used to plot these graphs; we used gimp to add labels and other graphics). However, graphing by hand is usually the best and quickest way.
We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by \(f(x)\) (red line), \(g(x)\) (blue line) and \(x=a\) (black line).
[Remember that an equation like \(x=a\) can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]
Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).
Vertically
Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.
Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing \(g(x)\), no matter where we draw it. Similarly, the arrow always exits the area by crossing \(f(x)\), no matter where we draw it. Do you see that?
But wait, how far to the right does it go? We are not given that information. What we need to do is find the xvalue where the functions \(f(x)\) and \(g(x)\) intersect. You should be able to do that. We will call that point \((b,f(b))\). Also, we will call the left boundary \(x=a\). So now we have everything we need to describe this area. We give the final results below.
Vertical Arrow  

\( g(x) \leq y \leq f(x) \) 
arrow leaves through \(f(x)\) and enters through \(g(x)\) 
\( a \leq x \leq b \) 
arrow sweeps from left (\(x=a\)) to right (\(x=b\)) 
Horizontally
We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of \(f(x)\) and \(g(x)\) in terms of \(y\). ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations \(f(x) \to F(y)\) and \(g(x) \to G(y)\).
Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line \(x=a\). However, there is something strange going on at the point \((b,f(b))\). Notice that when the arrow is below \(f(b)\), the arrow exits through \(g(x)\) but when the arrow is above \(f(b)\), the arrow exits through \(f(x)\). This is a problem. To overcome this, we need to break the area into two parts at \(f(b)\).
Lower Section   This section is described by the arrow leaving through \(g(x)\). So the arrow sweeps from \(g(a)\) to \(g(b)\).
Upper Section   This section is described by the arrow leaving through \(f(x)\). The arrow sweeps from \(f(b)\) to \(f(a)\).
The total area is the combination of these two areas. The results are summarized below.
Horizontal Arrow  

lower section  
\( a \leq x \leq G(y) \) 
arrow leaves through \(G(y)\) and enters through \(x=a\) 
\( g(a) \leq y \leq g(b) \) 
arrow sweeps from bottom (\(y=g(a)\)) to top (\(y=g(b)\)) 
upper section  
\( a \leq x \leq F(y) \) 
arrow leaves through \(F(y)\) and enters through \(x=a\) 
\( f(b) \leq y \leq f(a) \) 
arrow sweeps from bottom (\(y=f(b)\)) to top (\(y=f(a)\)) 
Type 1 and Type 2 Regions
Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.
Here is a quick video clip going into more detail on Type 1 and Type 2 regions.
video by Krista King Math 

[For some videos and practice problems dedicated to this topic, check out this page.]
Okay, so now that we have a description of the area in the plane, we need to set up the integral to calculate either the area or the volume. From the above panel, we know the limits of integration. The order of integration is as follows.
Outside Integral   The outside integral is the one with the constant limits of integration and the sweeping of the arrow.
Inside Integral   The inside integral is the other set of limits and the part of the arrow that is in the shaded region.
Here are the integrals.
if the arrow is vertical 
\(\displaystyle{\int_{a}^{b}{ \int_{g(x)}^{f(x)}{ h(x,y) ~dy } ~dx }}\)  
if the arrow is horizontal 
\(\displaystyle{\int_{c}^{d}{ \int_{g(y)}^{f(y)}{ h(x,y) ~dx } ~dy }}\) 
Notation
You may see the above integrals written in a shorthand notation like this. \[\displaystyle{ \iint\limits_R {h(x,y) ~ dA} }\] The two integral signs with the \(\text{R}\) underneath mean that we have a double integral over the region \(\text{R}\). The symbol \(dA\) can be either \(dx~dy\) or \(dy~dx\) and stands for a differential area.
Calculate Area in the Plane
Okay, so now the integral is almost complete. The only thing left is to determine the integrand, \(h(x,y)\).
to calculate area in the plane 

\(h(x,y) = 1\) 
In summary, for area in the plane, \(h(x,y) = 1\) giving you the integral \( \iint\limits_R {1 ~ dA} \). This makes sense since the limits of integration are already describing the area.
Time for some practice problems.
Practice
Basic
Calculate the area of the region bounded by \( y = x^2 \) and \( x = 4 \) in the first quadrant.
Problem Statement 

Calculate the area of the region bounded by \( y = x^2 \) and \( x = 4 \) in the first quadrant.
Final Answer 

\( 64/3\)
Problem Statement
Calculate the area of the region bounded by \( y = x^2 \) and \( x = 4 \) in the first quadrant.
Solution
video by Michel vanBiezen 

Final Answer
\( 64/3\)
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Calculate the area of the region bounded by \( y = 2x^2 \) and \( y = 1 + x^2\)
Problem Statement 

Calculate the area of the region bounded by \( y = 2x^2 \) and \( y = 1 + x^2\)
Final Answer 

\( 4/3 \)
Problem Statement
Calculate the area of the region bounded by \( y = 2x^2 \) and \( y = 1 + x^2\)
Solution
video by Michel vanBiezen 

Final Answer
\( 4/3 \)
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Calculate the area of the region bounded by \( y = \sin(x) \) and \( y = \sin(2x) \)
Problem Statement 

Calculate the area of the region bounded by \( y = \sin(x) \) and \( y = \sin(2x) \)
Hint 

The two curves intersect at \( x = \pi/3 \).
Problem Statement 

Calculate the area of the region bounded by \( y = \sin(x) \) and \( y = \sin(2x) \)
Final Answer 

\( 1/4 \)
Problem Statement
Calculate the area of the region bounded by \( y = \sin(x) \) and \( y = \sin(2x) \)
Hint
The two curves intersect at \( x = \pi/3 \).
Solution
video by Dr V's Mathematics Videos 

Final Answer
\( 1/4 \)
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Calculate the area bounded by \( y = (1/2)x^2 \) and \( y = 3x  x^2 \) by setting up and evaluating \( \iint{1 ~ dy ~ dx} \).
Problem Statement 

Calculate the area bounded by \( y = (1/2)x^2 \) and \( y = 3x  x^2 \) by setting up and evaluating \( \iint{1 ~ dy ~ dx} \).
Final Answer 

\( 2 \)
Problem Statement
Calculate the area bounded by \( y = (1/2)x^2 \) and \( y = 3x  x^2 \) by setting up and evaluating \( \iint{1 ~ dy ~ dx} \).
Solution
This is the easy direction for calculating the area. Practice 3537 asks you to calculate the area in the other direction.
video by Maths With Jay 

Final Answer
\( 2 \)
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Calculate the area of the region bounded by \( y = x \) and \( y = x^2 \)
Problem Statement
Calculate the area of the region bounded by \( y = x \) and \( y = x^2 \), giving your answer in exact terms.
Solution
video by Dr Chris Tisdell 

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Calculate the area of the region bounded by \( y = \sqrt{x}, y = x, y = x/2 \)
Problem Statement
Calculate the area of the region bounded by the curves \( y = \sqrt{x}, y = x, y = x/2 \), giving your answer in exact terms.
Solution
video by Dr Chris Tisdell 

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Intermediate
Calculate the area bounded by \( y = (1/2)x^2 \) and \( y = 3x  x^2 \) by setting up and evaluating \( \iint{1 ~ dx ~ dy} \).
Problem Statement 

Calculate the area bounded by \( y = (1/2)x^2 \) and \( y = 3x  x^2 \) by setting up and evaluating \( \iint{1 ~ dx ~ dy} \).
Final Answer 

\( 2 \)
Problem Statement
Calculate the area bounded by \( y = (1/2)x^2 \) and \( y = 3x  x^2 \) by setting up and evaluating \( \iint{1 ~ dx ~ dy} \).
Solution
This is the same area that you calculated in practice 3536 but this direction is more challenging.
video by Maths With Jay 

Final Answer
\( 2 \)
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Calculate the area bounded by \( 1 \leq x \leq \sqrt{2} \) and \( 0 \leq y \leq \ln(x + \sqrt{x^21}) \)
Problem Statement 

Calculate the area bounded by \( 1 \leq x \leq \sqrt{2} \) and \( 0 \leq y \leq \ln(x + \sqrt{x^21}) \)
Hint 

1. Here is the plot of the area to be calculated.
2. To evaluate the integral, try trig substitution.
Problem Statement
Calculate the area bounded by \( 1 \leq x \leq \sqrt{2} \) and \( 0 \leq y \leq \ln(x + \sqrt{x^21}) \)
Hint
1. Here is the plot of the area to be calculated.
2. To evaluate the integral, try trig substitution.
Solution
video by The Lazy Engineer 

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Practice Instructions
Unless otherwise instructed, use a double integral to calculate the specified area, giving your answers in exact terms.