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Calculating Area and Volume Using Double Integrals |
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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 shape that lie above a plane, usually the xy-plane. This page covers area and volume in rectangular coordinates. Double integrals in polar coordinates are covered on a separate page. |
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. Okay, let's get started.
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 winplot 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 first panel explains how to do that.
Describing A Region In The xy-Plane
To describe an area in the xy-plane, 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 |
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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 x-value 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 | |
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\( g(x) \leq y \leq f(x) \) | 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 |
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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 | ||||
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lower section | upper section | |||
\( a \leq x \leq G(y) \) | arrow leaves through \(G(y)\) and enters through \(x=a\) |
\( a \leq x \leq F(y) \) | arrow leaves through \(F(y)\) and enters through \(x=a\) | |
\( g(a) \leq y \leq g(b) \) | arrow sweeps from bottom (\(y=g(a)\)) to top (\(y=g(b)\)) |
\( f(b) \leq y \leq f(a) \) | arrow sweeps from bottom (\(y=f(b)\)) to top (\(y=f(a)\)) |
Type 1 and Type 2 Regions |
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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.
Krista King Math - type I and type 2 regions | |
[ 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.
Specificially, if the arrow is vertical, the integral is \(\displaystyle{\int_{a}^{b}{ \int_{g(x)}^{f(x)}{ h(x,y) ~dy } ~dx }}\)
Similarly for a horizontal arrow.
Okay, so now the integral is almost completely done. The only thing left is to determine the integrand, \(h(x,y)\) in the above example.
\(h(x,y)\) | integral | |
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to calculate area in the plane | \(h(x,y) = 1\) |
\(\displaystyle{ \iint\limits_R {1 ~ dA} }\) |
to calculate surface area |
\(h(x,y) = \sqrt{ 1+(\partial z/ \partial x)^2 + (\partial z/ \partial y)^2 }\) |
\(\displaystyle{ \iint\limits_R {\sqrt{ 1+(\partial z/ \partial x)^2 + (\partial z/ \partial y)^2 } ~ dA} }\) |
to calculate volume | \(h(x,y)\) is the height above the xy-plane |
\(\displaystyle{ \iint\limits_R {h(x,y) ~ dA} }\) |
This video clip explains calculation of volumes with double integrals in more detail.
Dr Chris Tisdell - calculate volume with double integrals | |
Time for some practice problems. After that, you are ready for triple integrals. |
next: triple integrals → |
Search 17Calculus
Practice Problems |
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Instructions - - Unless otherwise instructed, use a double integral to calculate the volume above the specified area, giving your answers in exact terms.
Level A - Basic |
Practice A01 | |
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\(f(x,y) = x+y\) \( R = { (x,y): ~0 \leq x \leq 1, ~x^2 \leq y \leq x} \) | |
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Practice A02 | |
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\(z = 4-x-y\); \(R = \{(x,y) : ~0 \leq x \leq 2, ~0 \leq y \leq 1\}\) | |
solution |
Practice A03 | |
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\( f(x,y)=1-6x^2y \); \( R = \{ (x,y) : 0 \leq x \leq 2, -1 \leq y \leq 1 \} \) | |
solution |
Practice A04 | |
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\( x^2+(1/2)y^2 + z = 12 \) \( 0 \leq x \leq 2, 0 \leq y \leq 3 \) | |
solution |
Practice A05 | |
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\( f(x,y) = xy^2 \) | |
solution |
Practice A06 | |
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\( f(x,y) = x\cos(xy) \) | |
solution |
Practice A07 | |
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\( z=x+y \) | |
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Practice A08 | |
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\( f(x,y) = xy \) | |
solution |
Practice A09 | |
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\(z=xy\) | |
solution |
Level B - Intermediate |
Practice B01 | |
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\( h(x,y)=3-x-y \); region bounded by \( y=0, ~x=1, ~y=x \) | |
solution |
Practice B02 | |
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\(\sin(x^2)\); region bounded by \(x=y,~x=2\) and the y-axis | |
solution |
Practice B03 | |
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Calculate the area of the region bounded by \( y=x \) and \( y=x^2 \) | |
solution |
Practice B04 | |
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Calculate the area of the region bounded by the curves \( y=\sqrt{x}, \) \( y = x, \) \( y = x/2 \) | |
solution |
Practice B05 | |
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Calculate the volume bounded by the coordinate planes and \( z = 2-2x-y \) | |
solution |
Practice B06 | |
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\( f(x,y) = y^2 \); region bounded by \( x \geq 0, ~ y = x-1, ~ y = x/2 \) | |
solution |
Practice B07 | |
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\( z = 3xy - x^3 \); region bounded by \( y = x^3 \) and \( y = \sqrt{x} \) | |
solution |
Practice B08 | |
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Calculate the volume enclosed by \( x^2+y^2 = 16 \); \( (x^2+y^2)/8 \leq z \leq 5 \). | |
solution |